这是用户在 2025-3-22 14:05 为 https://app.immersivetranslate.com/pdf-pro/1bf04502-97d2-4c86-a08a-f696776c74a6 保存的双语快照页面,由 沉浸式翻译 提供双语支持。了解如何保存?
modern wind turbines can generate electricity (see Figure 7.4).
现代风力涡轮机可以发电(见图 7.4)。

Figure 7.4: These giant turbines are part of a wind farm at Xinjiang in China. They produce as much electricity as a medium-sized coal-fired power station.
图 7.4:这些巨型涡轮机是中国新疆风电场的一部分。它们的发电量相当于一座中型燃煤发电站。
Wind is renewable and does not contribute to global warming. However, it is unreliable because the speed of the wind can vary and on calm days no power is produced. Wind turbines need a minimum wind speed of about 5 m / s 5 m / s 5m//s5 \mathrm{~m} / \mathrm{s} and are switched off when wind speeds exceed 25 m / s 25 m / s 25m//s25 \mathrm{~m} / \mathrm{s} to prevent them being damaged.
风能是可再生能源,不会造成全球变暖。然而,风力发电并不可靠,因为风速会有变化,在风平浪静的日子里,风力发电不会产生电力。风力涡轮机需要的最低风速约为 5 m / s 5 m / s 5m//s5 \mathrm{~m} / \mathrm{s} ,当风速超过 25 m / s 25 m / s 25m//s25 \mathrm{~m} / \mathrm{s} 时,涡轮机就会关闭,以防损坏。
Wind is a dilute energy resource. It would take a ‘wind farm’ of several hundred wind turbines (spread over several square kilometres) to produce the same energy as a typical fossil fuel power station. Wind turbines are also noisy and many people think they spoil the appearance of places where they are located.
风能是一种稀释能源。一个由几百台风力涡轮机(分布在几平方公里的土地上)组成的 "风力农场 "所产生的能量与一个典型的化石燃料发电站所产生的能量相同。风力涡轮机还会发出噪音,许多人认为它们会破坏所在地区的外观。

Wave power  波浪发电

Most of the energy of winds is transferred to the sea as waves are formed by friction between wind and water. Like wind, wave power is renewable and does not contribute to global warming. However, it is unreliable because the height of
风的大部分能量通过风和水的摩擦形成的波浪传递到海洋。与风能一样,波浪能也是可再生能源,不会造成全球变暖。但是,波浪能并不可靠,因为波浪的高度会影响风力的强度。

waves can vary and, when there are no waves, no power is produced. It is also difficult to convert the up-and-down motion of waves into the spinning motion required for a turbine in the wave energy converters, which float on the water. The cost is high because these machines corrode in the saltwater and can be damaged in storms.
波浪可能会变化,当没有波浪时,就无法产生电力。此外,也很难将波浪的上下运动转换成漂浮在水面上的波浪能转换器中涡轮机所需的旋转运动。这些机器在海水中会腐蚀,在风暴中也会损坏,因此成本很高。

Questions  问题

1 Explain why wind power can be traced back to sunlight.
1 解释为什么风能可以追溯到太阳光。

2 What is the difference between a solar panel and a solar cell?
2 太阳能电池板和太阳能电池有什么区别?

3 Describe the advantages and disadvantages of solar power.
3 描述太阳能的优缺点。

Hydroelectric power  水力发电

One of the smallest contributions to the chart in Figure 7.2 is hydroelectric power. For centuries, people have used the kinetic energy of moving water to turn water wheels, which then drive machinery. For example, they are used to grind corn and other crops, pump water and weave textiles. Today we have hydroelectric power stations (see Figure 7.5). Water stored behind a dam is released to turn turbines, which make generators spin. This is a very safe, clean and reliable way of producing electricity, but it is not without its problems. A new reservoir floods land that might otherwise have been used for hunting or farming. People may be made homeless, and wildlife habitats destroyed.
在图 7.2 的图表中,水力发电的贡献最小。几个世纪以来,人们一直利用水流的动能来转动水轮,然后驱动机械。例如,它们被用来碾磨玉米和其他农作物、抽水和纺织。如今,我们有了水力发电站(见图 7.5)。大坝后储存的水被释放出来,使涡轮机转动,从而带动发电机旋转。这是一种非常安全、清洁和可靠的发电方式,但也不是没有问题。新建的水库会淹没原本可用于狩猎或耕作的土地。人们可能无家可归,野生动物栖息地也会遭到破坏。

Figure 7.5: The giant Itaipú dam on the Paraná River in South America generates electricity for Brazil and Paraguay.
图 7.5:南美洲巴拉那河上的伊泰普巨型水坝为巴西和巴拉圭发电。
Hydroelectric power stations have a very short start up time (the time between switching on a power station and energy being delivered). This makes them very useful for storing energy until there is a sudden surge (increase) in demand. The demand for electricity varies during the day: it is highest during the daytime (when most people are awake) and lowest during the night. Power stations that use fossil fuels and nuclear fuels take a long time to start up and stop so, once started, they are allowed to continue running. It would take too long to stop them when demand is low and then start them again for the next rise in demand. This means that sometimes (usually at night) too much electricity is supplied and battery technology is not currently good enough to store large amounts of energy. In some hydroelectric power stations (called pumped storage systems), the turbines can be reversed so that water can be pumped back up a mountain to the reservoir so that energy can be stored as gravitational potential energy. This water can be allowed to fall back down the mountain to produce electricity when demand rises.
水电站的启动时间非常短(从开启电站到输送能源之间的时间)。因此,在需求突然激增(增加)之前,水电站非常适合储存能源。一天中对电力的需求各不相同:白天(大多数人都醒着)的需求最大,而夜间的需求最小。使用化石燃料和核燃料的发电站启动和停止都需要很长时间,因此一旦启动,它们就可以继续运行。如果在需求较低时停止发电,然后在下一次需求上升时再启动,则需要很长时间。这就意味着有时(通常是在夜间)会供应过多的电力,而目前的电池技术还不足以存储大量的能量。在一些水力发电站(称为抽水蓄能系统)中,涡轮机可以反转,以便将水从山上抽回水库,从而以重力势能的形式储存能量。当需求增加时,可让这些水流回山下发电。

Biomass fuels  生物质燃料

For many people in the world, wood is the most important fuel. It warms their homes and provides the heat necessary for cooking their food. Wood is made by
对世界上许多人来说,木材是最重要的燃料。它能温暖他们的家,提供烹饪食物所需的热量。木材是由

trees and shrubs. It stores energy that the plant has captured from sunlight in the process of photosynthesis. When we burn wood, we are releasing energy that came from the Sun in the recent past, perhaps ten or a hundred years ago.
是树木和灌木。它储存着植物在光合作用过程中从太阳光中获取的能量。当我们燃烧木材时,我们释放的能量来自不久前,也许是十年或一百年前的太阳。

Wood is just one example of a biofuel. Others include animal dung (Figure 7.6) and biogas, generated by rotting vegetable matter.
木材只是生物燃料的一个例子。其他生物燃料还包括动物粪便(图 7.6)和植物腐烂产生的沼气。

Figure 7.6: A Maasai blowing into elephant dung to make fire in a village in West Kilimanjaro, Tanzania.
图 7.6:在坦桑尼亚西乞力马扎罗山的一个村庄里,一名马赛人在向大象粪便吹气生火。
Biomass fuels account for roughly 10% of global energy consumption but, because no-one keeps track of all the wood consumed as fuel, it is a rough estimate. This means that it is rarely included in global figures and does not appear in Figure 7.2. About two-thirds of biomass fuel is used in developing countries for cooking and heating. The 25 % 25 % 25%25 \% of people in the developed (industrial) nations consume about six times as much energy as the other 75 % 75 % 75%75 \% of people living in the developing world.
生物质燃料约占全球能源消耗的 10%,但由于没有人记录所有作为燃料消耗的木材,这只是一个粗略的估计。这意味着生物质燃料很少被纳入全球数据,也不会出现在图 7.2 中。发展中国家约有三分之二的生物质燃料用于烹饪和取暖。发达(工业)国家 25 % 25 % 25%25 \% 人口的能源消耗大约是发展中国家 75 % 75 % 75%75 \% 人口的六倍。

Biomass has the advantage that it is renewable and does not contribute to global warming. It is reliable because it can be burned when needed. However, burning biofuels, particularly indoors, can lead to respiratory and other health problems.
生物质能的优点是可再生,不会造成全球变暖。生物质能在需要时燃烧,因此非常可靠。然而,燃烧生物燃料,尤其是在室内燃烧,可能会导致呼吸道和其他健康问题。

Fossil fuels  化石燃料

Oil, coal and gas are all examples of fossil fuels. These are usually hydrocarbons (compounds of hydrogen and carbon). When they are burned, they combine with oxygen from the air. In this process, the carbon becomes carbon dioxide. The hydrogen becomes dihydrogen monoxide, which we usually call water. Energy is released.
石油、煤和天然气都属于化石燃料。它们通常是碳氢化合物(氢和碳的化合物)。燃烧时,它们会与空气中的氧气结合。在这个过程中,碳变成二氧化碳。氢变成一氧化二氢,也就是我们通常所说的水。能量被释放出来。
We can write this as an equation:
我们可以把它写成一个等式:
hydrocarbon + oxygen carbon dioxide + water + energy  hydrocarbon + oxygen   carbon dioxide + water + energy  " hydrocarbon + oxygen "rarr" carbon dioxide + water + energy "\text { hydrocarbon + oxygen } \rightarrow \text { carbon dioxide + water + energy }
Hence, we can think of a fossil fuel as a store of chemical energy. Where has this energy come from?
因此,我们可以把化石燃料看作是一种化学能的储存。这些能量从何而来?
Fossil fuels (Figure 7.7) are the remains of organisms (plants and animals) that lived in the past. Many of the Earth’s coal reserves, for example, formed from trees that lived in the Carboniferous era, between 286 and 360 million years ago. (Carboniferous means coal-producing.) These trees captured energy from the Sun by photosynthesis. They grew and eventually they died. Their trunks fell into swampy ground, but they did not rot completely, because there was insufficient oxygen.
化石燃料(图 7.7)是生活在过去的生物(植物和动物)的残骸。例如,地球上的许多煤炭储量都是由生活在距今 2.86 亿至 3.6 亿年前石炭纪的树木形成的。 这些树木通过光合作用从太阳获取能量。它们不断生长,最终死亡。它们的树干掉进了沼泽地,但并没有完全腐烂,因为氧气不足。

Figure 7.7: Coal is a fossil fuel. A fossil is any living material that has been preserved for a long time. Usually, its chemical composition changes during the process. Coal sometimes shows evidence of the plant material from which it formed. Sometimes you can see fossilised creatures that lived in the swamps of the Carboniferous era. These creatures died along with the trees that eventually became coal.
图 7.7:煤是一种化石燃料。化石是指长期保存下来的任何有生命的物质。在这一过程中,其化学成分通常会发生变化。煤炭有时会显示出由植物材料形成的痕迹。有时,你可以看到生活在石炭纪沼泽中的生物化石。这些生物与树木一起死去,最终成为煤炭。
As material built up on top of these ancient trees, the pressure on them increased. Eventually, millions of years of compression turned them into underground reserves of coal (Figure 7.7). Today, when we burn coal, the light that we see and the warmth that we feel have their origins in the energy from the Sun trapped by trees hundreds of millions of years ago.
随着这些古树顶部物质的堆积,它们所承受的压力也随之增加。最终,经过数百万年的压缩,它们变成了地下储藏的煤(图 7.7)。今天,当我们燃烧煤炭时,我们所看到的光亮和所感受到的温暖都源于几亿年前被树木困住的太阳能量。

Oil and gas are usually found together. They are formed in a similar way to coal, but from the remains of tiny shrimp-like creatures called microplankton that lived in the oceans. The oilfields of the Arabian Gulf, North Africa and the Gulf of Mexico, which contain half of the world’s known oil reserves, all formed in the Cretaceous era, 75 to 120 million years ago.
石油和天然气通常是一起发现的。石油和天然气的形成方式与煤炭类似,都是由生活在海洋中的微型浮游生物(虾类)的残骸形成的。阿拉伯湾、北非和墨西哥湾的油田拥有世界已知石油储量的一半,它们都形成于距今 7500 万到 1.2 亿年前的白垩纪。
Burning fossil fuels releases carbon dioxide into the atmosphere. This enhances (increases) the greenhouse effect and is the cause of recent global warming. Coal produces more carbon dioxide than oil and natural gas. Burning coal and oil usually also produces sulfur dioxide, which leads to acid rain and damage to ecosystems and buildings.
燃烧化石燃料会向大气中释放二氧化碳。这增强(增加)了温室效应,是近期全球变暖的原因。煤炭比石油和天然气产生更多的二氧化碳。燃烧煤炭和石油通常还会产生二氧化硫,导致酸雨,破坏生态系统和建筑物。

Questions  问题

4 What form of energy is stored in fossil fuels and biofuels?
4 化石燃料和生物燃料中储存的是什么形式的能量?

5 What is the difference between biofuels and fossil fuels?
5 生物燃料和化石燃料有什么区别?

6 Describe how fossil fuels are formed.
6 描述化石燃料是如何形成的。

Nuclear fuels  核燃料

Nuclear power was developed in the second half of the 20th century. It is a very demanding technology, which requires very strict controls, because of the serious damage that can be caused by an accident.
核能是在 20 世纪下半叶发展起来的。这是一项要求非常高的技术,需要非常严格的控制,因为事故可能造成严重破坏。
The fuel for a nuclear power station (Figure 7.8) is usually uranium, sometimes plutonium. These are radioactive materials. Inside a nuclear reactor, the radioactive decay of these materials is speeded up so that the energy they store is released much more quickly. This is the process of nuclear fission.
核电站的燃料(图 7.8)通常是铀,有时是钚。这些都是放射性物质。在核反应堆内,这些材料的放射性衰变速度加快,因此它们储存的能量释放得更快。这就是核裂变过程。

Figure 7.8: Belleville nuclear power station in France.
图 7.8:法国 Belleville 核电站。
Uranium is a nuclear fuel. It is a very concentrated store of energy in the form of nuclear energy. A typical nuclear power station will receive about one truckload of new fuel each week. Coal is a less concentrated energy store. A similar coalfired power station is likely to need a whole trainload of coal every hour. A wind farm capable of generating electricity at the same rate would cover a large area of ground - perhaps 20 square kilometres.
铀是一种核燃料。它以核能的形式非常集中地储存能量。一个典型的核电站每周会收到大约一卡车的新燃料。煤炭是一种不太集中的能源储存。类似的燃煤发电站每小时可能需要一整列火车的煤炭。一个能够以同样速度发电的风力发电场需要覆盖很大的面积--可能是 20 平方公里。
In some countries that have few other resources for generating electricity, nuclear power provides a lot of energy. In France, for example, nuclear power stations generate three-quarters of the country’s electricity. Excess production is exported to neighbouring countries, including Spain, Switzerland and the UK.
在一些几乎没有其他发电资源的国家,核能提供了大量能源。例如,法国四分之三的电力来自核电站。多余的电力被出口到西班牙、瑞士和英国等邻国。

Nuclear fuel is a relatively cheap, concentrated energy resource. However, nuclear power has proved to be expensive because of the initial cost of building the power stations, and the costs of disposing of the radioactive spent fuel and decommissioning the stations at the end of their working lives. Also, accidents like those at Chernobyl in 1986 and Fukushima in 2011 can cause radioactive material to be spread over a wide area.
核燃料是一种相对廉价、集中的能源资源。但事实证明,核电成本高昂,因为建造发电站的初始成本、放射性乏燃料的处理成本以及发电站工作寿命结束后的退役成本。此外,1986 年切尔诺贝利事故和 2011 年福岛事故也会导致放射性物质大面积扩散。

Geothermal energy  地热能源

The interior of the Earth is hot. This would be a useful source of energy, if we could get at it. People do make use of this geothermal energy where hot rocks are found at a shallow depth below the Earth’s surface. These rocks are hot because of the presence of radioactive substances inside the Earth. To make use of this energy, water is pumped down into the rocks, where it boils. Highpressure steam returns to the surface, where it can be used to generate electricity.
地球内部很热。如果我们能够获取,这将是一种有用的能源。人们确实利用了这种地热能源,在地表下浅层发现了热岩石。这些岩石之所以热,是因为地球内部存在放射性物质。为了利用这种能量,人们将水泵入岩石,使其沸腾。高压蒸汽返回地面,可用于发电。

Suitable hot underground rocks are usually found in places where there are active volcanoes. Iceland, for example, has several geothermal power stations. These also supply hot water to heat nearby homes and buildings. While energy from geothermal resources has no obvious disadvantages, there are few places on Earth where it is available.
合适的地下热岩通常出现在有活火山的地方。例如,冰岛就有几座地热发电站。这些发电站还提供热水,为附近的家庭和建筑物供暖。虽然地热资源产生的能源没有明显的缺点,但在地球上可利用地热资源的地方却很少。

Tidal energy  潮汐能

A tidal power station is similar to a hydroelectric power station: electrical power is generated by moving water. A barrage (dam) is built across a river estuary (where a river meets the sea) creating a reservoir. As the tide goes in and out, water passes through turbines in the dam.
潮汐发电站与水力发电站类似:通过水的流动来发电。在河口(河流与大海交汇处)修建拦河坝(大坝),形成一个水库。随着潮水的涨落,水流通过大坝上的涡轮机。

Tidal power has the advantage of being renewable. Also, tides are predictable
潮汐发电具有可再生的优势。此外,潮汐是可预测的

making it a fairly reliable energy resource. However, by flooding estuaries, a tidal power station can destroy wetlands, an important habitat for wildlife, particularly migrating birds that use it to feed and rest before the next leg of their journey. The barrage can also block shipping routes.
使其成为相当可靠的能源资源。然而,潮汐发电站会淹没河口,从而破坏湿地,而湿地是野生动物的重要栖息地,尤其是候鸟,它们会在下一段旅程开始前在此觅食和休息。拦河坝还可能阻断航运路线。

Questions  问题

7 List the advantages and disadvantages of nuclear power.
7 列出核能的优缺点。

8 What energy is available from moving water?
8 流动的水能提供什么能量?

9 In what way is a hydroelectric power station like a cell (or battery) and how can it be adapted to act like a rechargeable cell?
9 水电站在哪些方面像电池(或蓄电池)?

Using energy resources to generate electricity
利用能源发电

Many of the energy resources in this chapter produce electricity so that it can be transferred to where it is needed. The thermal energy produced when fossil fuels are burned (Figure 7.9) or when nuclear fission takes place is used to heat water in a boiler to form steam. The steam turns the blades of a turbine, transferring thermal energy into kinetic energy. The turbine is linked by an axle to a generator where a voltage is induced in conducting wires when they move in a magnetic field. You will learn more about generators in Chapter 21.
本章中的许多能源都能发电,以便将电能输送到需要的地方。化石燃料燃烧(图 7.9)或核裂变产生的热能可用于加热锅炉中的水,形成蒸汽。蒸汽转动涡轮机的叶片,将热能转化为动能。涡轮机通过轮轴与发电机相连,当导电线在磁场中运动时,发电机就会产生感应电压。您将在第 21 章了解更多有关发电机的知识。

Figure 7.9: A schematic of coal-fired power station. The details of how thermal energy is produced in power stations that use fuel will vary but they will all have a boiler, turbine and generator.
图 7.9:燃煤发电站示意图。使用燃料的发电站产生热能的具体方式各不相同,但都有锅炉、涡轮机和发电机。
The details of how thermal energy is produced in power stations that use fuel will vary, but they will all have a boiler, turbine and generator. Those energy resources that do not use a fuel will not need a boiler, but they will still use a turbine linked to a generator to produce electricity. Moving air (wind) and moving water (hydroelectricity, wave and tidal) can turn a turbine directly.
使用燃料的发电站生产热能的细节各不相同,但都有锅炉、涡轮机和发电机。不使用燃料的能源则不需要锅炉,但仍使用与发电机相连的涡轮机发电。移动的空气(风)和移动的水(水力发电、波浪发电和潮汐发电)可以直接带动涡轮机转动。

Comparing energy resources
比较能源资源

We use fossil fuels a lot because they represent concentrated sources of energy. A modern gas-fired power station might occupy the space of a football ground and supply a town of 100000 people. To replace it with a wind farm might require 50 or more wind turbines spread over an area of several square kilometres. The wind is a much more dilute source of energy.
我们之所以大量使用化石燃料,是因为它们是集中的能源。一个现代化的燃气发电站可能占地相当于一个足球场,为一个拥有 10 万人的城镇供电。如果用一个风力发电场来替代,可能需要 50 台或更多的风力涡轮机,分布在几平方公里的区域内。风能是一种稀薄得多的能源。

This illustrates some of the ideas that we use when comparing different energy resources. Each has its advantages and disadvantages. We need to think about the following factors. 

Renewability  可再生性

As we have seen, there are limited reserves of fossil fuels. The same applies to uranium nuclear fuel. However, there are plentiful reserves of alternative nuclear fuels like thorium. All other energy resources are renewable, including all those that can be traced back to radiation from the Sun. The most important advantage of renewable resources is that, once installed, they do not contribute to global warming.
正如我们所看到的,化石燃料的储量有限。铀核燃料也是如此。然而,像钍这样的替代核燃料储量丰富。所有其他能源都是可再生的,包括所有可以追溯到太阳辐射的能源。可再生资源最重要的优点是,一旦安装,它们不会造成全球变暖。

Cost  费用

We should separate initial costs from running costs. A solar cell is expensive to buy but there are no costs for fuels - sunlight is free! While nuclear fuel is cheap, the costs of decommissioning nuclear power plants are high.
我们应该把初始成本和运行成本分开。太阳能电池的购买成本很高,但燃料没有成本,阳光是免费的!虽然核燃料很便宜,但核电站的退役成本却很高。

Availability  可用性

France uses nuclear power to produce about 75 % 75 % 75%75 \% of its electricity because it has few alternative energy resources. Norway has plenty of rainfall and mountains so generates about 95 % 95 % 95%95 \% of its electricity from hydroelectric power. Iceland uses geothermal energy, which is quite localised.
法国使用核电生产约 75 % 75 % 75%75 \% 的电力,因为它几乎没有替代能源。挪威有大量降雨和山脉,因此约 95 % 95 % 95%95 \% 的电力来自水力发电。冰岛使用地热能源,但这种能源相当局部。

Reliability  可靠性

Is the energy supply constantly available? The wind is variable, so wind power is
能源供应是否持续可用?风是可变的,因此风力发电

unreliable. Wars and trade disputes can interrupt fuel supplies.
不可靠。战争和贸易争端会中断燃料供应。

Scale  规模

A fossil fuel power station can be compact and still supply a large population. It would take several square metres of solar cells to supply a small household. Alternatively, people sometimes talk about how concentrated or dilute an energy resource is. When talking about fuels, they are comparing how much energy is stored in a certain mass of the fuel. Particularly when comparing wind turbines with other energy resources, people will talk about the land area required to generate the same amount of energy.
化石燃料发电站可以很紧凑,但仍能为大量人口供电。而为一个小家庭供电则需要几平方米的太阳能电池。另外,人们有时会谈论一种能源资源的浓缩或稀释程度。在谈到燃料时,他们是在比较一定质量的燃料能储存多少能量。特别是在比较风力涡轮机和其他能源时,人们会谈论产生相同能量所需的土地面积。

Environmental impact  环境影响

The use of fossil fuels leads to climate change. A hydroelectric dam may flood useful farmland. Every energy source has some effect on the environment.
使用化石燃料会导致气候变化。水电大坝可能会淹没有用的农田。每一种能源都会对环境产生一定的影响。

Questions  问题

10 Which of the following energy resources is renewable?
10 以下哪种能源是可再生能源?

A oil  A 油
B nuclear  B 核
C biofuels  C 生物燃料
D coal  D 煤
11 Which of the following energy resources is not renewable?
11 以下哪种能源不是可再生能源?

A hydroelectric power  水力发电
B wind  B 风
C tidal  C 潮汐
D nuclear  D 核

ACTIVITY 7.2  活动 7.2

Sanghera Island  桑盖拉岛
Part 1  第一部分
Sanghera Island (Figure 7.10) is a remote fictitious island. It has no fossil fuels. It is hot with jungle vegetation, though it can sometimes be cold at
桑盖拉岛(图 7.10)是一个偏远的虚构岛屿。岛上没有化石燃料。岛上气候炎热,有丛林植被,但有时也很寒冷。

night. High rainfall and mountainous terrain leads to fast-flowing streams. You are one of 25 members of a scientific expedition planning to study the island for three years. At the same time as minimising your impact (ecological footprint) on the island, you are the expert given the job of providing all the energy the team will need. 
Figure 7.10: Map of Sanghera Island. 
Answer the following questions to develop a plan that you will present to the team. 
1 Describe two ways of providing hot water and heat for the buildings (living accommodation and laboratory). 
2 Describe two possible ways of supplying heat for cooking food. 
3 How would you supply the electricity needed for lighting, PCs and 
machinery? 
4 There will be a refrigerator for life-saving medicines, as well as chemicals, that need to be stored at a constant temperature. How would you ensure a constant supply of electricity? 
5 Which natural resource on the island should be conserved? 
6 On a copy of the map, mark where you plan to locate buildings and the energy harvesters (devices that collect energy from the environment), showing how you would get the energy from the harvesters to the buildings, if necessary. Explain your reasons for choosing these particular sites. 
7 Would your answers be different if the team had a limited budget? Explain your answer. 
8 Describe what other information you might need before you can make a recommendation to the team. 

Part 2 

In a group of four, design your own island and a brief or set of questions for another group to decide how they will provide the required energy. Be prepared to answer questions and have your own solution in mind. 

REFLECTION 

Is there an effective method for memorising the different energy resources? For example, could you put them into categories? How will you learn their advantages and disadvantages? 

ACTIVITY 7.3 

Planning meeting 

Work in a group of three. Each group will be randomly allocated a ‘power station’ that uses a different energy resource (such as a wind farm). You will be given the role of either advocating (supporting) or opposing planning permission for your allocated power station to be built close to 
your community. 
You will need to identify the general advantages or disadvantages of your power station before identifying the specific advantages or disadvantages of locating it close to your community. 
Each group will present their case as a spoken presentation, with a whiteboard and pens the only resources permitted. 
  • You will be given a short time to agree, as a class, the criteria for judging each case. For example, you might decide to dismiss a case if the science is incorrect.
    你们将有短暂的时间在全班商定判断每个案例的标准。例如,如果科学依据不正确,你们可以决定驳回案件。
  • You will then be allocated your power station. Unless it is given as a homework task (when you could spend time doing more detailed research), spend about five minutes in your groups preparing your case.
    然后你们将被分配到自己的发电站。除非是作为家庭作业布置的任务(这时你们可以花时间做更详细的研究),否则请在小组内花大约五分钟准备你们的案例。
  • Those advocating and opposing each power station will be called to the front of the class and each given a maximum of one minute to put their case to the rest of the class. The class will have a maximum of 30 seconds to question each case. While listening to each case, note down the advantages and disadvantages of each power station in a table so that you have a summary. Questions from the class and answers from your teacher should identify any errors.
    每个发电站的支持者和反对者将被叫到教室前面,每人有最多一分钟的时间向全班其他同学陈述自己的理由。全班同学最多有 30 秒的时间对每个案例进行提问。在聆听每个案例的同时,将每个发电站的优缺点记录在表格中,以便进行总结。全班同学的提问和老师的回答应能找出任何错误。
  • While you are in the audience, you will score each presentation out of 10 according to the criteria agreed. Subtract your vote for the opposition from your vote for the advocates to get an overall score. Two examples are given in Table 7.1. A negative score suggests opposition but, if all proposals are negative, you might be forced to choose the least-worst option.
    当你在观众席上时,你将根据商定的标准为每个演讲打分,满分为 10 分。将你投给反对方的票减去你投给支持方的票,得出总分。表 7.1 给出了两个例子。负分表示反对,但如果所有提案都是负分,你可能被迫选择最差的方案。
Type of power station
发电站类型
For  对于 Against  反对 Overall score  总分
wind farm  风电场 2 4 -2
nuclear  核电 7 3 4
Type of power station For Against Overall score wind farm 2 4 -2 nuclear 7 3 4| Type of power station | For | Against | Overall score | | :--- | :---: | :---: | :---: | | wind farm | 2 | 4 | -2 | | nuclear | 7 | 3 | 4 |
Table 7.1: Example score sheet.
表 7.1:评分表范例。

REFLECTION  反思

Did you find Activity 7.3 helpful in summarising the advantages and disadvantages of the different energy resources? Is there a better method for learning this material?
活动 7.3 对你总结不同能源的优缺点有帮助吗?有没有更好的学习方法?

7.2 Energy from the Sun
7.2 来自太阳的能量

Most of the energy we use can be traced back to radiation from the Sun. To summarise:
我们使用的大部分能源都可以追溯到太阳辐射。总而言之
  • Fossil fuels are stores of energy that came from the Sun millions of years ago.
    化石燃料是数百万年前来自太阳的能量储存。
  • Radiation (light and heat) from the Sun can be absorbed by solar panels to provide hot water. Sunlight can also be absorbed by arrays of solar cells (photocells) to generate electricity. In some countries, you may see these on the roofs of houses.
    太阳能电池板可以吸收太阳辐射(光和热),提供热水。太阳能电池阵列(光电池)也可以吸收太阳光来发电。在一些国家,你可能会在房屋的屋顶上看到这些太阳能电池。
  • The wind is caused when air is heated by the Sun. Warm air rises; cool air flows in to replace it. This moving air can be used to generate electricity using wind turbines.
    风是空气被太阳加热后产生的。热空气上升,冷空气流入取而代之。这种流动的空气可以利用风力涡轮机发电。
  • Most hydroelectric power comes ultimately from the Sun. The Sun’s rays cause water to evaporate from the oceans and land surface. This water vapour in the atmosphere eventually forms clouds at high altitudes. Rain falls on high ground, and can then be trapped behind a dam. This is part of the water cycle. Without energy from the Sun, there would be no water cycle and no hydroelectric power.
    大多数水力发电最终都来自太阳。太阳光使海洋和陆地表面的水分蒸发。大气中的水蒸气最终在高海拔地区形成云。雨水落在高处,然后被困在水坝后面。这就是水循环的一部分。没有太阳的能量,就没有水循环和水力发电。

    However, we make use of a small amount of energy that does not come from the Sun as radiation. Here are three examples:
    不过,我们使用的少量能量并非来自太阳辐射。这里有三个例子:
  • The Moon and the Sun both contribute to the oceans’ tides. Their gravitational pull causes the level of the ocean’s surface to rise and fall every twelve-and-a-bit hours. At high tide, water can be trapped behind a dam. Later, at lower tides, it can be released to drive turbines and generators. Because this depends on gravity, and not the Sun’s heat and light, we can rely on tidal power even at night and when the Sun is hidden by the clouds.
    月亮和太阳都对海洋潮汐有影响。它们的引力导致海平面每隔十二个半小时上升和下降一次。涨潮时,海水会被困在水坝后面。之后,在潮汐降低时,水被释放出来,驱动涡轮机和发电机。由于潮汐发电依靠的是重力,而不是太阳的光和热,因此即使在夜晚或太阳被云层遮挡的时候,我们也可以依靠潮汐发电。
  • Nuclear power makes use of nuclear fuels - mostly uranium - mined from underground. Uranium is a slightly radioactive element, which has been in the ground ever since the Earth formed, together with the rest of the Solar System, 4.5 billion years ago.
    核能利用从地下开采的核燃料(主要是铀)。铀是一种略带放射性的元素,自 45 亿年前地球与太阳系其他部分一起形成以来,铀就一直存在于地下。
  • Geothermal energy also depends on the presence of radioactive substances inside the Earth. These have been there since the Earth formed; they have
    地热能还取决于地球内部放射性物质的存在。自地球形成以来,这些物质就一直存在;它们有

    been continuously releasing their store of energy ever since.
    从那时起,它们就一直在不断释放自己储存的能量。

The source of the Sun's energy
太阳的能量来源

The Sun releases vast amounts of energy, but it is not burning fuel in the same way as we have seen for fossil fuels. It is not a chemical reaction. The Sun consists largely of hydrogen, but there is no oxygen to burn this gas. Instead, energy is released in the Sun by the process of nuclear fusion. In nuclear fusion, four energetic hydrogen atoms collide and fuse (join together) to form an atom of helium.
太阳释放出大量能量,但它燃烧燃料的方式与我们看到的化石燃料不同。这不是化学反应。太阳主要由氢气组成,但没有氧气来燃烧这种气体。相反,太阳通过核聚变过程释放能量。在核聚变过程中,四个高能氢原子碰撞并融合(结合在一起)形成一个氦原子。
Nuclear fusion requires very high temperatures and pressures. The temperature inside the Sun is close to 15 million degrees. The pressure is also very high, so that hydrogen atoms are forced very close together, allowing them to fuse. At this temperature all the atoms are ionised. All the electrons have been removed from all the atoms, creating plasma of positive nuclei and negative electrons. Atomic nuclei all have a positive charge and like charges repel so a temperature of about 100 million degrees (and high pressure) is required to overcome this electrostatic repulsion and get the nuclei close enough to fuse. The mass of the final nucleus is slightly less than the combined mass of the initial nuclei and the difference in mass is turned into energy according to Einstein’s famous equation: E = m c 2 E = m c 2 E=mc^(2)E=m c^{2}. The energy, E E EE, released is big because the mass, m m mm, is multiplied by the speed of light, c c cc, squared (which is a big number).
核聚变需要非常高的温度和压力。太阳内部的温度接近 1500 万度。压力也非常高,因此氢原子被迫靠得非常近,使它们得以融合。在这个温度下,所有原子都被电离。所有原子中的电子都被清除,形成了由正原子核和负电子组成的等离子体。原子核都带有正电荷,同类电荷相互排斥,因此需要约 1 亿度的高温(和高压)来克服这种静电排斥力,使原子核靠近到足以熔合的程度。最终原子核的质量略小于初始原子核的总质量,根据爱因斯坦的著名方程,质量差转化为能量: E = m c 2 E = m c 2 E=mc^(2)E=m c^{2} 。由于质量 m m mm 乘以光速 c c cc 的平方(这是一个很大的数字),因此释放的能量 E E EE 很大。

Nuclear fusion reactors - artificial Suns on Earth?
核聚变反应堆--地球上的人造太阳?

Scientists have been trying to recreate the same process artificially here on Earth since the 1950s as it offers a clean source of almost unlimited energy. It will not produce greenhouse gases or nuclear waste.
自 20 世纪 50 年代以来,科学家们一直试图在地球上人工再造同样的过程,因为它提供了几乎无限的清洁能源。它不会产生温室气体或核废料。

It is essential to hold the hot plasma in place for long enough for fusion to take place. In the Sun, the star’s enormous gravitational field prevents the plasma escaping. In the 1950s, Soviet physicists came up with the idea of a tokamak (Figure 7.11) to contain the plasma.
必须将热等离子体保持在足够长的时间内,才能进行核聚变。在太阳中,恒星巨大的引力场会阻止等离子体逃逸。20 世纪 50 年代,苏联物理学家提出了用托卡马克(图 7.11)来控制等离子体的想法。

Figure 7.11: China’s new tokamak under construction in Chengdu.
图 7.11:中国正在成都建造的新托卡马克。
This is a container shaped like a torus (or doughnut) with a complicated arrangement of magnets to stop the plasma touching the walls. If the plasma were to touch the container walls, it would cool (and fusion would stop) and the container walls would be damaged. This is why fusion is a very challenging engineering problem.
这是一个形状像环形(或甜甜圈)的容器,里面有排列复杂的磁铁来阻止等离子体接触容器壁。如果等离子体接触到容器壁,就会冷却(聚变就会停止),容器壁就会损坏。这就是为什么核聚变是一个非常具有挑战性的工程问题。
Fusion reactors on Earth will fuse deuterium and tritium (two isotopes of hydrogen) to produce helium and a neutron. Only charged particles (that are moving) can experience a magnetic force and be confined by a magnetic field. Neutrons are neutral (have zero charge) so cannot be confined by the magnetic field and so they hit the walls of the tokamak. These collisions produce thermal energy. Heat exchangers in the walls conduct the thermal energy to heat up water to make steam to turn a turbine and produce electricity.
地球上的聚变反应堆将氘和氚(氢的两种同位素)熔化,产生氦和中子。只有带电粒子(运动着的)才能感受到磁力并被磁场束缚。中子是中性的(电荷为零),因此不能被磁场束缚,所以它们会撞击托卡马克的壁。这些碰撞会产生热能。箱壁中的热交换器将热能传导给水,加热水产生蒸汽,使涡轮机转动并发电。
So far, no fusion reactor has produced more energy than needs to be put in to keep the plasma hot. In 1997 Joint European Torus claimed the world record for getting out 67 % 67 % 67%67 \% of the input energy. It is hoped that reactors will get out ten times more energy than is put in. The International Thermonuclear Experimental
迄今为止,还没有一个聚变反应堆产生的能量超过为保持等离子体高温而需要输入的能量。1997 年,欧洲联合火炬反应堆(Joint European Torus)创下了输出 67 % 67 % 67%67 \% 输入能量的世界纪录。人们希望反应堆输出的能量是输入能量的十倍。国际热核实验
Reactor (ITER) project (Figure 7.12) is being built at Cadarache in France. This is an international collaboration that involves scientists from countries that represent half the world’s population.
热核实验堆(ITER)项目(图 7.12)正在法国卡达拉什建造。这是一项国际合作项目,有来自占世界人口一半的国家的科学家参与。

Figure 7.12: Employees in late 2018 building the tokamak inside the ITER (International Thermonuclear Experimental Reactor) construction site in France.
图 7.12:2018 年底,员工在法国 ITER(国际热核实验反应堆)建筑工地内建造托卡马克。

Questions  问题

12 What is plasma?  12 什么是血浆?
13 Compare how plasma is confined in a star with how we will achieve the same effect here on Earth.
13 将等离子体在恒星中的封闭方式与我们在地球上实现同样效果的方式进行比较。
14 Discuss the advantages and disadvantages of nuclear fusion.
14 讨论核聚变的优缺点。

15 What is the difference between nuclear fusion and nuclear fission?
15 核聚变和核裂变有什么区别?

PROJECT  项目

The future of energy resources
能源资源的未来

There is a huge variety of potential projects in this important area of physics. We need energy but getting it by burning fossil fuels contributes to
在这一重要的物理学领域,潜在的项目种类繁多。我们需要能源,但通过燃烧化石燃料获取能源会导致

global warming. Whatever topic you choose, you need to pose a question like the one describing thorium reactors at the start of the chapter. Your answer should be clear, concise and coherent. Write it in your own words and limit yourself to 1000 words. Use informative diagrams where you can. Whatever medium you choose to convey your answer, try to reach an audience beyond your own classroom. Or, you could promote (for example, to your friends and family on social media) good work that you have discovered during your research. You could:
全球变暖。无论你选择什么话题,你都需要提出一个类似本章开头描述钍反应堆的问题。你的答案应该清晰、简洁、连贯。用自己的话写,字数控制在 1000 字以内。尽可能使用信息丰富的图表。无论你选择何种媒介来表达你的答案,都要尽量让自己的课堂以外的人也能看到你的答案。或者,您可以宣传(例如,在社交媒体上向您的亲朋好友宣传)您在研究过程中发现的优秀作品。你可以
  • Investigate ways to reduce demand for energy. In cold countries this might include efforts to improve building insulation while, in hot countries, it might be worth thinking about the development of wind towers to reduce demand for air conditioning. Or you could focus on the development of more efficient transport (such as electric vehicles).
    研究减少能源需求的方法。在寒冷的国家,这可能包括努力改善建筑物的隔热性能;而在炎热的国家,可能值得考虑开发风塔,以减少对空调的需求。或者,您也可以关注开发更高效的交通工具(如电动汽车)。
  • Investigate ways to increase the supply of energy. This could focus on one of the energy resources you have already met in this chapter, or you might investigate one that is under development (for example, one based on algae).
    调查增加能源供应的方法。可以重点研究本章已经介绍过的一种能源,也可以研究正在开发的一种能源(例如,基于藻类的能源)。
  • Investigate the challenges facing development of the lifters you met at the start of the chapter. You should look at the process and make a comparison with nuclear power based on the uranium cycle.
    调查你在本章开始时遇到的升降机开发所面临的挑战。您应该研究这一过程,并与基于铀循环的核能进行比较。
  • Investigate fracking. You should explain the process itself and offer a balanced assessment of the advantages and disadvantages.
    调查压裂法。您应该解释这一过程本身,并对其利弊做出平衡的评估。
  • Investigate developments in energy storage, including improved battery technology, which might include the environmental impact of lithium mining.
    调查能源储存的发展情况,包括改进电池技术,其中可能包括锂矿开采对环境的影响。

SUMMARY  摘要

Solar panels are used to collect energy from the Sun.
太阳能电池板用于收集太阳能量。

Solar cells (also known as photocells) generate electricity using energy from the Sun.
太阳能电池(又称光电池)利用太阳的能量发电。
Oil, coal and natural gas are all examples of fossil fuels.
石油、煤炭和天然气都属于化石燃料。

Coal forms from trees, and oil and natural gas form from microplankton.
煤炭由树木形成,石油和天然气则由微型浮游生物形成。

Non-renewable energy resources will run out. This includes fossil fuels and nuclear fuel but not biofuel.
不可再生能源将枯竭。这包括化石燃料和核燃料,但不包括生物燃料。
Renewable energy resources are replenished (replaced) after they have been used.
可再生能源在使用后会得到补充(替换)。

Biofuels are renewable, reliable, cheap to set up and use, but are diffuse.
生物燃料是可再生的、可靠的、廉价的,但具有分散性。

Geothermal energy is harvested (collected) where hot rock is close to the Earth’s surface.
地热能是在靠近地球表面的热岩中采集(收集)的。
Wind power, wave power and solar power are renewable but unreliable and dilute energy resources. Running costs are low but they are expensive to set up.
风力发电、波浪发电和太阳能发电都是可再生能源,但并不可靠,而且会稀释能源资源。运行成本低,但安装费用高。
Hydroelectric power, tidal power and geothermal power are renewable, reliable and concentrated energy resources but suitable locations are limited, and they are expensive to set up.
水力发电、潮汐发电和地热发电都是可再生、可靠和集中的能源资源,但合适的地点有限,而且建设成本高昂。
Nuclear power stations use nuclear fuel, which produces thermal energy by nuclear fission when heavy nuclei break apart.
核电站使用核燃料,当重核破裂时,核裂变产生热能。

The Sun is the origin of all our energy resources except geothermal, nuclear and tidal.
除了地热、核能和潮汐能,太阳是我们所有能源的起源。

The source of the Sun’s energy is nuclear fusion, when hydrogen fuses (joins) together to form helium.
太阳的能量来源于核聚变,即氢聚变(结合)成氦。

We are trying to develop nuclear fusion reactors.
我们正在努力开发核聚变反应堆。

EXAM-STYLE QUESTIONS  考题

1 Which of the following energy resources does not depend on sunlight?
1 下列哪种能源不依赖阳光?
A hydroelectric power  水力发电
B tidal  B 潮汐
C coal  C 煤
D wind  D 风
2 Which of the following energy resources produces greenhouse gases?
2 以下哪种能源会产生温室气体?
A hydroelectric power  水力发电
B nuclear  B 核
C biofuels  C 生物燃料
D natural gas  D 天然气
3 Which of these power stations generates electricity from gravitational potential energy?
3 哪些发电站利用重力势能发电?
A tidal power station  潮汐发电站
B geothermal power station
B 地热发电站

C wind power  C 风能
D nuclear power station
D 核电站

4 Which of these power stations generates electricity from thermal energy?
4 哪些发电站利用热能发电?
A hydroelectric power station
水电站

B wind farm  B 风电场
C coal-fired power station
C 燃煤发电站

D solar farm  D 太阳能发电场
5 Here is a list of energy resources available to the world. Some of these are renewable and some are non-renewable.
5 以下是世界上可利用的能源资源清单。其中有些是可再生能源,有些是不可再生能源。
Energy resource  能源资源 Non-renewable  不可再生 Renewable  可再生
wave power  波浪发电
hydroelectricity  水力发电
geothermal  地热
coal  煤炭
nuclear energy  核能
oil  油类
solar energy  太阳能
natural gas  天然气
tidal energy  潮汐能
wind energy  风能
Energy resource Non-renewable Renewable wave power hydroelectricity geothermal coal nuclear energy oil solar energy natural gas tidal energy wind energy | Energy resource | Non-renewable | Renewable | | :--- | :--- | :--- | | wave power | | | | hydroelectricity | | | | geothermal | | | | coal | | | | nuclear energy | | | | oil | | | | solar energy | | | | natural gas | | | | tidal energy | | | | wind energy | | |
Copy the table. In the first blank column, put a tick by any three resources that are non-renewable.
复制表格。在第一空栏中,在任意三种不可再生资源旁打上"√"。
In the second blank column, put a tick by any three resources that are renewable.
在第二空栏中,在任意三种可再生资源旁打上"√"。
6 This question is about hydroelectric power stations. Water is stored behind a dam. Water released from the dam flows downhill and spins a turbine. The spinning turbine causes a generator to turn, which produces electricity.
6 本题是关于水电站的。水储存在大坝后面。从大坝放出的水流向山下,带动涡轮机旋转。旋转的涡轮机带动发电机转动,从而产生电力。

a Write down the two energy transfers that occur in a hydroelectric power station.
a 请写出水电站中发生的两种能量传递。

b Explain how electricity from a hydroelectric power station relies on energy from the Sun.
b 解释水力发电站的发电如何依赖于太阳的能量。

c The demand for electricity is not constant. Explain how a hydroelectric power station can help match supply to demand.
c 电力需求不是恒定的。请解释水电站如何帮助实现供需平衡。
7 A coal-fired power station and a wind turbine both produce electrical power. The power station produces 2500 MW and the wind turbine produces 2.0 MW .
7 一个燃煤发电站和一个风力涡轮机都能发电。发电站的发电量为 2500 兆瓦,风力涡轮机的发电量为 2.0 兆瓦。

a State one advantage of using wind turbines instead of a coal-fired power station to produce electricity.
a 请说明使用风力涡轮机而不是燃煤发电站发电的一个优势。

b Coal-fired power stations still account for one third of the world’s energy consumption. Explain why wind turbines have not replaced them.
b 燃煤发电站仍占世界能源消耗的三分之一。请解释为什么风力涡轮机没有取代燃煤发电站。

[Total: 3]  [共计:3]
8 An energy company is proposing a new power station but has to decide between a solar power station and a geothermal power station. Explain how the location and the climate might affect the decision.
8 某能源公司拟新建一座发电站,但必须在太阳能发电站和地热发电站之间做出选择。请解释地理位置和气候如何影响该决定。
9 Imagine that you are writing a science fiction story in which fossil fuels have run out and the Sun stops shining. Only power stations based on three different energy resources will continue producing power.
9 想象一下,你正在写一个化石燃料耗尽、太阳不再发光的科幻小说。只有基于三种不同能源的发电站才能继续发电。

a You want to make your story believable, so what power stations should people use if they want to survive?
a 你想让你的故事可信,那么如果人们想生存下去,应该使用什么发电站?

b As the temperature falls, explain why one of these power stations might stop working.
b 随着气温下降,请解释其中一个发电站可能停止工作的原因。

SELF-EVALUATION CHECKLIST
自我评估清单

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
学习完本章后,请思考自己对不同主题的掌握程度。这将帮助你看到自己在知识方面的差距,并帮助你更有效地学习。
I can  我可以
  查看主题...
See
Topic...
See Topic...| See | | :--- | | Topic... |
  需要更多努力
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
  快到了
Almost
there
Almost there| Almost | | :--- | | there |

有信心继续前进
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |

认识到电力是将能源转移到使用地点的一种便捷方式。
Appreciate that electricity is a
convenient way of moving energy to
where it is used.
Appreciate that electricity is a convenient way of moving energy to where it is used.| Appreciate that electricity is a | | :--- | | convenient way of moving energy to | | where it is used. |
7.1

回顾所有能源资源:化石燃料、生物燃料、核能、地热、太阳能、水电、潮汐能、风能和波浪能。
Recall all the energy resources: fossil
fuels, biofuels, nuclear, geothermal,
solar, hydroelectric, tidal, wind and
wave.
Recall all the energy resources: fossil fuels, biofuels, nuclear, geothermal, solar, hydroelectric, tidal, wind and wave.| Recall all the energy resources: fossil | | :--- | | fuels, biofuels, nuclear, geothermal, | | solar, hydroelectric, tidal, wind and | | wave. |
7.1

描述每种能源如何产生电力或其他有用的能源形式。
Describe how each energy resource
can produce electricity or other useful
forms of energy.
Describe how each energy resource can produce electricity or other useful forms of energy.| Describe how each energy resource | | :--- | | can produce electricity or other useful | | forms of energy. |
7.1

说明每种能源在可再生性、成本、可靠性、可用性、规模和环境影响方面的优缺点。
Give advantages and disadvantages of
each energy resource in terms of
renewability, cost, reliability,
availability, scale and environmental
impact.
Give advantages and disadvantages of each energy resource in terms of renewability, cost, reliability, availability, scale and environmental impact.| Give advantages and disadvantages of | | :--- | | each energy resource in terms of | | renewability, cost, reliability, | | availability, scale and environmental | | impact. |
7.1

了解太阳是除地热、核能和潮汐能之外所有能源的起源。
Understand that the Sun is the origin
of all our energy resources except
geothermal, nuclear and tidal.
Understand that the Sun is the origin of all our energy resources except geothermal, nuclear and tidal.| Understand that the Sun is the origin | | :--- | | of all our energy resources except | | geothermal, nuclear and tidal. |
7.2

了解核聚变是太阳的能量来源。
Understand that nuclear fusion is the
source of the Sun’s energy.
Understand that nuclear fusion is the source of the Sun’s energy.| Understand that nuclear fusion is the | | :--- | | source of the Sun’s energy. |
7.2

要知道,在开发核聚变反应堆方面已经进行了大量研究。
Understand that there has been a lot of
research into developing a nuclear
fusion reactor.
Understand that there has been a lot of research into developing a nuclear fusion reactor.| Understand that there has been a lot of | | :--- | | research into developing a nuclear | | fusion reactor. |
7.2
I can "See Topic..." "Needs more work" "Almost there" "Confident to move on" "Appreciate that electricity is a convenient way of moving energy to where it is used." 7.1 "Recall all the energy resources: fossil fuels, biofuels, nuclear, geothermal, solar, hydroelectric, tidal, wind and wave." 7.1 "Describe how each energy resource can produce electricity or other useful forms of energy." 7.1 "Give advantages and disadvantages of each energy resource in terms of renewability, cost, reliability, availability, scale and environmental impact." 7.1 "Understand that the Sun is the origin of all our energy resources except geothermal, nuclear and tidal." 7.2 "Understand that nuclear fusion is the source of the Sun’s energy." 7.2 "Understand that there has been a lot of research into developing a nuclear fusion reactor." 7.2 | I can | See <br> Topic... | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :---: | :---: | :---: | :--- | | Appreciate that electricity is a <br> convenient way of moving energy to <br> where it is used. | 7.1 | | | | | Recall all the energy resources: fossil <br> fuels, biofuels, nuclear, geothermal, <br> solar, hydroelectric, tidal, wind and <br> wave. | 7.1 | | | | | Describe how each energy resource <br> can produce electricity or other useful <br> forms of energy. | 7.1 | | | | | Give advantages and disadvantages of <br> each energy resource in terms of <br> renewability, cost, reliability, <br> availability, scale and environmental <br> impact. | 7.1 | | | | | Understand that the Sun is the origin <br> of all our energy resources except <br> geothermal, nuclear and tidal. | 7.2 | | | | | Understand that nuclear fusion is the <br> source of the Sun’s energy. | 7.2 | | | | | Understand that there has been a lot of <br> research into developing a nuclear <br> fusion reactor. | 7.2 | | | |

> Chapter 8  > 第 8 章

Work and power  工作与权力

IN THIS CHAPTER YOU WILL:
在本章中,您将

  • learn that work done equals energy transferred, or the force multiplied by the distance moved in the direction of the force
    了解做功等于传递的能量,或力乘以沿力的方向移动的距离
  • relate power to work done and time taken
    将功率与所做的工作和所用的时间联系起来
  • calculate work done and power ( W = F d = Δ E W = F d = Δ E (W=Fd=Delta E:}\left(W=F d=\Delta E\right. and P = Δ E t ) P = Δ E t {:P=(Delta E)/(t))\left.P=\frac{\Delta E}{t}\right).
    计算做功和功率 ( W = F d = Δ E W = F d = Δ E (W=Fd=Delta E:}\left(W=F d=\Delta E\right. P = Δ E t ) P = Δ E t {:P=(Delta E)/(t))\left.P=\frac{\Delta E}{t}\right)

GETTING STARTED  入门

What is the everyday meaning of ‘work’? Do you know what the word
工作 "的日常含义是什么?您知道

‘work’ means in physics?
功 "在物理学中是什么意思?

What is the everyday meaning of ‘power’? Do you know what the word ‘power’ means in physics?
功率 "的日常含义是什么?您知道 "功率 "一词在物理学中的含义吗?

THE MACHINE AGE  机器时代

Scottish engineer, James Watt (Figure 8.1), developed the steam engine just over two centuries ago. It delivered the power for Britain to enter the industrial age. According to legend, Watt’s first customer wanted a steam engine only if it could pump at least as much water from a well to his brewery as one of his horses. The brewer picked his strongest horse and worked him hard for eight hours. The horse pumped 33000 lb (1 lb = 0.453592 kg lb = 0.453592 kg lb=0.453592kg\mathrm{lb}=0.453592 \mathrm{~kg} ) of water to a height of one foot ( 0.3048 metres) per minute; this is the origin of the unit of horsepower. As you will see, this is equivalent to about 750 watts. The watt, W, is the SI unit for power, named after James Watt. Long before steam power, humans developed mechanical machines to perform tasks that might otherwise be impossible (such as the ramp to build the pyramids). Electric motors have been developed since.
两个多世纪前,苏格兰工程师詹姆斯-瓦特(图 8.1)发明了蒸汽机。它为英国进入工业时代提供了动力。据传说,瓦特的第一位客户想要一台蒸汽机,条件是蒸汽机从井里抽出的水至少要和他的一匹马抽出的水一样多。酿酒师挑选了他最强壮的一匹马,并让它辛勤工作了八个小时。这匹马每分钟将 33000 磅(1 lb = 0.453592 kg lb = 0.453592 kg lb=0.453592kg\mathrm{lb}=0.453592 \mathrm{~kg} )的水抽到一英尺(0.3048 米)的高度;这就是马力单位的由来。如您所见,这相当于约 750 瓦特。瓦特(W)是表示功率的 SI 单位,以詹姆斯-瓦特(James Watt)命名。早在蒸汽动力之前,人类就开发出了机械设备来完成原本不可能完成的任务(例如建造金字塔的斜坡)。此后,人们又开发出了电动机。

Figure 8.1: The Golden Boys statue in the UK, which shows Matthew Boulton, James Watt (centre) and William Murdoch, who all played a part in improving the steam engine.
图 8.1:英国的 "金童 "雕像,雕像中的马修-布尔顿、詹姆斯-瓦特(中)和威廉-默多克都曾参与改进蒸汽机。

Discussion questions  讨论问题

1 What are the advantages and disadvantages of using steam power?
1 使用蒸汽动力有哪些优缺点?

2 Can we produce power without any of the disadvantages of steam power?
2 我们能否在没有蒸汽动力任何缺点的情况下发电?
3 The steam engine is a machine that changes thermal energy (from a difference in temperature) into mechanical work. What other machines are mentioned in this section, and what energy transfers take place?
3 蒸汽机是一种将热能(来自温度差)转化为机械功的机器。本节还提到了哪些其他机器,发生了哪些能量转移?

8.1 Doing work  8.1 工作

Figure 8.2 shows one way of lifting a heavy object. Pulling on the rope raises the heavy box. As you pull, the force moves the box upwards.
图 8.2 展示了一种提升重物的方法。拉动绳子可以将重物举起。拉动时,力会使箱子向上移动。

Figure 8.2: Lifting an object requires an upward force, pulling against gravity. As the box rises upwards, the force also moves upwards. Energy is being transferred by the force to the box.
图 8.2:提升物体需要一个向上的力,与重力对抗。当箱子向上升起时,力也向上移动。力将能量传递给了盒子。
To lift an object, you need a store of energy (as chemical energy, in your muscles). You give the object more gravitational potential energy (g.p.e.). The force is your means of transferring energy from you to the object. The name given to this type of energy transfer by a force is doing work.
要举起一个物体,你需要储存能量(如肌肉中的化学能)。你赋予物体更多的重力势能(g.p.e.)。力是将能量从你传递给物体的手段。这种通过力进行的能量转移被称为做功。

The more work that a force does, the more energy it transfers. The amount of work done is simply the amount of energy transferred:
力做的功越大,传递的能量就越多。所做的功就是传递的能量:

work done = energy transferred
做功 = 能量传递

Three further examples of forces doing work are shown in Figure 8.3.
图 8.3 还显示了三个力做功的例子。

Figure 8.3: Three examples of forces doing work. In each case, the force moves as it transfers energy. a: Pushing a shopping trolley to start it moving. The pushing force does work. It transfers energy to the trolley, and the trolley’s kinetic energy (k.e.) increases. b: An apple falling from a tree. Gravity pulls the apple downwards. Gravity does work, and the apple’s k.e. increases. c: Braking to stop a bicycle. The brakes produce a force of friction, which slows down the bicycle. The friction does work, and the bicycle’s k.e. is transferred to the internal energy of the brakes, which get hot.
图 8.3:力做功的三个例子。在每个例子中,力在传递能量的同时也在移动。 a: 推购物车使其开始移动。推力做功。b: 苹果从树上掉下来。重力将苹果向下拉。重力做功,苹果的动能增加:刹车让自行车停下来。刹车产生摩擦力,使自行车减速。摩擦力做功,自行车的 k.e. 转化为刹车的内能,刹车变热。
Mechanical or electrical work is equal to energy transferred. In this chapter, we are focussing on mechanical work. Electrical work is discussed in Chapter 18.
机械功或电功等于传递的能量。本章重点讨论机械功。电功将在第 18 章中讨论。

How much work?  工作量有多大?

Think about lifting a heavy object, as shown in Figure 8.2. A heavy object needs a big force to lift it. The heavier the object is, and the higher it is lifted, the more its g.p.e. increases. This suggests that the amount of energy transferred by a force depends on two things:
如图 8.2 所示,想一想举起重物的过程。提升重物需要很大的力。物体越重,举得越高,其 g.p.e. 就越大。这表明,力所传递的能量取决于两个方面:
  • the size of the force - the greater the force, the more work it does
    力的大小--力越大,做功越多
  • the distance moved in the direction of the force - the further it moves, the more work it does.
    沿力的方向移动的距离--移动得越远,做功越多。
So a big force moving through a big distance does more work than a small force moving through a small distance.
因此,通过大距离运动的大力比通过小距离运动的小力做功更多。

Words in physics  物理学中的词汇

You will by now understand that ‘work’ is a word that has a specialised meaning in physics, different from its meaning in everyday life. When physicists think about the idea of work, they think about forces causing movement.
你现在应该明白,"功 "这个词在物理学中有着特殊的含义,与日常生活中的含义不同。物理学家在思考 "功 "的概念时,会想到引起运动的力。
If you are sitting thinking about your homework, no forces are causing movement and you are doing no work. It is only when you start to write that you are doing work in the physics sense. To make the ink flow from your pen, you must push against the force of friction, and then you really are working. Similarly, you are doing work (in the sense of physics) when you lift up this heavy book.
如果你正坐着思考作业,没有任何力会导致运动,你也没有做功。只有当你开始写字时,你才在做物理学意义上的功。为了让墨水从你的笔中流出,你必须顶住摩擦力,然后你才真正在做功。同样,当你举起这本沉重的书时,你也在做功(物理学意义上的功)。
Many words have specialised meanings in science. In earlier chapters, we used these words: ‘force’, ‘mass’, ‘weight’, ‘velocity’, ‘moment’ and ‘energy’. Each has a carefully defined meaning in physics. This is important because physicists have to agree on the terms they are using. However, if you look these words up in a dictionary, you will find that they have a range of everyday meanings, as well as their specialised scientific meaning. This is not a problem, provided you know whether you are using a particular word in its scientific sense or in a more everyday sense. (Some physicists get very upset if they hear shopkeepers talking about weights in kilograms, for example, but no-one will understand you if you ask for 10 newtons of oranges!)
许多词在科学中都有专门的含义。在前面的章节中,我们使用了这些词:力"、"质量"、"重量"、"速度"、"力矩 "和 "能量"。每个词在物理学中都有明确的含义。这一点很重要,因为物理学家必须就他们使用的术语达成一致。但是,如果你在字典中查找这些词,你会发现它们除了专业的科学含义外,还有一系列日常含义。这不是问题,只要你知道你是在科学意义上还是在更日常的意义上使用某个词。(例如,一些物理学家听到店主说重量是以千克为单位时会非常生气,但如果你要 10 牛顿的橘子,没人会明白你的意思!)。

8.2 Calculating work done
8.2 计算所做功

When a force does work, it transfers energy to the object it is acting on. The amount of energy transferred is equal to the amount of work done. We can write this as a simple equation:
力做功时,会将能量传递给它作用的物体。传递的能量等于做的功。我们可以将其写成一个简单的等式:
W = Δ E W = Δ E W=Delta EW=\Delta E
In this equation, we use the symbol Δ Δ Delta\Delta (Greek capital letter delta) to mean ‘amount of’ or ‘change in’. So,
在这个等式中,我们使用符号 Δ Δ Delta\Delta (希腊文大写字母 delta)来表示 "量 "或 "变化"。那么

Δ E = Δ E = Delta E=\Delta E= change in energy
Δ E = Δ E = Delta E=\Delta E= 能量变化

How can we calculate the work done by a force? The work done depends on two things:
如何计算力做的功?所做的功取决于两点:
  • the size of the force, F F FF
    力的大小, F F FF
  • the distance, d d dd, moved by the force.
    力移动的距离, d d dd
We can then write an equation for this:
然后,我们就可以写出一个等式:

work done by a force = = == force × × xx\times distance moved by the force in the direction of the force.
力做的功 = = == × × xx\times 力沿力的方向移动的距离。
In symbols:  用符号表示:
W = F d = Δ E W = F d = Δ E W=Fd=Delta EW=F d=\Delta E

KEY EQUATION  关键方程

work done by a force = = == force × × xx\times distance moved by the force in the c
力做的功 = = == × × xx\times 力在 c 中移动的距离
W = F d = Δ E W = F d = Δ E W=Fd=Delta EW=F d=\Delta E
The phrase ‘in the direction of the force’ will be explained shortly. As amount of work done is the same as the amount of energy transferred, it is measured in joulles ( J ), the unit of energy.
在力的方向上 "这句话稍后会解释。由于所做的功等同于传递的能量,因此用能量单位焦耳(J)来衡量。

Joules and newtons 

The equation for the work done by a force ( W = F × d W = F × d W=F xx dW=F \times d ) shows us the relationship between joules and newtons. If we replace each quantity in the equation by its SI unit, we get 1 J = 1 N × 1 m = 1 Nm 1 J = 1 N × 1 m = 1 Nm 1J=1Nxx1m=1Nm1 \mathrm{~J}=1 \mathrm{~N} \times 1 \mathrm{~m}=1 \mathrm{Nm}. So, a joule is a newton metre. 

WORKED EXAMPLE 8.1 

A crane lifts a crate upwards through a height of 20 metres. The lifting force provided by the crane is 5.0 kN , as shown in Figure 8.4. 

Figure 8.4: 

a How much work is done by the force? 
b How much energy is transferred to the crate? 
Step 1: Write down what you know, and what you want to know. 
F = 5.0 kN = 5000 N d = 20 m W = ? F = 5.0 kN = 5000 N d = 20 m W = ? {:[F=5.0kN=5000N],[d=20m],[W=?]:}\begin{aligned} & F=5.0 \mathrm{kN}=5000 \mathrm{~N} \\ & d=20 \mathrm{~m} \\ & W=? \end{aligned}
Step 2: Write down the equation for work done, substitute values and solve. 
W = F × d = 5000 N × 20 m 100000 Nm 100000 J W = F × d = 5000 N × 20 m 100000 Nm 100000 J {:[W=F xx d=5000Nxx20m],[100000Nm-=100000J]:}\begin{aligned} W= & F \times d=5000 \mathrm{~N} \times 20 \mathrm{~m} \\ & 100000 \mathrm{Nm} \equiv 100000 \mathrm{~J} \end{aligned}

Answer 

a The work done by the force is 100000 J , or 100 kJ . 
b Since work done = energy transferred, 100 kJ of energy is transferred to the crate. 

Work done and mgh 

Worked Example 8.1 illustrates an important idea. The force provided by the crane to lift the crate must equal the crate’s weight, mg . It lifts the crate through a height, h h hh. Then the work it does is force × × xx\times distance, or m g × h m g × h mg xx hm g \times h. The gain in g.p.e. of the crate is mgh . This explains where the equation for g.p.e. comes from. 
In Figure 8.5, the child slides down the ramp. Gravity pulls her downwards, and makes her speed up. To calculate the work done by gravity, we need to know the vertical distance, h h hh, because this is the distance moved in the direction of the force. If we calculated the work done as weight × × xx\times distance moved down the ramp, we would get an answer that was too large. Now you should understand why we write the definition of work done like this: 
Figure 8.5: It is important to use the correct distance when calculating work done by a force. Gravity makes the child slide down the slope. However, to calculate the energy transferred by gravity, we must use the vertical height moved. 

Forces doing no work 

If you sit still on a chair (Figure 8.6), there are two forces acting on you. These are your weight, mg , acting downwards, and the upward contact force, C C CC, of the chair, which stops you from falling through the bottom of the chair. 
Figure 8.6: When you sit still in a chair, there are two forces acting on you. Neither transfers energy to you. 
Neither of these forces is doing any work on you. The reason is that neither of the forces is causing movement, so you do not move through any distance, d d dd. From W = F × d W = F × d W=F xx dW=F \times d, the amount of work done by each force is zero. When you sit still on a chair, your energy does not increase or decrease as a result of the forces acting on you. 
Figure 8.7 shows another example of a force that is doing no work. A spacecraft is travelling around the Earth in a circular orbit. The Earth’s gravity pulls on the spacecraft to keep it in its orbit. The force is directed towards the centre of the Earth. However, since the spacecraft’s orbit is circular, it does not get any closer to the centre of the Earth. There is no movement in the direction of the force, and so gravity does no work. The spacecraft continues at a steady speed (its k.e. is constant) and at a constant height above the Earth’s surface (its g.p.e. is constant). Of course, although the force is doing no work, this does not mean that it is not having an effect. Without the force, the spacecraft would escape from the Earth and disappear into the depths of space. 
Figure 8.7: The spacecraft stays at a constant distance from the Earth. Gravity keeps it in its orbit without transferring any energy to it. 

WORKED EXAMPLE 8.2 

A girl can provide a maximum pushing force of 200 N . To move a box weighing 400 N onto a platform, she uses a plank as a ramp, as shown in Figure 8.8. 

Figure 8.8 

a How much work does she do in raising the box? 
b How much g.p.e. does the box gain? 
Step 1: Write down what you know, and what you want to know. 
pushing force along the ramp F = 200 N F = 200 N F=200NF=200 \mathrm{~N} distance moved along ramp d = 2.5 m d = 2.5 m d=2.5md=2.5 \mathrm{~m} 
weight of box downwards m g = 400 N m g = 400 N mg=400Nm g=400 \mathrm{~N} vertical distance moved h = 0.75 m h = 0.75 m h=0.75mh=0.75 \mathrm{~m} 
work done along the ramp W = W = W=W= ? 
work done against gravity W = W = W=W= ? 
Step 2: Calculate the work done, W W WW, by the pushing force along the ramp. 
W = pushing force along ramp × distance moved along ras = F × d = 200 N × 2.5 m = 500 J W =  pushing force along ramp  ×  distance moved along ras  = F × d = 200 N × 2.5 m = 500 J {:[W=" pushing force along ramp "xx" distance moved along ras "],[=F xx d],[=200Nxx2.5m],[=500J]:}\begin{aligned} W & =\text { pushing force along ramp } \times \text { distance moved along ras } \\ & =F \times d \\ & =200 \mathrm{~N} \times 2.5 \mathrm{~m} \\ & =500 \mathrm{~J} \end{aligned}
Step 3: Calculate the gain in g.p.e. of the box. This is the same as the work done against gravity, W W W^(')W^{\prime}. 
W = weight of box vertical distance moved = m g × h = 400 N × 0.75 m = 300 J W =  weight of box vertical distance moved  = m g × h = 400 N × 0.75 m = 300 J {:[W^(')=" weight of box vertical distance moved "],[=mg xx h],[=400Nxx0.75m],[=300J]:}\begin{aligned} W^{\prime} & =\text { weight of box vertical distance moved } \\ & =m g \times h \\ & =400 \mathrm{~N} \times 0.75 \mathrm{~m} \\ & =300 \mathrm{~J} \end{aligned}

Answer 

a The girl does 500 J of work in raising the box.
a 女孩为抬起箱子做了 500 J 的工作。

b The box gains 300 J in g.p.e.
b 箱体获得 300 焦耳的能量密度。

Note: only 300 J is transferred to the box. The remaining 200 J is the work done against friction as the box is pushed along the slope.
注意:只有 300 焦耳传给了盒子。剩下的 200 焦耳是盒子在斜坡上被推动时摩擦力做的功。

ACTIVITY 8.1  活动 8.1

The efficiency of a ramp
斜坡的效率

The Ancient Egyptians are famous for their pyramids. The pyramids are made from stone blocks with a mass of at least one tonne. The useful work done in lifting a block is the weight of the block multiplied by the vertical distance it is moved. Whether the block is dragged or lifted, work has to be done against the force of gravity.
古埃及人以金字塔闻名于世。金字塔由质量至少为一吨的石块砌成。提升石块所做的有用功是石块的重量乘以垂直移动的距离。无论是拖动还是提升石块,都必须对重力做功。

The Ancient Egyptians did not have cranes. It is believed they used ramps to drag the blocks into place. The ramp is an example of a simple machine.
古埃及人没有起重机。据说他们用斜坡把木块拖到合适的位置。斜坡就是简单机械的一个例子。
It reduces the required force (sometimes called the effort) but this force has to be applied over a bigger distance than if it were moved vertically. If there was no friction, the ramp would be 100 % 100 % 100%100 \% efficient. However, additional work needs to be done against friction.
它减小了所需的力(有时称为努力),但与垂直移动相比,这个力需要施加的距离更大。如果没有摩擦,斜面的效率就会 100 % 100 % 100%100 \% 高。但是,需要对摩擦做额外的功。
Your task is to design an experiment to work out how the efficiency of a ramp depends on the angle of the slope.
您的任务是设计一个实验,以确定斜坡的效率如何取决于斜坡的角度。

You need to recall what efficiency is and how to calculate it.
您需要记住什么是效率以及如何计算效率。

You need to decide on:
您需要决定
  • the equipment you will need (sketch a labelled diagram of your assembled apparatus)
    你需要的设备(勾画出你组装的设备的标示图)
  • your independent, dependent and control variables (that is, what you will change, what you will measure and what you will keep the same) 
  • the measurements that you will take (design a table to record them)
    您要进行的测量(设计一张表格记录测量结果)
  • what you need to calculate (what equations you will need)
    您需要计算的内容(您需要哪些公式)
  • the graph that you will plot and predict what your graph should look like.
    您要绘制的图形,并预测您的图形应该是什么样子。
Then join together to form design teams of three. First, peer assess each other’s work. Then develop a combined plan, putting together the best of the individual plans.
然后一起组成三人设计小组。首先,互评彼此的作品。然后制定一个综合计划,将各个计划中的精华融合在一起。

Discuss your plan with a neighbouring group and be prepared to share your ideas with the class.
与邻近小组讨论您的计划,并准备在全班分享您的想法。

PEER ASSESSMENT  同侪评估

When you check the plan in Activity 8.1, go through this list.
当您检查活动 8.1 中的计划时,请查看此列表。
  • Does the plan include the correct equipment? Is anything missing?
    计划是否包括正确的设备?是否有遗漏?
  • Does the plan include a clear sketch of the experimental setup?
    计划是否包括清晰的实验装置草图?
  • Are the independent, dependent and control variables identified correctly?
    自变量、因变量和控制变量的确定是否正确?
  • The experiment requires that one end of the ramp is raised. Is there a sensible range of heights? Does the plan explain how the angle of the
    实验要求将斜坡的一端升高。是否有合理的高度范围?计划是否解释了坡道的角度

    slope is calculated?  斜率是如何计算的?
  • Does it suggest repeating measurements so that an average value of the dependent variable can be calculated for each value of the independent variable?
    是否建议重复测量,以便为自变量的每个值计算出因变量的平均值?
  • Does the plan explain how to do any calculations?
    计划是否解释了如何进行计算?
  • Does the plan explain what to plot on each axis of the graph?
    计划是否解释了在图表的每个轴上绘制什么?
  • Does the plan predict what the graph should look like?
    计划是否预测了图表应该是什么样子?
  • Can you carry out the experiment successfully based on what your classmates have written?
    你能根据同学所写的内容成功进行实验吗?
Improve your own work using the feedback on it from another group.
利用其他小组的反馈意见改进自己的作品。

Questions  问题

1 Explain why no work is being done by the Sun’s gravity on the Earth.
1 解释为什么太阳引力没有对地球做功。

2 A boy pulls a sled (at a constant speed) with a force of 50 N for a distance of 250 metres. How much work does the boy do?
2 一个男孩用 50 牛的力拉着一个雪橇(匀速)走了 250 米。男孩做了多少功?
3 A 100 g apple falls from a tree and lands on the ground 6 metres below.
3 一个 100 克重的苹果从树上掉下来,落在 6 米以下的地面上。

a What is the force that is pulling the apple, and how large is the force?
a 拉动苹果的力是什么,有多大?

b Calculate how much work gravity does on the apple as it falls.
b 计算苹果下落时重力做了多少功。

c What energy transfer is taking place?
c 发生了什么能量转移?

4 A man is trying to move his washing machine into the back of a removal van. The washing machine has a weight of 650 N and the floor of the removal van is 0.80 metres above the ground (Figure 8.9).
4 某人正试图将他的洗衣机搬到一辆搬运车的后部。洗衣机重 650 牛顿,搬运车的地板离地面 0.80 米(图 8.9)。

Figure 8.9: Moving a washing machine.
图 8.9:移动洗衣机。

a Calculate the work the man does when he pushes the washing machine up the ramp into the back of the van against a frictional force of 440 N .
a 计算该男子在 440 N 的摩擦力作用下将洗衣机从斜坡上推到货车后部时所做的功。

b Superman happens to be passing and he lifts the washing machine vertically into the van (without using the plank). Calculate the work Superman does. 
c Explain why a and b b b\mathbf{b} give different answers. 
d Calculate the efficiency of the ramp. 

8.3 Power 

Exercising in the gym (Figure 8.10) can put great demands on your muscles. Speeding up a treadmill means that you have to work harder to keep up. 
Equally, your trainer might ask you to find out how many times you can lift a set of weights in one minute. These exercises are a test of how powerful you are. The faster you work, the greater your power. 
Figure 8.10: At the gym, it is easier to lift small loads, and to lift them slowly. The greater the load you lift and the faster you lift it, the greater the power required. It is the same with running on a treadmill. The faster you have to run, the greater the rate at which you do work. 
In physics, the word power is used with a special meaning. It means the rate at which you do work (that is, how fast you work). The more work you do, and the shorter the time in which you do it, the greater your power. Power is the rate at which energy is transferred, or the rate at which work is done. 

Fast working 

Power tells you about the rate at which a force does work, that is the rate at which it transfers energy. When you lift an object up, you are transferring energy to it. Its potential energy increases. You can increase your power by: 
  • lifting a heavier object in the same time 
  • lifting the object more quickly. 
It is not just people who do work. Machines also do work, and we can talk about their power in the same way. 
  • A crane does work when it lifts a load. The bigger the load and the faster it lifts the load, the greater the power of the crane. 
  • A locomotive pulling a train of coaches or wagons does work. The greater the force with which it pulls and the greater the speed at which it pulls, the greater the power of the locomotive. 

Question 

5 Your neighbour is lifting bricks and placing them on top of a wall. He lifts them slowly, one at a time. State two ways in which he could increase his power (the rate at which he is transferring energy to the bricks). 

8.4 Calculating power 

We know from Section 8.3, that power is the rate at which work is done. Since work done is equal to energy transferred, we can write these ideas about power as equations, as shown. 

KEY EQUATIONS 

power = work done time taken p = w t  power  =  work done   time taken  p = w t {:[" power "=(" work done ")/(" time taken ")],[p=(w)/(t)]:}\begin{aligned} \text { power } & =\frac{\text { work done }}{\text { time taken }} \\ p & =\frac{w}{t} \end{aligned}
power = energy transferred time taken p = Δ E t  power  =  energy transferred   time taken  p = Δ E t {:[" power "=(" energy transferred ")/(" time taken ")],[p=(Delta E)/(t)]:}\begin{aligned} \text { power } & =\frac{\text { energy transferred }}{\text { time taken }} \\ p & =\frac{\Delta E}{t} \end{aligned}

ACTIVITY 8.2 

The meaning of work and power 
Work and power have very specific meanings in physics. You will create a resource that helps people understand their correct meanings in physics. You could start by collecting definitions that are wrong in physics. For example, collect images of powerful people or cars, or an image of a gymnast performing the ‘iron cross’ (Figure 8.11). If you are feeling creative, you could write a poem or create a song or podcast. 
Figure 8.11: This gymnast is performing the ‘iron cross’ on rings. It is a move that requires tremendous strength in the core, arms and wrists. But is he doing any work from a physics point of view? 
  • Work for a few minutes in small groups to discuss ideas and choose one idea to develop. 
  • Develop your idea within the timeframe given by your teacher. 
  • If the class is divided into two or three large groups, you will perform or present your idea to the other pairs or threes in your group. 
  • Vote on the other presentations. A pair or three cannot represent their group unless the physics is correct so help correct any physics mistakes. 
  • The pair or three chosen by each group will present to the rest of the class. 

Units of power 

Power is measured in watts ( W ). One watt ( 1 W ) ( 1 W ) (1W)(1 \mathrm{~W}) is the power when one joule ( 1 J J JJ ) of work is done per unit time. So one watt is one joule per second. 
  • 1 W = 1 J / s 1 W = 1 J / s 1W=1J//s1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}
  • 1000 W = 1 kW 1000 W = 1 kW 1000W=1kW1000 \mathrm{~W}=1 \mathrm{~kW} (kilowatt) 
  • 1000000 W = 1 MW 1000000 W = 1 MW 1000000W=1MW1000000 \mathrm{~W}=1 \mathrm{MW} (megawatt) 
Take care not to confuse W W WW for work done (or energy transferred) with W for watts. In books, the first of these is shown in italic type (as here), but you cannot usually tell the difference when they are handwritten. SI units are often related to each other. It is useful to remember some of the connections, such as 1 J = 1 Nm 1 J = 1 Nm 1J=1Nm1 \mathrm{~J}=1 \mathrm{Nm} and 1 W = 1 J / s 1 W = 1 J / s 1W=1J//s1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}. 

Power in general 

We can apply the idea of power to any transfer of energy. For example, electric light bulbs transfer energy supplied to them by electricity. They transfer energy by light and heating. Most light bulbs are labelled with their power rating - for example, 40 W , 60 W , 100 W 40 W , 60 W , 100 W 40W,60W,100W40 \mathrm{~W}, 60 \mathrm{~W}, 100 \mathrm{~W} - to tell the user about the rate at which the bulb transfers energy. 
We can express the efficiency of a light bulb or any other energy-changing device in terms of the power it supplies: 
percentage efficiency = useful power output power input × 100 %  percentage efficiency  =  useful power output   power input  × 100 % " percentage efficiency "=(" useful power output ")/(" power input ")xx100%\text { percentage efficiency }=\frac{\text { useful power output }}{\text { power input }} \times 100 \%
Remember, from Chapter 6, how this compares with the equation for energy efficiency in terms of energy: 
percentage efficiency = useful energy output energy input × 100 %  percentage efficiency  =  useful energy output   energy input  × 100 % " percentage efficiency "=(" useful energy output ")/(" energy input ")xx100%\text { percentage efficiency }=\frac{\text { useful energy output }}{\text { energy input }} \times 100 \%
There is more about electrical power in Chapter 18. 

WORKED EXAMPLE 8.3 

A car of mass 800 kg accelerates from rest to a speed of 25 m / s 25 m / s 25m//s25 \mathrm{~m} / \mathrm{s} in 10 s . What is its power? 
Step 1: Calculate the work done. This is the increase in the car’s kinetic energy. 
k.e. = 1 2 m v 2 = 1 2 × 800 kg × ( 25 m / s ) 2 = 250000 J  k.e.  = 1 2 m v 2 = 1 2 × 800 kg × ( 25 m / s ) 2 = 250000 J {:[" k.e. "=(1)/(2)mv^(2)],[=(1)/(2)xx800kgxx(25m//s)^(2)],[=250000J]:}\begin{aligned} \text { k.e. } & =\frac{1}{2} m v^{2} \\ & =\frac{1}{2} \times 800 \mathrm{~kg} \times(25 \mathrm{~m} / \mathrm{s})^{2} \\ & =250000 \mathrm{~J} \end{aligned}
Step 2: Calculate the power. 
power = work done time taken = 250000 J 10 s = 25000 W 25 kW  power  =  work done   time taken  = 250000 J 10 s = 25000 W 25 kW {:[" power "=(" work done ")/(" time taken ")],[=(250000(J))/(10(s))],[=25000W-=25kW]:}\begin{aligned} \text { power } & =\frac{\text { work done }}{\text { time taken }} \\ & =\frac{250000 \mathrm{~J}}{10 \mathrm{~s}} \\ & =25000 \mathrm{~W} \equiv 25 \mathrm{~kW} \end{aligned}

Answer 

The energy is being transferred to the car (from its engine) at a rate of 25 kW , or 25 kJ per second. 
Car engines are not very efficient. In this example, the car’s engine may transfer energy at the rate of 100 kW or so, although most of this is wasted as thermal energy. 

REFLECTION 

Do you find the problem-solving strategy in the worked examples useful? 
Do you use it all the time? 
Do you have a better approach? 

Questions 

6 a How many watts are there in a kilowatt? 
b How many watts are there in a megawatt? 
7 One horsepower is the power output of one horse when it lifts a mass of 33000 lb of water through a height of one foot in one minute as shown in Figure 8.12. 
Figure 8.12: A graphical representation of horse power. 
You need to know that 1 lb = 0.453592 kg 1 lb = 0.453592 kg 1lb=0.453592kg1 \mathrm{lb}=0.453592 \mathrm{~kg} and one foot = 0.3048 = 0.3048 =0.3048=0.3048 metres. 
a Calculate the mass of water the horse lifts in one minute. 
b Calculate the weight of water the horse lifts in one minute. (Assume that g = 10 N / kg g = 10 N / kg g=10N//kgg=10 \mathrm{~N} / \mathrm{kg} ) 
c Calculate the work the horse does in one minute. 
d Calculate the power output of the horse. 
8 An average man needs to eat food containing about 2500 kcal of chemical potential energy per day ( 1 kcal = 4.18 kJ 1 kcal = 4.18 kJ 1kcal=4.18kJ1 \mathrm{kcal}=4.18 \mathrm{~kJ} ). 
a What is 2500 kcal expressed in joules? 
b Calculate the power output of the average man, even when he is doing no work. 
9 It is estimated that the human brain has a power requirement of 40 W . How much energy does it use in an hour? 
10 A light bulb transfers 1000 J of energy in 10 s . What is its power?
10 一只灯泡在 10 秒内传递了 1000 焦耳的能量,它的功率是多少?

11 An electric motor transfers 100 J in 8.0 s . It then transfers the same amount of energy in 6.0 s . Has its power increased or decreased?
11 电动机在 8.0 秒内传输 100 焦耳的能量,然后在 6.0 秒内传输相同数量的能量,其功率是增加了还是减小了?

ACTIVITY 8.3  活动 8.3

What is the power of a world-class sprinter?
世界级短跑运动员的力量是什么?

Work on this problem on your own for three minutes. Share your ideas with
自己研究这个问题三分钟。与他人分享你的想法

a partner for a further two minutes. Be prepared to share your results and thinking with the rest of the class. 
The 100 metre sprint world record of 9.58 seconds was set in the final of the World Athletics Championships in Berlin in 2009.
9.58 秒的 100 米短跑世界纪录是在 2009 年柏林世界田径锦标赛决赛中创造的。
  • Work out the average speed for a sprinter running 100 m in 9.58 s .
    计算短跑运动员在 9.58 秒内跑完 100 米的平均速度。
  • Using the average speed, work out the sprinter’s kinetic energy for the race. Assume that he had a mass of 86 kg .
    利用平均速度,计算出短跑运动员在比赛中的动能。假设他的质量为 86 千克。
  • Work out his power output.
    计算出他的功率输出。
This power output is much smaller than the figure worked out by a team of Mexican scientists. They worked out that he had a maximum power output of about 2600 W and his total mechanical work was 80 kJ .
这个输出功率比墨西哥科学家小组计算出的数字要小得多。他们计算出,他的最大输出功率约为 2600 W,总机械功为 80 kJ。
  • What is missing from your calculation?
    你的计算中缺少了什么?
  • Apart from the fact that he did not run at a constant speed, are there any other assumptions you made in your calculation?
    除了他没有匀速奔跑这一事实外,你在计算中还有其他假设吗?
  • To reach a more realistic value, think about the forces involved when he is accelerating and even when he is running at a constant speed.
    为了得出一个更切合实际的数值,请考虑一下他在加速甚至匀速奔跑时所受到的力。

ACTIVITY 8.4  活动 8.4

Revision boomerang  修订回旋镖

You will be in a group of three. Your teacher will let you know how much time you have to draw a mind map on a sheet of A3. It should include all the key terms, concepts and equations related to energy (from Chapters 6, 7 and 8). Use pictures to illustrate the ideas where possible, but your mind map should include:
你们将三人一组。老师会告诉你们有多少时间在一张 A3 纸上绘制思维导图。思维导图应包括与能量有关的所有关键术语、概念和方程式(第 6、7 和 8 章)。尽可能使用图片说明观点,但思维导图应包括以下内容:
  • the principle of conservation of energy and the various energy stores and energy transfers
    能量守恒原理以及各种能量储存和能量传递
  • the various energy resources (which ones depend on sunlight, which are renewable, and so on)
    各种能源(哪些依靠阳光,哪些是可再生能源等)
  • any equations (both word and symbol) you need to know (include units in pencil or a different colour)
    您需要知道的任何方程式(包括文字和符号)(用铅笔或不同颜色标明单位)
When your teacher tells you to, pass your mind map to the person on your left. You receive a mind map from the person on your right. Your job is to correct any mistakes and add missing information.
当老师告诉你时,把思维导图传给左边的人。你从右边的人那里收到思维导图。你的任务是纠正任何错误并添加遗漏的信息。
When your teacher tells you, the mind maps will change hands a second time. Correct and add as before.
当老师告诉您时,思维导图将第二次易手。像之前一样更正和补充。
The next time your teacher asks you to pass on the mind maps, your own mind map should arrive back with you.
下一次老师让你传思维导图时,你自己的思维导图也要一起传回来。

SELF ASSESSMENT  自我评估

Did you know most of the information for Activity 8.4, or did your classmates need to add a lot of missing information?
你是否知道活动 8.4 的大部分信息,或者你的同学是否需要补充很多遗漏的信息?
If you could not remember as much as you thought you would, you need to develop a strategy to help you learn the material (for example, by developing a set of flash cards).
如果你记不住的东西没有你想象的那么多,你就需要制定一种策略来帮助你学习这些材料(例如,编制一套闪存卡)。

PROJECT  项目

How does the stopping distance of a vehicle vary with its speed?
车辆的停车距离随车速有何变化?

Figure 8.13: A collision between a car and a crash test dummy in order to raise awareness of road safety.
图 8.13:汽车与碰撞试验假人之间的碰撞,以提高道路安全意识。
You are going to apply some of the physics you have learned to keep you and other motorists safer. Table 8.1 shows how the stopping distance for an emergency stop is affected by the speed of the car. An emergency stop is when a driver attempts to stop in the shortest possible distance in order to avoid an accident.
您将应用所学的一些物理知识,使自己和其他驾驶者更加安全。表 8.1 显示了紧急停车的停车距离受车速的影响。紧急停车是指驾驶员试图在尽可能短的距离内停车,以避免发生事故。
Note that stopping distance = = == thinking distance + braking distance. The driver cannot apply the brakes instantly. Thinking distance is the speed of the car multiplied by the driver’s reaction time. This is the distance travelled between the driver becoming aware of a hazard and applying the brakes.
请注意,停车距离 = = == 思考距离 + 制动距离。驾驶员不能立即踩下刹车。思考距离是车速乘以驾驶员的反应时间。这是驾驶员从意识到危险到踩下刹车之间的距离。
Speed / m / s / m / s //m//s/ \mathrm{m} / \mathrm{s}  速度 / m / s / m / s //m//s/ \mathrm{m} / \mathrm{s}

思考距离 / m / m //m/ \mathrm{m}
Thinking
distance / m / m //m/ \mathrm{m}
Thinking distance //m| Thinking | | :---: | | distance $/ \mathrm{m}$ |

制动距离 / m / m //m/ \mathrm{m}
Braking distance
/ m / m //m/ \mathrm{m}
Braking distance //m| Braking distance | | :---: | | $/ \mathrm{m}$ |
 
Stopping
distance / m / m //m/ \mathrm{m}
Stopping distance //m| Stopping | | :---: | | distance $/ \mathrm{m}$ |
8.9 6 6 12
13.4 9 14 23
17.9 12 24 36
22.3 15 38 53
26.8 18 55 73
31.3 21 75 96
Speed //m//s "Thinking distance //m" "Braking distance //m" "Stopping distance //m" 8.9 6 6 12 13.4 9 14 23 17.9 12 24 36 22.3 15 38 53 26.8 18 55 73 31.3 21 75 96| Speed $/ \mathrm{m} / \mathrm{s}$ | Thinking <br> distance $/ \mathrm{m}$ | Braking distance <br> $/ \mathrm{m}$ | Stopping <br> distance $/ \mathrm{m}$ | | :---: | :---: | :---: | :---: | | 8.9 | 6 | 6 | 12 | | 13.4 | 9 | 14 | 23 | | 17.9 | 12 | 24 | 36 | | 22.3 | 15 | 38 | 53 | | 26.8 | 18 | 55 | 73 | | 31.3 | 21 | 75 | 96 |
Table 8.1: Stopping distances for different speeds. 
When a car comes to an emergency stop all its kinetic energy is transformed into thermal energy because of the work done by the brakes, W W WW, which apply a braking (frictional) force, F F FF, throughout the braking distance. So, W = F d W = F d W=FdW=F d, where F F FF is the frictional force and d is the stopping 
distance. We can write: 1 2 m v 2 = F d 1 2 m v 2 = F d (1)/(2)mv^(2)=Fd\frac{1}{2} m v^{2}=F d. 
1 Assume that the mass of the car is 1500 kg and the braking force is 10 000 N. Show that you get nearly the same braking distance as in Table 8.1. 
2 Find the guidelines published in your country. Though the physics is the same, different assumptions might have been made to arrive at different numbers. If you know how to, develop a spreadsheet so that calculations for the different speeds can be done at the same time. 
3 What could reduce the braking force? 
4 Work out the braking distance when the braking force is halved. 
5 Work out the reaction time for the thinking distances in Table 8.1. 
6 Work out the thinking distance when the thinking time is doubled. 
7 List the ways in which reaction time for a driver could increase. 
8 Design one of two safety campaign posters. Use physics (and graphs) to back up your claims. 
  • Design a poster urging people to drive more slowly. A longer stopping distance reduces the chance that an accident can be avoided and increases the impact speed. Emphasise that damage (and injuries) depends on the kinetic energy of the car, not its speed. 
  • Design a safety campaign poster urging people not to use their phone when driving as it could increase their reaction time. 

SUMMARY 

Energy transferred by a force is called work done. 
The amount of work done is the amount of energy transferred. 
The amount of work done depends on the size of the force - the greater the force, the more work is done. 
The amount of work done depends the distance moved in the direction of the force - the further it moves, the more work is done. 
To calculate the energy transferred by the force of gravity, we must use the vertical height moved. 
Work done = = == force × × xx\times distance in the direction of the force. 
The work done to move a weight vertically upwards is equal to g.p.e. 
Power is the rate at which energy is transferred, or the rate at which work is done. 
Power can be increased by increasing the work done in a given time. 
Power can be increased by reducing the time over which the same work is done. 
Recall and use the equation P = Δ E t P = Δ E t P=(Delta E)/(t)P=\frac{\Delta E}{t} in simple systems. 

EXAM-STYLE QUESTIONS 

1 Work done is measured in which of the following units? 
A newton 
B watt 
C kilogram 
D joule 
2 Power is measured in which of the following units? 
A newton 
B watt 
C kilogram 
D joule 
3 A man pushes a 25 kg pram up a slope as shown in the diagram. 
He pushes with a force of 150 N along the 20 m slope. How much energy is dissipated as thermal energy? 
A 250 J 
B 500 J 
C 2000 J 
D 2500 J 
4 For each sentence, select the correct word from the list. work energy more less 
a When it moves an object, a smaller force does qquad\qquad work than a bigger force. 
b The greater the distance an object is moved by the force, the qquad\qquad work it does. 
c Power is the rate at which ___ is transferred. 
d Power is the rate at which ___ is done. 
[Total: 4] 
5 The Empire State Building in New York is the venue for an annual running competition. Competitors race up 86 floors ( 1576 stairs) or 320 metres to finish close to the top. The record fastest time is 9 minutes 33 seconds by Paul Crake in 2003. His mass was 62.5 kg . 
a Calculate Paul Crake’s weight. 
b State the relationship between work done, force and distance. 
c Calculate the work done by Paul Crake. Express your answer in kJ ( 1 kJ = 1000 J ) kJ ( 1 kJ = 1000 J ) kJ(1kJ=1000J)\mathrm{kJ}(1 \mathrm{~kJ}=1000 \mathrm{~J}).
c 计算保罗-克拉克所做的功。用 kJ ( 1 kJ = 1000 J ) kJ ( 1 kJ = 1000 J ) kJ(1kJ=1000J)\mathrm{kJ}(1 \mathrm{~kJ}=1000 \mathrm{~J}) 表示答案。

d State the relationship between power, work done, and time.
d 说明功率、做功和时间之间的关系。

e Calculate Paul Crake’s average power output during his record-breaking run.
e 计算保罗-克雷克破纪录时的平均功率输出。
6 a The diagram shows a section through a subway station. The track at the station platform is designed to be higher up than the track in the tunnels. The driver uses brakes to stop the train at the platform and a motor to make the train set off. What is the advantage of having the platform higher up than the track in the tunnels? Explain your answer in terms of the work done by the brakes and the motor.
6 a 图中显示的是地铁站的一段。车站站台的轨道设计得比隧道内的轨道高。司机使用制动器将列车停在站台上,并使用电机使列车启动。站台比隧道轨道高的好处是什么?请根据制动器和电机所做的功来解释答案。


b A subway train has a mass of 270000 kg (including passengers) and a maximum kinetic energy of 81.7 MJ.
b 一列地铁列车的质量为 270000 千克(包括乘客),最大动能为 81.7 兆焦耳。

i State the relationship between kinetic energy (k.e.), mass and velocity.
i 说明动能(k.e.)、质量和速度之间的关系。

ii Calculate the maximum velocity of the train.
ii 计算列车的最大速度。

SELF-EVALUATION CHECKLIST
自我评估清单

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
学习完本章后,请思考自己对不同主题的掌握程度。这将帮助你看到自己在知识方面的差距,并帮助你更有效地学习。
I can  我可以
  查看主题...
See
Topic...
See Topic...| See | | :--- | | Topic... |
  需要更多努力
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
  快到了
Almost
there
Almost there| Almost | | :--- | | there |

有信心继续前进
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |

理解所做的功等于传递的能量。
Understand that work done equals
energy transferred.
Understand that work done equals energy transferred.| Understand that work done equals | | :--- | | energy transferred. |
8.1

将做功与力乘以沿力的方向移动的距离联系起来。
Relate work done to the force
multiplied by the distance moved in
the direction of the force.
Relate work done to the force multiplied by the distance moved in the direction of the force.| Relate work done to the force | | :--- | | multiplied by the distance moved in | | the direction of the force. |
8.2
Recall and use W = F d = Δ E W = F d = Δ E W=Fd=Delta EW=F d=\Delta E.
回顾并使用 W = F d = Δ E W = F d = Δ E W=Fd=Delta EW=F d=\Delta E
8.2

了解功率、完成的工作和时间之间的关系。
Understand the relationship between
power, work done and time.
Understand the relationship between power, work done and time.| Understand the relationship between | | :--- | | power, work done and time. |
8.3

回忆并使用方程 P = Δ E t P = Δ E t P=(Delta E)/(t)P=\frac{\Delta E}{t} 解决简单问题。
Recall and use the equation P = Δ E t P = Δ E t P=(Delta E)/(t)P=\frac{\Delta E}{t}
for simple problems.
Recall and use the equation P=(Delta E)/(t) for simple problems.| Recall and use the equation $P=\frac{\Delta E}{t}$ | | :--- | | for simple problems. |
8.4
I can "See Topic..." "Needs more work" "Almost there" "Confident to move on" "Understand that work done equals energy transferred." 8.1 "Relate work done to the force multiplied by the distance moved in the direction of the force." 8.2 Recall and use W=Fd=Delta E. 8.2 "Understand the relationship between power, work done and time." 8.3 "Recall and use the equation P=(Delta E)/(t) for simple problems." 8.4 | I can | See <br> Topic... | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :---: | :---: | :---: | :--- | | Understand that work done equals <br> energy transferred. | 8.1 | | | | | Relate work done to the force <br> multiplied by the distance moved in <br> the direction of the force. | 8.2 | | | | | Recall and use $W=F d=\Delta E$. | 8.2 | | | | | Understand the relationship between <br> power, work done and time. | 8.3 | | | | | Recall and use the equation $P=\frac{\Delta E}{t}$ <br> for simple problems. | 8.4 | | | |

> Chapter 9  > 第 9 章

The kinetic particle model of matter
物质的动力学粒子模型

IN THIS CHAPTER YOU WILL:
在本章中,您将

  • describe the three states of matter (solid, liquid and gas)
    描述物质的三种状态(固态、液态和气态)
  • investigate changes of state
    探赜索隐
  • use the kinetic model to explain changes of state and the behaviour of gases
    使用动力学模型解释状态变化和气体行为

    > > >> explain the kinetic particle model in terms of the forces between particles
    > > >> 用粒子间的作用力解释动粒子模型

GETTING STARTED  入门

Work with a partner. Take a large sheet of paper and write the words ‘solid’, ‘liquid’ and ‘gas’ on the paper.
与伙伴合作。拿一张大纸,在纸上写下 "固体"、"液体 "和 "气体"。

Around each word, write as much as you can about that state of matter. You can include drawings.
围绕每个单词,尽可能多地写出有关该物质状态的内容。您可以附上图画。
Using a different coloured pen, make as many links as you can between the three words.
用不同颜色的笔,尽可能多地将这三个词联系起来。

Figure 9.1: Making links between words.
图 9.1:在单词之间建立联系。

MELTING ICE LEADS TO MAGMA FLOW
融冰导致岩浆流动

Figure 9.2: Scientists in Arizona and Oxford have shown a correlation between the melting of glaciers and an increase in volcanic eruptions such as this one in Iceland which erupted in 2010.
图 9.2:亚利桑那州和牛津大学的科学家研究表明,冰川融化与火山喷发增加之间存在相关性,例如 2010 年冰岛的火山喷发。
The volcano in Figure 9.2 is erupting. High temperatures inside the earth have melted the rock creating magma. Scientists believe the eruption may have been triggered by the melting of the glacier situated above the volcano. The glacier melting meant there was less ice pressing down on the rocks. This reduced the pressure on the magma underneath the rocks. This made it easier for magma to flow. The melting of the glaciers is linked to global warming, which is caused by the changes to gases in the atmosphere. Changes of state, such as the melting of glaciers, can have dramatic effects.
图 9.2 中的火山正在喷发。地球内部的高温融化了岩石,产生了岩浆。科学家认为,火山爆发可能是由于火山上方的冰川融化引发的。冰川融化意味着压在岩石上的冰减少了。这就减少了岩石下岩浆的压力。这使得岩浆更容易流动。冰川融化与全球变暖有关,全球变暖是由大气中的气体变化引起的。冰川融化等状态变化会产生巨大影响。
We are familiar with the changes that happen when ice melts. A glass-like solid changes into a transparent, colourless, runny liquid. Heat the liquid and it ‘vanishes’ into thin air. Although this sounds like a magic trick, it is so familiar that it does not surprise us. It is more surprising when we see solid rock heat up and become magma.
我们熟悉冰融化时发生的变化。玻璃状固体变成透明、无色、流动的液体。加热液体,它就会 "消失 "得无影无踪。虽然这听起来像魔术,但它是如此熟悉,以至于我们并不感到惊讶。当我们看到固体岩石受热变成岩浆时,才会感到更加惊讶。
In this chapter, we will look at materials and their different states - solid, liquid and gas. We will consider how the particles in matter behave and how this can help us explain some of the things we observe when materials change from one state to another. 

Discussion questions 

1 List ten solids, ten liquids and ten gases. Are there any substances which are hard to categorise? 
2 The Earth is distinctive among the planets of the Solar System in being the only planet on which water is found to exist naturally in all three of its physical states. Discuss how life on Earth would be affected if one of the states of matter did not exist. 

9.1 States of matter
9.1 物质状态

Figure 9.3: Ice cubes have a fixed shape.
图 9.3:冰块有固定的形状。
Figure 9.4: This coloured water takes
图 9.4:这种彩色水需要
Figure 9.5: The steam leaving the pot quickly condenses to form water droplets.
图 9.5:离开锅的蒸汽很快凝结成水滴。
https://cdn.mathpix.com/cropped/2025_03_22_a681e41e77bc06e0ad29g-063.jpg?height=313&width=478&top_left_y=515&top_left_x=1388 Figure 9.5: The steam leaving the pot quickly condenses to form water droplets.| ![](https://cdn.mathpix.com/cropped/2025_03_22_a681e41e77bc06e0ad29g-063.jpg?height=313&width=478&top_left_y=515&top_left_x=1388) | | :--- | | Figure 9.5: The steam leaving the pot quickly condenses to form water droplets. |
State  国家 solid  固体 liquid  液态 gas  燃气
Size  rigid, fixed shape, fixed volume; cannot be squashed  not rigid, no fixed shape, fixed volume; cannot be squashed  not rigid, no fixed shape, no fixed volume; can be squashed 
Shape  takes the shape of its container  takes the shape of its container  expands to fill its container 
Figure 9.3: Ice cubes have a fixed shape. Figure 9.4: This coloured water takes "https://cdn.mathpix.com/cropped/2025_03_22_a681e41e77bc06e0ad29g-063.jpg?height=313&width=478&top_left_y=515&top_left_x=1388 Figure 9.5: The steam leaving the pot quickly condenses to form water droplets." State solid liquid gas Size rigid, fixed shape, fixed volume; cannot be squashed not rigid, no fixed shape, fixed volume; cannot be squashed not rigid, no fixed shape, no fixed volume; can be squashed Shape takes the shape of its container takes the shape of its container expands to fill its container| | Figure 9.3: Ice cubes have a fixed shape. | Figure 9.4: This coloured water takes | ![](https://cdn.mathpix.com/cropped/2025_03_22_a681e41e77bc06e0ad29g-063.jpg?height=313&width=478&top_left_y=515&top_left_x=1388) <br> Figure 9.5: The steam leaving the pot quickly condenses to form water droplets. | | :---: | :---: | :---: | :---: | | State | solid | liquid | gas | | Size | rigid, fixed shape, fixed volume; cannot be squashed | not rigid, no fixed shape, fixed volume; cannot be squashed | not rigid, no fixed shape, no fixed volume; can be squashed | | Shape | takes the shape of its container | takes the shape of its container | expands to fill its container |
Table 9.1: The distinguishing properties of the three states of matter. 
Matter exists in three states: solid, liquid and gas. An example of this is water which can exist as solid ice, liquid water or steam which is an invisible gas. Steam quickly condenses in air to form tiny water droplets, which are what we see. We can describe these states by describing their shape and volume (size). Table 9.1 shows how these help us to distinguish between solids, liquids and gases. 
Figure 9.6: The volume of a liquid stays the same. 
Figure 9.6 shows a famous psychology experiment. A young child will usually think the taller glass holds more water even when they see it being poured from the wider glass. The child does not realise yet that a liquid has a fixed volume. Although the drink changes its shape when you pour it from one glass to the other, its volume stays the same. 

Changes of state 

Heat a solid and it melts to become a liquid. Heat the liquid and it boils to become a gas. Cool the gas and it becomes first a liquid and then a solid. These are changes of state. The names for these changes are shown in Figure 9.7. 
Figure 9.7: Naming changes of state. 
Another word for a liquid changing to a gas is evaporation. We will see the difference between evaporation and boiling later. 

Questions 

1 Copy and complete these sentences: 
The three states of matter are qquad\qquad , qquad\qquad and qquad\qquad . 
A solid has a definite shape and qquad\qquad . A liquid has a definite qquad\qquad but takes the shape of its container. A gas will expand to fill all the qquad\qquad available. 
When a solid is heated it qquad\qquad to make a qquad\qquad . The temperature this happens at is called the qquad\qquad —. 
The boiling point is the temperature at which a qquad\qquad turns to a qquad\qquad . 
2 a What name is given to the temperature at which a gas condenses to form a liquid? 
b What name is given to the process during which a liquid changes into a solid? 
c What name is given to the temperature at which this happens? 
3 To measure the volume of a liquid, you can pour it into a measuring cylinder. Measuring cylinders come in different shapes and sizes - tall, short, wide, narrow. Explain why the shape of the cylinder does not affect the measurement of volume. 

9.2 The kinetic particle model of matter 

In this section, we will use a model to help us explain the behaviour of materials. Scientists often use models to explain things which we cannot see directly. Using this model will help us answer questions such as: 
  • why does an ice cube change shape as it melts? 
  • how can we smell perfume from across a room? 
  • why does it take time to melt a solid? 
The model is called the kinetic particle model of matter. The word ‘kinetic’ means related to movement. All matter is made up of tiny particles - atoms, moleculles or ions. When a substance is heated, its particles gain energy and move faster. The higher the temperature, the faster the particles move. 
There are many different types of particles with different chemical properties. In this chapter we will look at how these particles move rather than how they react. We will draw all the atoms and molecules as spheres and refer to them all as particles. 
Figure 9.8a: Air is a mixture of elements and compounds. 
Figure 9.8b: For this model we will assume all atoms and molecules are identical spherical particles. 
The idea that matter is made up of identical, spherical molecules is a great simplification, but using this model will help to explain the behaviour of materials. 

Describing the particle structure of solid, liquids and gases 

Figure 9.9 shows how we picture the particles in a solid, a liquid and a gas. Each state of matter can be described by describing the arrangement, separation and motion of its particles. 

Particle movement and temperature 

Kinetic theory uses the idea that as particles heat up they gain more kinetic energy and so move faster. As a substance cools down its particles lose kinetic energy and slow down. This suggests that there is a theoretical lower limit to how cold anything can be. If the particles lose all their kinetic energy and stop moving, it is not possible for the substance to cool any further. 
The lowest possible temperature anything can reach is 273 C 273 C -273^(@)C-273^{\circ} \mathrm{C}. This is also known as absolute zero. 
Figure 9.9: Representations of a: gas, b: liquid, c: solid. The arrangement, separation and motion of the particles change as the gas cools to become a liquid and then a solid. 
State 
 
Arrangement and separation of
particles
Arrangement and separation of particles| Arrangement and separation of | | :--- | | particles |
Motion of particles 
Solid 
 
The particles are packed
closely together, in a regular
pattern. Notice that each
particle is in close contact with
all its neighbours.
The particles are packed closely together, in a regular pattern. Notice that each particle is in close contact with all its neighbours.| The particles are packed | | :--- | | closely together, in a regular | | pattern. Notice that each | | particle is in close contact with | | all its neighbours. |
 
Because the particles are so
tightly packed, they cannot
move around. However, they
do move a bit. They can
vibrate about a fixed position.
The hotter the solid, the more
they vibrate.
Because the particles are so tightly packed, they cannot move around. However, they do move a bit. They can vibrate about a fixed position. The hotter the solid, the more they vibrate.| Because the particles are so | | :--- | | tightly packed, they cannot | | move around. However, they | | do move a bit. They can | | vibrate about a fixed position. | | The hotter the solid, the more | | they vibrate. |
Liquid 
 
The particles are packed
slightly less closely together
than in a solid. The particles
are arranged randomly rather
than in a fixed pattern.
The particles are packed slightly less closely together than in a solid. The particles are arranged randomly rather than in a fixed pattern.| The particles are packed | | :--- | | slightly less closely together | | than in a solid. The particles | | are arranged randomly rather | | than in a fixed pattern. |
 
Because the particles are
slightly less tightly packed
than in a solid, they can move
around. So the particles are
both vibrating and moving
from place to place. The hotter
the liquid is, the faster its
molecules move.
Because the particles are slightly less tightly packed than in a solid, they can move around. So the particles are both vibrating and moving from place to place. The hotter the liquid is, the faster its molecules move.| Because the particles are | | :--- | | slightly less tightly packed | | than in a solid, they can move | | around. So the particles are | | both vibrating and moving | | from place to place. The hotter | | the liquid is, the faster its | | molecules move. |
State "Arrangement and separation of particles" Motion of particles Solid "The particles are packed closely together, in a regular pattern. Notice that each particle is in close contact with all its neighbours." "Because the particles are so tightly packed, they cannot move around. However, they do move a bit. They can vibrate about a fixed position. The hotter the solid, the more they vibrate." Liquid "The particles are packed slightly less closely together than in a solid. The particles are arranged randomly rather than in a fixed pattern." "Because the particles are slightly less tightly packed than in a solid, they can move around. So the particles are both vibrating and moving from place to place. The hotter the liquid is, the faster its molecules move."| State | Arrangement and separation of <br> particles | Motion of particles | | :--- | :--- | :--- | | Solid | The particles are packed <br> closely together, in a regular <br> pattern. Notice that each <br> particle is in close contact with <br> all its neighbours. | Because the particles are so <br> tightly packed, they cannot <br> move around. However, they <br> do move a bit. They can <br> vibrate about a fixed position. <br> The hotter the solid, the more <br> they vibrate. | | Liquid | The particles are packed <br> slightly less closely together <br> than in a solid. The particles <br> are arranged randomly rather <br> than in a fixed pattern. | Because the particles are <br> slightly less tightly packed <br> than in a solid, they can move <br> around. So the particles are <br> both vibrating and moving <br> from place to place. The hotter <br> the liquid is, the faster its <br> molecules move. |
Gas 
 
The particles are widely
separated from one another.
They are no longer in contact,
unless they collide with each
other. In air, the average
separation between the
particles is about ten times
their diameter.
The particles are widely separated from one another. They are no longer in contact, unless they collide with each other. In air, the average separation between the particles is about ten times their diameter.| The particles are widely | | :--- | | separated from one another. | | They are no longer in contact, | | unless they collide with each | | other. In air, the average | | separation between the | | particles is about ten times | | their diameter. |
 
The particles move freely
about, bouncing off one
another and off the walls of the
container. In air at room
temperature, the average speed
of the particles is about 500
m / s m / s m//s\mathrm{m} / \mathrm{s} and this increases with
temperature.
The particles move freely about, bouncing off one another and off the walls of the container. In air at room temperature, the average speed of the particles is about 500 m//s and this increases with temperature.| The particles move freely | | :--- | | about, bouncing off one | | another and off the walls of the | | container. In air at room | | temperature, the average speed | | of the particles is about 500 | | $\mathrm{m} / \mathrm{s}$ and this increases with | | temperature. |
Gas "The particles are widely separated from one another. They are no longer in contact, unless they collide with each other. In air, the average separation between the particles is about ten times their diameter." "The particles move freely about, bouncing off one another and off the walls of the container. In air at room temperature, the average speed of the particles is about 500 m//s and this increases with temperature."| Gas | The particles are widely <br> separated from one another. <br> They are no longer in contact, <br> unless they collide with each <br> other. In air, the average <br> separation between the <br> particles is about ten times <br> their diameter. | The particles move freely <br> about, bouncing off one <br> another and off the walls of the <br> container. In air at room <br> temperature, the average speed <br> of the particles is about 500 <br> $\mathrm{m} / \mathrm{s}$ and this increases with <br> temperature. | | :--- | :--- | :--- |
Table 9.2: The arrangement and motion of particles in the three different states of matter. Compare these statements with the diagrams shown in Figure 9.9. 

Evidence for the kinetic model 

Atoms and molecules are far too small to see, even with a microscope, but experiments show the effects of moving atoms and molecules. These experiments do not prove there are moving particles, but they do provide support for the idea. 
In 1827, a scientist called Robert Brown was using a microscope to study pollen grains when he noticed tiny particles jiggling about. At first he thought that they might be alive, but when he repeated his experiment with tiny grains of dust suspended in water, he saw that the dust also moved around. This motion is now known as Brownian motion, and it happens because the moving particles are constantly knocked about by the fast-moving particles of the air. 
We can do a similar experiment using smoke particles. The oxygen and nitrogen molecules that make up the air are far too small to see, so we have to look at something bigger, and look for the effect of the air molecules. 
We can use a smoke cell (Figure 9.10a). This is a small glass box which contains air with a small amount of smoke. The cell is lit from the side, and the microscope is used to view the smoke particles.
我们可以使用烟雾箱(图 9.10a)。这是一个装有空气和少量烟雾的小玻璃盒。从侧面照亮玻璃盒,用显微镜观察烟雾颗粒。

Figure 9.10a: An experimental arrangement for observing Brownian motion. The smoke particles are just large enough to show up under the microscope. The air molecules that collide with them are much too small to see. b: The invisibly small air molecules repeatedly hit the smoke particle making it change direction. The dotted line shows the path of the smoke particle.
图 9.10a:观察布朗运动的实验装置。烟雾粒子大到足以在显微镜下显示出来。b: 微小的空气分子反复撞击烟雾粒子,使其改变方向。虚线表示烟雾粒子的运动轨迹。
The smoke particles show up as tiny specks of light, but they are too small to see any detail of their shape. What is noticeable is the way they move. If you watch a single particle, you will see that it follows random path, frequently changing direction. This is because air molecules repeatedly hit the smoke particle.
烟雾颗粒显示为微小的光斑,但它们太小了,无法看到任何形状细节。值得注意的是它们的移动方式。如果你观察单个烟雾微粒,就会发现它的运动轨迹是随机的,而且经常改变方向。这是因为空气分子反复撞击烟雾颗粒。

Explanations using the kinetic model
使用动力学模型进行解释

The kinetic model of matter can be used to explain many observations. Here are some of them: 
  • Liquids take up the shape of their container because their particles are free to move about within the liquid. 
  • Gases fill their container because their particles can move about with complete freedom. 
  • Solids keep their shape because the particles are packed tightly together. 
  • Gases diffuse (spread out) from place to place, so that, for example, we can smell perfume across the room. The perfume particles spread about because they are free to move. 
  • Dissolved substances diffuse throughout a liquid. Sugar crystals in a drink dissolve and molecules spread throughout the liquid, carried by the mobile particles. In a hotter drink, the particles are moving faster and the sugar 

diffuses more quickly. 

  • Most solids expand when they melt. The particles are slightly further apart in a liquid than in a solid. 
  • Liquids expand a lot when they boil. The particles of a gas are much further apart than in a liquid. We can think about this the other way round. Gases contract a lot when they condense. If all of the air in the room you are now in was cooled enough, it would condense to form a thin layer of liquid, two or three millimetres deep, on the floor. 

More on Brownian motion 

The molecules in air have an average diameter of about 4 × 10 10 4 × 10 10 4xx10^(-10)4 \times 10^{-10} metres. This makes them impossible to see with a laboratory microscope. The smoke particles consist of many molecules and so are many times larger than the air molecules. A smoke particle has a diameter of about 1 × 10 7 1 × 10 7 1xx10^(-7)1 \times 10^{-7} metres, which is about 250 times the diameter of the air molecules. The air molecules are light but fast moving and so have enough kinetic energy to cause the smoke particles to change direction on impact. 

Forces and the kinetic model 

We have seen that the kinetic model of matter can explain the differences between solids, liquids and gases. We can explain some other observations if we add another scientific idea to the kinetic theory: we need to consider the forces between the particles that make up matter. 
Why do the particles that make up a solid or a liquid stick together? There must be attractive forces (forces pulling them together) between them. Without attractive forces to hold together the particles, there would be no solids or liquids, only gases. No matter how much we cooled matter down, it would remain as a gas. 
Another way to refer to these forces is to say that there are bonds between the particles. Each particle of a solid is strongly bonded to its neighbours. This is because the forces between particles are strongest when the particles are close together. In a liquid, the particles are slightly further apart and so the forces between them are slightly weaker. In a gas, the particles are far apart, so that the particles do not attract each other and can move freely about. 

ACTIVITY 9.1 

Demonstrating the kinetic model 

The kinetic theory can be modelled using small balls or marbles in a tray. 
Figure 9.11: Modelling kinetic theory using marbles in a tray. 
1 Place some identical small balls or marbles on a shallow tray. They should cover about one-quarter of the area of the tray. 
2 Tip the tray slightly so that the balls all roll to the lower end. The pattern they form is like the arrangement of particles in a solid. 
3 Keep the tray slightly tipped and shake it gently so that the balls can move about. This is like a liquid. 
4 Keep shaking the tray and tip it so that it becomes horizontal. The balls move around freely, colliding with each other and the sides of the tray. This is like the particles in a gas. 
Your task is to make a video, or a presentation to explain the kinetic model, using the tray of balls. You could: 
  • show the arrangement and movement of particles in each of the three 

states 

  • demonstrate absolute zero 
  • demonstrate and explain changes of state - melting, boiling, condensing and freezing 
  • model Brownian motion (hint: use a larger, different-coloured ball for the smoke particle) 
  • add some different-coloured balls to one corner of the tray to represent perfume; show how the perfume diffuses (spreads out). 

Questions 

4 Sketch three diagrams to show the arrangement of molecules in a solid, a liquid and a gas. 
5 a In which state of matter are the particles most closely packed? 
b In which state of matter are they most widely separated? 
c In which state do the particles move fastest? 
6 a Describe what is meant by Brownian motion. 
b How can the kinetic model be used to explain Brownian motion? 
c A student did an experiment to observe Brownian motion. She then repeated the experiment in a much colder room. Describe and explain how her observations would change. 
7 Use the kinetic model of matter to explain why we can walk through air and swim through water but we cannot walk through a solid wall. 

9.3 Gases and the kinetic model 

The kinetic model can help us understand how gases behave. Thinking about the particles in a gas helps us answer questions about the gas. 
Why does a gas cause pressure on the walls of its container? Figure 9.12 shows the particles that make up a gas. The particles of a gas move around inside its container, bumping into the sides. The gas causes pressure on the walls of the container because the gas particles are constantly colliding with the walls. 
Figure 9.12: Moving gas particles in a container. 
What happens to a gas when it is heated? Figure 9.13 shows the same gas at a higher temperature. The higher the temperature of a gas, the faster its particles are moving. The particles will hit the walls more often and with more force. This increases the pressure. 
Figure 9.13: Heating the gas in a container. 
What happens when a gas is compressed (squashed)? Figure 9.14 shows the same gas again. This time the volume of the container has been decreased. The gas has been compressed into a smaller space. The particles don’t move as far between collisions, so they collide with the walls more often. Decreasing the volume of a gas increases its pressure. 
Figure 9.14: Decreasing the volume of a gas in a container. 
Figure 9.15: Pumping up tyres means squashing a lot more air particles into a fixed space. This increases the pressure. 
The cyclist in Figure 9.15 needs to understand gas pressure. He pumps air into his tyres to make them hard. If the temperature rises, the pressure in his tyres will increase even more. This could burst his tyres. 
Figure 9.16: You need air pressure to play the trumpet. 
To produce a loud sound the girl in Figure 9.16 needs to blow out with as much pressure as she can. Musicians learn to breathe from their diaphragm. Figure 9.17 shows how the girl’s diaphragm moves down so a large volume of air can enter her lungs. Her diaphragm then moves up, reducing the volume and so increasing the pressure of the air she blows out and the loudness of her music. 
Figure 9.17: The diaphragm moves down as you breathe in and up as you breathe out. 

Questions 

8 Copy and complete the paragraph. 
The molecules in a qquad\qquad are constantly moving freely and rapidly and so hit the walls of the container. This creates qquad\qquad on the walls. When the gas is heated, the molecules move qquad\qquad and the pressure qquad\qquad . When the gas is compressed, the molecules hit the walls more often and the pressure qquad\qquad . 
9 A balloon is inflated by blowing air into it. Explain what would happen if 
an inflated balloon was put in a freezer. 
10 A tin can containing air is tightly sealed so no air can escape. The can is then heated. Describe what happens to: 
a the speed of the air molecules inside the can 
b how often the air molecules hit the walls of the can 
c the force with which the air molecules hit the walls 
d the pressure on the walls of the can. 

9.4 Temperature and the Celsius scale 

Temperature is a measure of how hot or cold something is. Temperature is measured using a thermometer. Most thermometers take a minute or so to measure temperature. This is because thermal energy has to be transferred to or from the thermometer until it is at the same temperature as the thing it is measuring. 
Figure 9.18a: Thermal energy transfers from the girl to the thermometer until they are at the same temperature. b: Thermal energy transfers from the thermometer to the ice until they are at the same temperature. 

Temperature and internal energy 

A thermometer tells us about the average energy of the particles in the object whose temperature we are measuring. It does this by sharing the energy of the particles. If they are moving rapidly, the thermometer will indicate a higher temperature. Placing a thermometer into an object to measure its temperature is rather like putting your finger into some bath water to detect how hot it is. Your finger does not have a scale from 0 to 100, but it can tell you how hot or cold the water is, from uncomfortably cold to comfortably warm to painfully hot. 
The temperature of an object is a measure of the average kinetic energy of its particles. Because it is the average kinetic energy of a particle, it does not depend on the size of the object. 
  • Internal energy is the total energy of all of the particles. 
  • Temperature is a measure of the average kinetic energy of the individual particles. 
    A bath of water at 50 C 50 C 50^(@)C50^{\circ} \mathrm{C} can have the same temperature as a cup of tea, but it has more internal energy than the cup of tea because it has far more molecules. 
    The temperature of a firework may reach 1500 C 1500 C 1500^(@)C1500^{\circ} \mathrm{C}, which means its particles have very high kinetic energy, but it has less total energy than the bath because it has very few particles. 
Figure 9.19: The firework has the fastest particles and so the highest 
temperature. The bath water has many more particles and much more internal energy than the tea. 

The Celsius temperature scale 

The melting and boiling points of water are used to define the Celsius temperature scale. 
Figure 9.20: A modern liquid-in-glass thermometer. 
Thermometers like the one shown in Figure 9.20 are used in school laboratories. The bulb contains a liquid which expands when it gets hot. The liquid moves into the tube and we can read the temperature on the scale. 
The liquid used is usually alcohol. This expands a lot when heated and it is safe. 
Some thermometers use mercury. Mercury thermometers can be used at very low temperatures. Mercury is poisonous so mercury thermometers are not used in schools. 
The scale was devised by the Swedish scientist Anders Celsius. 
His scale is known as the Celsius scale. It has two fixed points: 
  • 0 C 0 C quad0^(@)C\quad 0^{\circ} \mathrm{C} : the melting point of pure ice at atmospheric pressure 
  • 100 C 100 C quad100^(@)C\quad 100^{\circ} \mathrm{C} : the boiling point of pure water at atmospheric pressure. 
Each time he made a new thermometer, Celsius would calibrate it. He put the 
thermometer in melting ice and marked 0 C 0 C 0^(@)C0^{\circ} \mathrm{C} on the thermometer. He then put it in boiling water and marked 100 C 100 C 100^(@)C100^{\circ} \mathrm{C}. He then divided the space between these points into 100 parts. Each part represents one degree Celsius. 

The Kelvin temperature scale 

Kinetic theory suggests there is a limit to how low temperatures can go. The lowest possible temperature (the point at which molecules have no kinetic energy) is 273 C 273 C -273^(@)C-273^{\circ} \mathrm{C}. This is called absolute zero. 
The Kelvin temperature scale starts from absolute zero, or 273 C 273 C -273^(@)C-273^{\circ} \mathrm{C}. Temperatures measured on this scale are called absolute temperatures. Scientists often measure temperatures using the Kelvin scale. A change in temperature of one degree is the same for both scales. The Kelvin temperature ( T T TT ) can be calculated from the Celsius temperature ( θ ) ( θ ) (theta)(\theta) using the equation: 
T ( K ) = θ ( C ) + 273 T ( K ) = θ C + 273 T(K)=theta(^(@)C)+273T(\mathrm{~K})=\theta\left({ }^{\circ} \mathrm{C}\right)+273

KEY EQUATION 

conversion between Kelvin temperature and degrees Celsius: 
T ( K ) = θ ( C ) + 273 T ( K ) = θ C + 273 T(K)=theta(^(@)C)+273T(\mathrm{~K})=\theta\left({ }^{\circ} \mathrm{C}\right)+273

WORKED EXAMPLE 9.1 

Calculate the absolute temperature of the human body. Assume that the temperature of the human body is 37 C 37 C 37^(@)C37^{\circ} \mathrm{C}. 
T ( K ) = θ ( C ) + 273 T ( K ) = θ C + 273 T(K)=theta(^(@)C)+273T(\mathrm{~K})=\theta\left({ }^{\circ} \mathrm{C}\right)+273
T = 37 + 273 = 310 K T = 37 + 273 = 310 K T=37+273=310KT=37+273=310 \mathrm{~K}.

Questions

11 Copy and complete these sentences.
A thermometer is used to measure qquad\qquad , which is a measure of how hot something is.
It is measured in qquad\qquad or qquad\qquad .
Temperature depends on how fast the qquad\qquad are moving.
12 A laboratory thermometer has no temperature markings. Describe how you could use ice and boiling water to calibrate the thermometer.
13 Calculate the temperature in kelvin of:
a a classroom at 20 C 20 C 20^(@)C20^{\circ} \mathrm{C}
b lava at 800 C 800 C 800^(@)C800^{\circ} \mathrm{C}
c the surface temperature of minor planet Pluto at 233 C 233 C -233^(@)C-233^{\circ} \mathrm{C}.

9.5 The gas laws

The temperature, pressure and volume of a gas all affect each other. The gas laws explain mathematically how the three affect each other. There are laws describing how each of these quantities affects the others, but we will look just at the law connecting the pressure of a gas and its volume.
It is important to be clear about what the terms ‘temperature’, ‘pressure’ and ‘volume’ mean. The gas laws all refer to a fixed mass of gas. Imagine the gas in a sealed container which can be squashed or heated. The number of molecules does not change.
The temperature of a gas is a measure of the average kinetic energy of the molecules. In a hot gas, the molecules move faster than in a cold gas.
The pressure of a gas is caused by atoms or molecules hitting the walls, changing momentum and so causing a force. The pressure is the force per unit area on the walls of the container.

Pressure and volume

In 1662, Robert Boyle, a physicist and chemist, investigated the relationship between the pressure on a gas, p p pp, and its volume, V V VV.
Figure 9.21: Increasing the pressure on a gas decreases its volume.
The effect of increasing the pressure on a gas can be investigated using the simple apparatus shown in Figure 9.21. Adding weights increases the pressure and causes the volume of the gas to decrease.
Figure 9.22 shows a more accurate method. In this apparatus, some air is trapped inside the vertical glass tube. The oil in the bottom of the apparatus can be compressed with a pump, so that it pushes up inside the tube, compressing the air. The volume of the air can be read from the scale. The pressure exerted on it by the oil can be read from the dial gauge.
Increasing the pressure on the gas decreases its volume. Table 9.3 shows some typical results. Boyle found a mathematical relationship between the pressure, p p pp, and the volume, V V VV, of the gas.
Figure 9.22: Apparatus for increasing the pressure on a gas. A fixed mass of air is trapped inside the tube, and the pressure on it is increased.
Pressure, p / Pa p / Pa p//Pap / \mathrm{Pa} Volume, V / cm 3 V / cm 3 V//cm^(3)V / \mathrm{cm}^{3}
Pressure × × xx\times volume, p V / Pa p V / Pa pV//Pap V / \mathrm{Pa}
cm 3 cm 3 cm^(3)\mathrm{cm}^{3}
Pressure xx volume, pV//Pa cm^(3)| Pressure $\times$ volume, $p V / \mathrm{Pa}$ | | :--- | | $\mathrm{cm}^{3}$ |
Pressure, p//Pa Volume, V//cm^(3) "Pressure xx volume, pV//Pa cm^(3)"| Pressure, $p / \mathrm{Pa}$ | Volume, $V / \mathrm{cm}^{3}$ | Pressure $\times$ volume, $p V / \mathrm{Pa}$ <br> $\mathrm{cm}^{3}$ | | :--- | :--- | :--- |
100 60 6000
125 48 6000
150 40 6000
200 30 6000
250 24 6000
300 20 6000
100 60 6000 125 48 6000 150 40 6000 200 30 6000 250 24 6000 300 20 6000| 100 | 60 | 6000 | | :---: | :---: | :---: | | 125 | 48 | 6000 | | 150 | 40 | 6000 | | 200 | 30 | 6000 | | 250 | 24 | 6000 | | 300 | 20 | 6000 |
Table 9.3: Typical results for an experiment into pressure and volume of gas. The temperature of the gas does not change.
Boyle’s experiments showed that increasing the pressure decreased the volume. This is shown in Figure 9.23a.

Figure 9.23: Two graphs to represent the results of a Boyle’s law experiment. a: The graph of pressure against volume shows that increasing the pressure causes a decrease in the volume. b: The mathematical relationship between p p pp and 1 V 1 V (1)/(V)\frac{1}{V} can be seen from this graph. It is a straight line through the origin, which means that pressure is inversely proportional to volume.
Boyle also found that when he multiplied pressure by volume, he always got the same result. You can see this in the last column of Table 9.3.
This can be written mathematically.as:
p V = constant p V =  constant  pV=" constant "p V=\text { constant }

KEY EQUATION

relationship between pressure and volume for gas at a constant temperature:
p V = p V = pV=p V= constant
This relationship can be expressed in different ways. We can write the same idea in a way that is useful for doing calculations:
initial pressure × × xx\times initial volume = = == final pressure × × xx\times final volume Or: p 1 V 1 = p 2 V 2 p 1 V 1 = p 2 V 2 p_(1)V_(1)=p_(2)V_(2)p_{1} V_{1}=p_{2} V_{2}
where p 1 p 1 p_(1)p_{1} and V 1 V 1 V_(1)V_{1} are one pair of readings of pressure and volume, and p 2 p 2 p_(2)p_{2} and V 2 V 2 V_(2)V_{2} are another pair.
Doubling the pressure halves the volume. This means that pressure is inversely proportional to volume. Using the symbol !=\neq (‘is proportional to’), we can write:
p 1 v or v 1 p p 1 v  or  v 1 p p!=(1)/(v)quad" or "quad v!=(1)/(p)p \neq \frac{1}{v} \quad \text { or } \quad v \neq \frac{1}{p}
Figure 9.23b shows that plotting p p pp against 1 V 1 V (1)/(V)\frac{1}{V} gives a straight-line graph passing through the origin.
Finally, we can write the relationship in words: the volume of a fixed mass of gas is inversely proportional to its pressure, provided its temperature remains constant.

WORKED EXAMPLE 9.2

A scuba diver releases a bubble of air. The bubble has a volume of 2 cm 3 2 cm 3 2cm^(3)2 \mathrm{~cm}^{3}. He watches it rise to the surface, expanding as it rises. The diver is at a depth where the pressure is 5 atmospheres. What will the volume of the bubble be when it reaches the surface, where the volume is 1 atmosphere? Assume that the temperature does not change.
Step 1: Write down the initial and final values of the quantities that we know.
p 1 = 5 atmospheres V 1 = 2 cm 3 p 1 = 1 atmosphere V 2 = ? p 1 = 5  atmospheres  V 1 = 2 cm 3 p 1 = 1  atmosphere  V 2 = ? {:[p_(1)=5" atmospheres "],[V_(1)=2cm^(3)],[p_(1)=1" atmosphere "],[V_(2)=?]:}\begin{aligned} p_{1} & =5 \text { atmospheres } \\ V_{1} & =2 \mathrm{~cm}^{3} \\ p_{1} & =1 \text { atmosphere } \\ V_{2} & =? \end{aligned}
Step 2: Write down the Boyle’s law equation and substitute values.
p 1 V 1 = p 2 V 2 p 1 V 1 = p 2 V 2 p_(1)V_(1)=p_(2)V_(2)p_{1} V_{1}=p_{2} V_{2}
5 atmospheres × 2 cm 3 = 1 × 2 cm 3 = 1 xx2cm^(3)=1\times 2 \mathrm{~cm}^{3}=1 atmosphere × V 2 × V 2 xxV_(2)\times V_{2}
Step 3: There is only one unknown quantity in this equation ( V 2 ) V 2 (V_(2))\left(V_{2}\right).
Rearrange it and solve.
V 2 = p 1 V 1 p 2 = 5 atmospheres × 2 cm 3 1 atmosphere = 10 cm 3 V 2 = p 1 V 1 p 2 = 5  atmospheres  × 2 cm 3 1  atmosphere  = 10 cm 3 {:[V_(2)=(p_(1)V_(1))/(p_(2))],[=(5" atmospheres "xx2cm^(3))/(1" atmosphere ")=10cm^(3)]:}\begin{aligned} V_{2} & =\frac{p_{1} V_{1}}{p_{2}} \\ & =\frac{5 \text { atmospheres } \times 2 \mathrm{~cm}^{3}}{1 \text { atmosphere }}=10 \mathrm{~cm}^{3} \end{aligned}

Answer

The volume of the air increases to 10 cm 3 10 cm 3 10cm^(3)10 \mathrm{~cm}^{3}.
Worked Example 9.2 shows how to use the equation p 1 V 1 = p 2 V 2 p 1 V 1 = p 2 V 2 p_(1)V_(1)=p_(2)V_(2)p_{1} V_{1}=p_{2} V_{2} to find how the volume of a gas changes when the pressure on it is changed. You can use the same equation to work out how the pressure changes when the volume is changed.

Units

In the equation p 1 V 1 = p 2 V 2 p 1 V 1 = p 2 V 2 p_(1)V_(1)=p_(2)V_(2)p_{1} V_{1}=p_{2} V_{2}, it does not matter what units we use for p p pp and V V VV, as long as we use the same units for both values of p p pp and the same units for both values of V V VV.
The standard unit of pressure is the Pascal (Pa). 1 Pa = 1 N / m 2 1 Pa = 1 N / m 2 1Pa=1N//m^(2)1 \mathrm{~Pa}=1 \mathrm{~N} / \mathrm{m}^{2}. Pressure can also be measured in kPa , N / cm 2 kPa , N / cm 2 kPa,N//cm^(2)\mathrm{kPa}, \mathrm{N} / \mathrm{cm}^{2} or atmospheres. One atmosphere is approximately 100 . kPa 100 . kPa 100.kPa100 . \mathrm{kPa}.
Volume is usually measured in m 3 , dm 3 , cm 3 m 3 , dm 3 , cm 3 m^(3),dm^(3),cm^(3)\mathrm{m}^{3}, \mathrm{dm}^{3}, \mathrm{~cm}^{3} or litres.

Questions

14 What do each of the terms in the equation p 1 V 1 = p 2 V 2 p 1 V 1 = p 2 V 2 p_(1)V_(1)=p_(2)V_(2)p_{1} V_{1}=p_{2} V_{2}.represent?
15 The pressure on 6 dm 3 6 dm 3 6dm^(3)6 \mathrm{dm}^{3} of nitrogen gas is doubled at a fixed temperature. What will its volume become?
16 A flask holds 6 litres of air at a pressure of 2 atmospheres. Calculate the volume when the gas is compressed by increasing the pressure to 6 atmospheres. Assume that the temperature remains constant.

REFLECTION

Physics involves a lot of calculations. Presenting your calculations clearly helps you check your work. It also helps you identify any errors.
Look back over your calculations. Which steps are you using?
  • Identify which values are given in the question.
  • Check that the units are consistent and change them if necessary.
  • Write down the equation.
  • Substitute values into the equation.
  • Rearrange the equation if needed.
  • Calculate the answer.
  • Give units for the answer.
Can you make your calculation answers more clear? Make a note of what you could improve and refer to this next time you are doing calculations.

PROJECT

Making a game
Your task is to make a game to check how well you have understood this topic. You can invent your own game or use one of the ideas below.

Pairs

Make pairs of cards, one with a key word and the other with its definition. Do this for as many key words or ideas as you can find. Be creative - for example, you could draw Brownian motion rather than write a definition.
Put all the cards face down on a table, making sure they are mixed up.
Players take turns to turn over two cards. If they match, the player keeps them and takes another go. If they don’t match, the player turns them face down again and the next player takes a turn.
The winner is the player with most cards at the end.

Banned words

In this game a player takes a card which has a key word on it and a list of banned words.
An example is shown here.

Condensation

  • liquid
  • gas
  • COOl
  • particle
The player has to describe the word ‘condensation’ to their team without using any of the banned words. They could say, ‘it is the change of state that happens when water vapour turns to water’.
You can make your game harder or easier by the words you choose for your banned list.
Make a set of cards for the following words and phrases: solid, liquid, gas, vapour, melting, evaporation, condensation, freezing, melting point, boiling point, kinetic model of matter, Brownian motion, temperature, diffuse, expand, change of state, Boyle’s law, Pascal.
When you have made your cards, swap with another group and play the game. Players take turns choosing a card and describing the word to their team. The aim is to get through all the cards as quickly as possible.

SUMMARY

Matter can exist in three states - solid, liquid and gas.
Solids have a fixed shape and volume. Liquids have a fixed volume but take the shape of their container. Gases expand to fill their container.
The kinetic model explains the behaviour of materials by describing what is happening to the particles of which they are made.
According to the kinetic model, matter is made of moving particles that are close together in solids and liquids, and far apart in gases.
There are attractive forces between particles that act strongly when the particles are close together.
In Brownian motion, the movement of water or air particles is revealed by their effect on visible grains of pollen or smoke particles.
As the temperature of a substance increases, the kinetic energy of its particles increases.
The particles of a gas bombard the walls of its container. This causes pressure. Increasing temperature increases pressure. Decreasing the volume increases the pressure.
Temperature can be measured using the Kelvin temperature scale which has its zero at 273 C 273 C -273^(@)C-273^{\circ} \mathrm{C}.
Pressure and volume are related by the equation p V = p V = pV=p V= constant.

EXAM-STYLE QUESTIONS

1 Which one of these statements is describing a gas?
A It expands to fill the volume of its container.
B It has a fixed size and shape.
C It has a fixed volume but takes up the shape of its container.
D It has strong forces between the particles.
2 Brownian motion can be observed by looking at smoke particles under a microscope. What causes the movement of the smoke particles?
A The smoke particles are moved by air currents.
B The smoke particles hit the walls of the container.
C Air molecules hit the smoke particles.
D There are strong forces between the smoke particles.
3 Describe the diff erences between solids, liquids and gases in terms of the arrangement and movement of their particles.
[6]
4 A student observes smoke particles under a microscope. The smoke is lit up so she sees the smoke particles as dots of light.
a Describe the movement of the smoke particles.
b Name the particles which are causing this movement.
c Explain why observation of this movement provides evidence for the kinetic model of matter.
5 The diagram shows a syringe containing a small amount of air. The end of the syringe is sealed so no air can enter or leave.

SELF-EVALUATION CHECKLIST

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
I can
See
Topic...
See Topic...| See | | :--- | | Topic... |
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
Almost
there
Almost there| Almost | | :--- | | there |
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |
Describe the shape and volume of
solids, liquids and gases
Describe the shape and volume of solids, liquids and gases| Describe the shape and volume of | | :--- | | solids, liquids and gases |
9.1
Draw and explain particle diagrams
showing the separation and
arrangement of particles in solids,
liquids and gases.
Draw and explain particle diagrams showing the separation and arrangement of particles in solids, liquids and gases.| Draw and explain particle diagrams | | :--- | | showing the separation and | | arrangement of particles in solids, | | liquids and gases. |
9.2
Describe the movement of particles in
all three states of matter.
Describe the movement of particles in all three states of matter.| Describe the movement of particles in | | :--- | | all three states of matter. |
9.2
Describe and explain Brownian
motion.
Describe and explain Brownian motion.| Describe and explain Brownian | | :--- | | motion. |
9.2
Use the kinetic model to explain gas
pressure.
Use the kinetic model to explain gas pressure.| Use the kinetic model to explain gas | | :--- | | pressure. |
9.3
Describe and use the Celsius
temperature scale.
Describe and use the Celsius temperature scale.| Describe and use the Celsius | | :--- | | temperature scale. |
9.4
Describe and use the Kelvin
temperature scale.
Describe and use the Kelvin temperature scale.| Describe and use the Kelvin | | :--- | | temperature scale. |
9.4
Convert temperatures from Celsius to
Kelvin and from Kelvin to Celsius
Convert temperatures from Celsius to Kelvin and from Kelvin to Celsius| Convert temperatures from Celsius to | | :--- | | Kelvin and from Kelvin to Celsius |
9.4
Relate gas pressure to the change in
force created per unit area of particles
hitting the wall of a container.
Relate gas pressure to the change in force created per unit area of particles hitting the wall of a container.| Relate gas pressure to the change in | | :--- | | force created per unit area of particles | | hitting the wall of a container. |
9.5
Recall and use the equation p V = p V = pV=p V=
constant.
Recall and use the equation pV= constant.| Recall and use the equation $p V=$ | | :--- | | constant. |
9.5
I can "See Topic..." "Needs more work" "Almost there" "Confident to move on" "Describe the shape and volume of solids, liquids and gases" 9.1 "Draw and explain particle diagrams showing the separation and arrangement of particles in solids, liquids and gases." 9.2 "Describe the movement of particles in all three states of matter." 9.2 "Describe and explain Brownian motion." 9.2 "Use the kinetic model to explain gas pressure." 9.3 "Describe and use the Celsius temperature scale." 9.4 "Describe and use the Kelvin temperature scale." 9.4 "Convert temperatures from Celsius to Kelvin and from Kelvin to Celsius" 9.4 "Relate gas pressure to the change in force created per unit area of particles hitting the wall of a container." 9.5 "Recall and use the equation pV= constant." 9.5 | I can | See <br> Topic... | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :--- | :--- | :--- | :--- | | Describe the shape and volume of <br> solids, liquids and gases | 9.1 | | | | | Draw and explain particle diagrams <br> showing the separation and <br> arrangement of particles in solids, <br> liquids and gases. | 9.2 | | | | | Describe the movement of particles in <br> all three states of matter. | 9.2 | | | | | Describe and explain Brownian <br> motion. | 9.2 | | | | | Use the kinetic model to explain gas <br> pressure. | 9.3 | | | | | Describe and use the Celsius <br> temperature scale. | 9.4 | | | | | Describe and use the Kelvin <br> temperature scale. | 9.4 | | | | | Convert temperatures from Celsius to <br> Kelvin and from Kelvin to Celsius | 9.4 | | | | | Relate gas pressure to the change in <br> force created per unit area of particles <br> hitting the wall of a container. | 9.5 | | | | | Recall and use the equation $p V=$ <br> constant. | 9.5 | | | |

IN THIS CHAPTER YOU WILL:

  • describe how and why solids, liquids and gases expand when their temperatures rise
  • explain some everyday uses and consequences of thermal expansion
  • relate energy supplied to increase in temperature when an object is heated
    > > >> measure the specific heat capacity of some materials
  • explain changes of state using the kinetic model of matter.

GETTING STARTED

Each of these pictures relates to a change of temperature or a change of state.
Look at the pictures and discuss them with a partner.
What change is shown?
Describe the physics of what is happening, using as many key words as you can.
Why is the chameleon included?
Figure 10.1: Changes involving thermal energy.

THE STRANGE BEHAVIOUR OF WATER

Figure 10.2: The water below the ice is cold, but not cold enough to freeze.
Most substances expand when they get hot and contract when they get cold. Water does not always follow this rule. Think about how a pond cools down in cold weather. As the air temperature drops, the water at the top of the pond is cooled by the cold air above it and contracts. This makes the water at the top of the pond more dense, so it sinks to the bottom. The coldest water will be at the bottom of the pond. If this continued, ice would form at the bottom of the pond.
Fortunately for the fish, water does not keep contracting. As it cools from 4 C C ^(@)C{ }^{\circ} \mathrm{C} to 0 C 0 C 0^(@)C0^{\circ} \mathrm{C}, the water expands. This means that the colder water is less dense than the water below it and so it remains at the top. This water continues to cool and eventually freezes. The ice forms at the top of the pond, not the bottom. This means fish can survive in the water beneath the ice. This property of water is called the anomalous expansion of water.
Unlike most substances, water is less dense as a solid (ice) than it is as a liquid. This means that ice floats in water. Most substances are more dense when they are solid and so do not float. This strange behaviour is caused by the hydrogen bonds in water which cause it to form a crystal structure with the particles in ice more spaced out than in water.

Discussion questions

1 Sketch a graph to show how the density of water changes from 0 to 10
C C ^(@)C{ }^{\circ} \mathrm{C}.
2 Discuss what would happen to life on Earth if water froze like other liquids so that ponds froze from the bottom up.

10.1 Thermal expansion

Most substances - solids, liquids and gases - expand when their temperature rises. This is called thermal expansion. Thermal expansion happens because the particles gain energy and move faster, pushing each other further apart.

The expansion of solids

Figure 10.3 shows an experiment that demonstrates that a metal ball expands when it is heated.
  • When the ball is cold, it just fits through the ring.
  • The ball, but not the ring, is heated strongly. It now will not pass through the ring. It has expanded.
  • When the ball cools down, it contracts and returns to its original size and will once again pass through the ring.
Figure 10.3a: The metal ball is cold and has passed through the ring. b: The metal ball is hot. It has expanded and will no longer fit through the ring.

Uses of expansion

Rivets are used in shipbuilding and other industries to join metal plates. A redhot rivet is passed through holes in two metal plates and then hammered until the ends are rounded (Figure 10.4). As the rivet cools, it contracts and pulls the two plates together tightly.
A metal lid or cap may stick on a glass jar or bottle, and be hard to unscrew. Heating the lid (for example, by running hot water over it) causes it to expand. The glass expands much less than the metal lid, meaning that the lid loosens and can be removed.
hot rivet

Figure 10.4: Joining two metal plates using a rivet.
Figure 10.5: Steel tyres are heated so they expand and can be fitted to train wheels. They contract and fit very tightly.
A steel ‘tyre’ can be fitted on to the wheel of a train while the tyre is very hot. It then cools and contracts, so that it fits tightly on to the wheel.
Figure 10.6: A bimetallic strip. Invar is a metal alloy which expands very little when heated. Copper expands more. This difference in expansion causes the strip to bend.
A bimetallic strip (Figure 10.6) is designed to bend as it gets hot. The strip is made of two metals joined firmly together. One metal expands much more than the other. As the strip is heated, this metal expands, causing the strip to bend. The metal that expands more is on the outside of the curve, because the outer curve is longer than the inner one. These strips are used in devices such as fire alarms and thermostats. Thermostats are used to control the temperature of devices such as ovens and irons.

ACTIVITY 10.1

Uses of a bimetallic strip

A bimetallic strip can be used in an electric circuit to complete or break a circuit depending on the temperature.
  • Design a fire alarm circuit which will sound a buzzer when it gets hot. Design a poster to promote your invention. Include a circuit diagram and an explanation of how the device works.
  • Consider how you could change your circuit to make an alarm to warn gardeners when temperatures are dropping and allow them to protect delicate plants from frost. Design a poster for a frost alarm.

Consequences of expansion

The expansion of materials can cause problems. For example, metal bridges and railway lines expand on hot days, and there is a danger that they might bend. To avoid this, bridges are made in sections, with expansion joints between the sections (Figure 10.7). On a hot day, the bridge expands and the gaps between sections decrease. Railway lines are now usually made from a metal alloy that expands very little. On a concrete roadway, you may notice that the road surface is in short sections. The gaps between are filled with soft tar, which becomes squashed as the road expands.
Figure 10.7: This lorry is about to cross an expansion joint on a road bridge. On a hot day, the bridge expands and the interlocking teeth of the joint move closer together.

The expansion of liquids

Many thermometers use the expansion of a liquid to measure temperature. As the temperature of the liquid rises, it expands and the level of liquid in the tube rises.
Figure 10.8: As the temperature drops, so does the volume of the liquid in the thermometer. The alcohol in this thermometer remains liquid at very low temperatures.
Glass containers may crack when hot liquid is placed in them. This is because
the inner surface of the glass expands rapidly, before the thermal energy has passed through to the outer surface. The force of expansion cracks the glass. To avoid this, glass such as Pyrex has been developed that expands very little on heating. An alternative is toughened glass, which has been treated with chemicals to reduce the chance of cracking.

The expansion of gases

Gases expand when they are heated, just like solids and liquids. We can explain this using the kinetic model of matter (see Chapter 9). Figure 10.9 shows some gas in a cylinder fitted with a piston. At first, the gas is cold and its particles press weakly on the piston. When the gas is heated, its particles move faster. Now they push with greater force on the piston and push it upwards. The gas has expanded.
Figure 10.9: A gas expands when it is heated at constant pressure.
The expansion of gases as they heat up means that the density of a hot gas is lower than the density of the same gas when it is cold. This is why hot air rises (you will learn more about this in Chapter 11).
Figure 10.10: The expansion of air as it heats up creates thermal currents, which this hang glider is using.

Comparing solids, liquids and gases

Which expands most. a solid, a liquid or a gas, for a given rise in temperature?
  • Solids expand least when they are heated. Some, such as Pyrex glass and invar metal alloy, have been designed to expand as little as possible.
  • Liquids generally expand more than solids.
  • Gases expand even more than liquids.
There are some exceptions to this. For example, liquid paraffin expands very rapidly on heating. Petrol (gasoline) is stored in cool underground tanks. If a motorist fills their tank on a hot day, the petrol will heat up and expand. This can cause the fuel to overflow when it expands.
When a material expands, its particles (atoms or molecules) do not get any bigger. However, they have more energy, so they can move around more and take up more space. It is difficult for the particles of a solid to push their neighbours aside, so a solid does not expand much. When a gas is heated, its particles move about more rapidly, and it is easy for them to push the walls of their container further apart, so that the gas takes up more space.

Questions

1 Copy and complete these sentences:
When an object is heated it qquad\qquad . When it cools it qquad\qquad .
Liquids expand more than qquad\qquad but less than qquad\qquad .
A bimetallic strip is made of two qquad\qquad which expand by different amounts when heated.
This causes the strip to qquad\qquad .
2 Look at the bridge in Figure 10.11. One end of the bridge is on rollers rather than being fixed. Explain how this helps it cope with extreme temperatures.
Figure 10.11: A bridge.
3 Alcohol has a freezing point of 115 C 115 C -115^(@)C-115{ }^{\circ} \mathrm{C}. Explain why coloured water could not be used in the thermometer in Figure 10.8.
4 Figure 10.12 shows an experiment to compare the expansion of liquids. Equal volumes of ethanol, water and mercury were heated in a water bath.
Figure 10.12: Experiment to compare expansion of liquids.
a Which liquid expands the most?
b Explain why this liquid is used in most school laboratory thermometers.

10.2 Specific heat capacity

Energy and temperature

All objects store energy, called internal energy. Internal energy is a measure of the total energy of all the particles in the object. This includes both the kinetic energy of the particles and chemical potential energy of the bonds between them. Supplying thermal energy to an object will raise its temperature. The particles will move faster as they gain kinetic energy.
However, energy and temperature are not the same thing. The internal energy of an object is the total energy of all its particles. A cup of tea may have a higher temperature than a bath of warm water, but the energy stored in the bath water is much greater as there are many more water particles.
C
Figure 10.13: The tea is hotter than the bath so its particles move faster. However, the bath has far more particles and stores more energy.
takes more energy to raise the temperature of a large amount of water because more particles need to have their speed changed. This section deals with increasing the kinetic energy of particles. We will look at the potential energy in Section 10.3.
Suppose that you want to make a hot drink for yourself and some friends. You need to boil some water. You will be wasting energy if you put too much water in the kettle or pan. It is sensible to boil just the right amount. Also, if the water from your tap is really cold, it will take longer, and require more energy, to reach boiling point.
So, the amount of energy you need to supply to boil the water will depend on two facts:
  • the mass of the water
  • the increase in temperature.
In order to calculate how much energy must be supplied to boil a certain mass of water, we need to know a third fact:
  • it takes 4200 J to raise the temperature of 1 kg of water by 1 C 1 C 1^(@)C1^{\circ} \mathrm{C}.
Let us assume that the cold water from your tap is at 20 C 20 C 20^(@)C20^{\circ} \mathrm{C}. You have to provide enough energy to heat it to 100 C 100 C 100^(@)C100^{\circ} \mathrm{C}, so its temperature must increase by 80 C 80 C 80^(@)C80^{\circ} \mathrm{C}. Let us also assume that you need 2 kg of water for all the drinks.
The amount of energy required to heat 2 kg of water by 80 C 80 C 80^(@)C80^{\circ} \mathrm{C} is, therefore:
energy required = 2 kg × 4200 J / ( kg C ) × 80 C = 672000 J = 672 kJ  energy required  = 2 kg × 4200 J / kg C × 80 C = 672000 J = 672 kJ {:[" energy required "=2kgxx4200J//(kg^(@)C)xx80^(@)C],[=672000J=672kJ]:}\begin{aligned} \text { energy required } & =2 \mathrm{~kg} \times 4200 \mathrm{~J} /\left(\mathrm{kg}{ }^{\circ} \mathrm{C}\right) \times 80^{\circ} \mathrm{C} \\ & =672000 \mathrm{~J}=672 \mathrm{~kJ} \end{aligned}
Another way to express the third fact above is to say that the specific heat capacity of water is 4200 J per kg per C C ^(@)C{ }^{\circ} \mathrm{C} or 4200 J / ( kg C ) 4200 J / kg C 4200J//(kg^(@)C)4200 \mathrm{~J} /\left(\mathrm{kg}{ }^{\circ} \mathrm{C}\right). In general, the specific heat capacity of a substance is defined as the energy needed to raise the temperature of 1 kg of the substance by 1 C 1 C 1^(@)C1^{\circ} \mathrm{C}. It is defined by the equation:
c = Δ E m Δ θ c = Δ E m Δ θ c=(Delta E)/(m Delta theta)c=\frac{\Delta E}{m \Delta \theta}

KEY EQUATION

specific heat capacity = energy required mass × temperature increase c = Δ E m Δ θ  specific heat capacity  =  energy required   mass  ×  temperature increase  c = Δ E m Δ θ {:[" specific heat capacity "=(" energy required ")/(" mass "xx" temperature increase ")],[c=(Delta E)/(m Delta theta)]:}\begin{aligned} \text { specific heat capacity } & =\frac{\text { energy required }}{\text { mass } \times \text { temperature increase }} \\ c & =\frac{\Delta E}{m \Delta \theta} \end{aligned}
where c = c = c=c= specific heat capacity, Δ E = Δ E = Delta E=\Delta E= energy required, m = m = m=m= mass and Δ θ = Δ θ = Delta theta=\Delta \theta= increase in temperature. is the Greek letter delta; it means ‘change in’.
The energy needed can be calculated by rearranging the equation: energy required = = == mass × × xx\times specific heat capacity × × xx\times increase in temper: Δ E = m c Δ θ Δ E = m c Δ θ Delta E=mc Delta theta\Delta E=m c \Delta \theta
Worked Example 10.1 shows how to use this equation in more detail.

WORKED EXAMPLE 10.1

A kettle heats 1.5 kg of water. How much energy is needed to raise the temperature of the water from 20 C 20 C 20^(@)C20^{\circ} \mathrm{C} to 100 C 100 C 100^(@)C100^{\circ} \mathrm{C} ? Assume that specific heat capacity of water = 4200 J / ( kg C ) = 4200 J / kg C =4200J//(kg^(@)C)=4200 \mathrm{~J} /\left(\mathrm{kg}{ }^{\circ} \mathrm{C}\right).
Step 1: Calculate the required increase in temperature.
increase in temperature = 100 C 20 C = 80 C  increase in temperature  = 100 C 20 C = 80 C {:[" increase in temperature "=100^(@)C-20^(@)C],[=80^(@)C]:}\begin{aligned} \text { increase in temperature } & =100^{\circ} \mathrm{C}-20^{\circ} \mathrm{C} \\ & =80^{\circ} \mathrm{C} \end{aligned}
Step 2: Write down the other quantities needed to calculate the energy.
mass of water = 1.5 kg specific heat capacity of water = 4200 J / ( kg C )  mass of water  = 1.5 kg  specific heat capacity of water  = 4200 J / kg C {:[" mass of water "=1.5kg],[" specific heat capacity of water "],[=4200J//(kg^(@)C)]:}\begin{aligned} & \text { mass of water }=1.5 \mathrm{~kg} \\ & \text { specific heat capacity of water } \\ & =4200 \mathrm{~J} /\left(\mathrm{kg}{ }^{\circ} \mathrm{C}\right) \end{aligned}
Step 3: Write down the equation for energy required, substitute values, and calculate the result.
energy required = = == mass × × xx\times specific heat capacity × × xx\times increas
= 1.5 kg × 4200 J / ( kg C ) × 80 C = 1.5 kg × 4200 J / kg C × 80 C =1.5kgxx4200J//(kg^(@)C)xx80^(@)C=1.5 \mathrm{~kg} \times 4200 \mathrm{~J} /\left(\mathrm{kg}{ }^{\circ} \mathrm{C}\right) \times 80^{\circ} \mathrm{C}
= 504000 = 504000 =504000=504000.
= 504 kJ = 504 kJ =504kJ=504 \mathrm{~kJ}
Answer
504 kJ of energy is required to boil the water.

The meaning of specific heat capacity

Energy is needed to raise the temperature of any material. The energy is needed to increase the kinetic energy of the particles of the material. In solids, they vibrate more. In gases, they move about faster. In liquids, it is a bit of both.
We can compare different materials by considering standard amounts ( 1 kg ), and a standard increase in temperature ( 1 C 1 C 1^(@)C1^{\circ} \mathrm{C} ). Different materials require different amounts of energy to raise the temperature of 1 kg by 1 C 1 C 1^(@)C1^{\circ} \mathrm{C}. In other words, they have different specific heat capacities. Table 10.1 shows the values of specific heat capacity for a variety of materials.
Table 10.1 shows that there is quite a wide range of values. The specific heat capacity of steel, for example, is one-tenth that of water. This means that, when you supply equal amounts of energy to 1 kg of steel and to 1 kg of water, the temperature of the steel rises ten times as much as the temperature of the water.
Type of material Material
Specific heat capacity / J/(kg
C ) C {:^(@)C)\left.{ }^{\circ} \mathrm{C}\right)
Specific heat capacity / J/(kg {:^(@)C)| Specific heat capacity / J/(kg | | :--- | | $\left.{ }^{\circ} \mathrm{C}\right)$ |
metals steel 420
aluminium 910
385
gold 300
lead 130
liquids glass 670
nylon 1700
polythene 2300
ice 2100
gases water 4200
sea water 3900
ethanol 2500
olive oil 1970
air 1000
water vapour 2020 (at 100 C 100 C 100^(@)C100{ }^{\circ} \mathrm{C} )
Type of material Material "Specific heat capacity / J/(kg {:^(@)C)" metals steel 420 aluminium 910 385 gold 300 lead 130 liquids glass 670 nylon 1700 polythene 2300 ice 2100 gases water 4200 sea water 3900 ethanol 2500 olive oil 1970 air 1000 water vapour 2020 (at 100^(@)C ) | Type of material | Material | Specific heat capacity / J/(kg <br> $\left.{ }^{\circ} \mathrm{C}\right)$ | | :--- | :--- | :--- | | metals | steel | 420 | | | aluminium | 910 | | | 385 | | | | gold | 300 | | | lead | 130 | | liquids | glass | 670 | | | nylon | 1700 | | | polythene | 2300 | | | ice | 2100 | | gases | water | 4200 | | | sea water | 3900 | | | ethanol | 2500 | | | olive oil | 1970 | | | air | 1000 | | | water vapour | 2020 (at $100{ }^{\circ} \mathrm{C}$ ) | | | | |
Table 10.1: Specific heat capacities of a variety of materials.

The specific heat capacity of water

Water is an unusual substance. As you can see from Table 10.1, it has a high value of specific heat capacity (s.h.c.) compared with other materials. This has important consequences:
  • it takes a lot of energy to heat up water
  • hot water takes a long time to cool down.
The consequences of this can be seen in our climates. In the hot months of summer, the land warms up quickly (low specific heat capacity) while the sea warms up only slowly. In the winter, the sea cools gradually while the land cools rapidly. People who live a long way from the sea (in the continental interior of North America or Eurasia, for example) experience freezing winters and very hot summers. People who live on islands and in coastal areas (such as western Europe) are protected from climatic extremes because the sea acts as a store of heat in the winter, and stays relatively cool in the summer.

EXPERIMENTAL SKILLS 10.1

Measuring the specific heat capacity of a metal
In this experiment you will heat a metal block. You will measure the energy you supply, the mass of the metal and the change in temperature. You will then be able to calculate the s.h.c. of the metal.

You will need:

  • a block of metal with holes for the thermometer and heater
  • insulation for the block
  • electric heater
  • power pack
  • joulemeter (your teacher may show you an alternative way to find the energy used)
  • thermometer
  • access to balance.
Safety: The electric heater can become extremely hot. Leave it to cool inside the block when you have finished your experiment.

Getting started

1 Should you insulate the metal block? Explain your answer.
2 Why should you measure the mass of the block at the start of the experiment rather than the end?

Method

1 Measure and record the mass, m m mm, of the block in kg.
2 Set up the experiment as shown in Figure 10.14.
3 Measure and record the initial temperature ( θ 1 θ 1 theta_(1)\theta_{1} ) of the block.
4 Turn on the power supply.
5 When there has been a temperature rise of 10 C 10 C 10^(@)C10^{\circ} \mathrm{C}, turn off the power supply and record the Joulemeter reading.
6 Watch the thermometer for a few minutes and record the highest temperature ( θ 2 ) θ 2 (theta_(2))\left(\theta_{2}\right) it reaches.
7 Calculate the change in temperature using Δ θ = θ 2 θ 1 Δ θ = θ 2 θ 1 Delta theta=theta_(2)-theta_(1)\Delta \theta=\theta_{2}-\theta_{1}.
8 Calculate the specific heat capacity of the metal using the equation:
c = Δ E m Δ θ c = Δ E m Δ θ c=(Delta E)/(m Delta theta)c=\frac{\Delta E}{m \Delta \theta}

EXPERIMENTAL SKILLS 10.2

Measuring the specific heat capacity of water
In this experiment you will heat a mass of water and find its specific heat capacity

You will need:

  • a beaker of water with a lid with holes for the thermometer and heater
  • insulation for the beaker
  • electric heater
  • power pack
  • joulemeter (your teacher may show you an alternative way to find the energy used)
  • thermometer
  • access to balance.
Safety: The heater and water will get hot. Allow them to cool before clearing away your apparatus.

Getting started

1 Why is it important to have a lid on the beaker?
2 How will you measure the mass of the water?
Method
power pack
Figure 10.15
1 Put 0.25 kg of water into the beaker.
2 Set up the experiment as shown in Figure 10.15.
3 Measure and record the initial temperature ( θ 1 θ 1 theta_(1)\theta_{1} ) of the block.
4 Turn on the power supply and leave until the temperature changes by about 50 C 50 C 50^(@)C50^{\circ} \mathrm{C}.
5 Turn off the power supply. Record the final temperature ( θ 2 ) θ 2 (theta_(2))\left(\theta_{2}\right)
6 Calculate the change in temperature using Δ θ = θ 2 θ 1 Δ θ = θ 2 θ 1 Delta theta=theta_(2)-theta_(1)\Delta \theta=\theta_{2}-\theta_{1}.
7 Record the joulemeter reading.
8 Calculate the specific heat capacity of the metal using the equation:
c = Δ E m Δ θ c = Δ E m Δ θ c=(Delta E)/(m Delta theta)c=\frac{\Delta E}{m \Delta \theta}

Questions

1 Compare your answers with the values of specific heat capacity given in Table 10.1. Are they higher or lower?
2 The thermal energy supplied by the heater heats the heater itself, the thermometer and the surrounding air as well as the block. Explain what effect this will have on the accuracy of your answer.

Questions

5 Table 10.2 gives the specific heat capacities of some materials.
Material Specific heat capacity
water 4200
gold 300
glass 840
air 1000
Material Specific heat capacity water 4200 gold 300 glass 840 air 1000| Material | Specific heat capacity | | :--- | ---: | | water | 4200 | | gold | 300 | | glass | 840 | | air | 1000 |
Table 10.2
a Name the substance that requires the least amount of energy to raise its temperature.
b The table does not give units. What is the unit for specific heat capacity?
c State the amount of energy needed to heat 1 kg of glass by 1 C 1 C 1^(@)C1^{\circ} \mathrm{C}.
d Calculate the energy needed to heat 1 kg of glass by 10 C 10 C 10^(@)C10^{\circ} \mathrm{C}.
6 The specific heat capacity of copper is 385 J / kg C 385 J / kg C 385J//kg^(@)C385 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C}. Calculate the energy needed to heat 3 kg of copper:
a by 1 C 1 C 1^(@)C1^{\circ} \mathrm{C}
b from 20 C 20 C 20^(@)C20^{\circ} \mathrm{C} to 50 C 50 C 50^(@)C50^{\circ} \mathrm{C}.
7 A cook heats 500 g of olive oil in a steel pan which has a mass of 300 g . The oil needs to be heated from 20 C 20 C 20^(@)C20^{\circ} \mathrm{C} to 190 C 190 C 190^(@)C190^{\circ} \mathrm{C}. Using data from Table 10.1, calculate the thermal energy needed:
a to heat the pan
b to heat the oil
c in total.
8 The electric kettle in Figure 10.16 has a power rating of 2000 W. It takes 90 seconds to heat 500 g water from 20 C 20 C 20^(@)C20^{\circ} \mathrm{C} to boiling.
Figure 10.16: An electric kettle.
a Use this information to calculate an approximate value for the specific heat capacity of water.
b Compare your answer to the specific heat capacity of water given in Table 10.1. Comment on why it is different.

10.3 Changing state

Figure 10.17 shows what happens to the temperature when you take some ice from the freezer and heat it at a steady rate. In a freezer, ice is at a temperature well below its freezing point, maybe as low as 20 C 20 C -20^(@)C-20^{\circ} \mathrm{C}. From the graph, you can see that the ice warms up to 0 C 0 C 0^(@)C0^{\circ} \mathrm{C}, then stays at this temperature while it melts. As lumps of ice float in water, both are at 0 C 0 C 0^(@)C0^{\circ} \mathrm{C}. When all of the ice has melted, the water’s temperature starts to rise again. At 100 C 100 C 100^(@)C100^{\circ} \mathrm{C}, the boiling point of water, the temperature again remains steady. The water is boiling to form steam. Eventually, all of the water has turned to steam. If you continue to heat the steam, the temperature of the steam will rise again.
Figure 10.17: A temperature against time graph to show the changes that occur when ice is heated until it eventually becomes steam. A: ice is heating up. B: ice is melting. C: water is heating up. D: water is boiling. E: steam is heating up.
It takes energy to change a solid into a liquid. The temperature stays the same as the ice melts. Similarly, when a liquid becomes a gas, its temperature stays the same even though energy is being supplied to it. This is due to changes in the chemical potential energy in the bonds between the molecules or atoms. Energy must be provided to break bonds and change a substance from solid to liquid. Energy is also needed to overcome the attraction between the particles when a substance changes from liquid to gas. When these changes are reversed, energy is given out.
Condensation is when a gas changes to a liquid. The particles slow down and the particles are drawn together. Solidification, or freezing, is when a liquid changes to a solid. As a liquid loses energy, its particles slow down and the bonds holding the particles together re-form.
The boiling and melting points of a substance change if the air pressure changes. Water has a melting point of 0 C 0 C 0^(@)C0{ }^{\circ} \mathrm{C} and a boiling point of 100 C 100 C 100^(@)C100^{\circ} \mathrm{C} at standard atmospheric pressure. This is the pressure at sea level which is about 101.32 kPa or 1 atmosphere. At altitude, water boils at a lower temperature.
Figure 10.18: On top of Mount Everest water boils at 71 C 71 C 71^(@)C71^{\circ} \mathrm{C}.

Investigating a change of state

Figure 10.19a: The thermometer measures the temperature of the wax as it cools. b: The graph shows how the temperature changes.
Figure 10.19a shows one way to investigate the behaviour of a liquid material as it solidifies. The test tube contains a waxy substance called stearic acid. This is
warmed up, and it becomes a clear, colourless liquid. It is then left to cool down, and its temperature is monitored using a thermometer (or an electronic temperature probe) and recorded.
Figure 10.19b shows the results.
At first, the liquid wax cools down. Its temperature drops gradually. The wax is hotter than its surroundings, so thermal energy is transferred to the surroundings. The graph is slightly curved. As the temperature drops, there is less difference between the temperature of the wax and its surroundings, so it cools more slowly.
Now the temperature of the wax stays the same for a few minutes. The tube contains a mixture of clear liquid and white solid. The wax is solidifying. The wax is still transferring thermal energy to the surroundings, because it is still warmer than its surroundings, but its temperature does not decrease.
The wax’s temperature starts to drop again. All the wax is now solid. It continues cooling until it reaches the temperature of its surroundings.
The dashed line on the graph has been drawn to find the temperature at which the stearic acid changes from a liquid to a solid.
The stearic acid experiment (Figure 10.19) shows that a pure substance changes from solid to liquid at a fixed temperature, known as the melting point.
Similarly, a liquid changes to a gas at a fixed temperature, known as its boiling point. Table 10.3 shows the melting and boiling points of some pure substances.
Substance Melting point / C / C //^(@)C/{ }^{\circ} \mathrm{C} Boiling point / C / C //^(@)C/{ }^{\circ} \mathrm{C}
helium -272 -269
oxygen -218 -183
nitrogen -191 -177
mercury -39 257
water 0 100
iron 2080 3570
Substance Melting point //^(@)C Boiling point //^(@)C helium -272 -269 oxygen -218 -183 nitrogen -191 -177 mercury -39 257 water 0 100 iron 2080 3570| Substance | Melting point $/{ }^{\circ} \mathrm{C}$ | Boiling point $/{ }^{\circ} \mathrm{C}$ | | :--- | :---: | :---: | | helium | -272 | -269 | | oxygen | -218 | -183 | | nitrogen | -191 | -177 | | mercury | -39 | 257 | | water | 0 | 100 | | iron | 2080 | 3570 |
Table 10.3: The melting and boiling points of some pure substances. Mercury is interesting because it is the only metal that is not solid at room temperature. Tungsten is a metal, and it has the highest boiling point of any substance.
Helium has the lowest melting and boiling points of any element. In fact, helium
will only solidify if it is compressed (squashed) as well as cooled.
A pure substance has a clear melting point and a clear and boiling point. A mixture of substances may melt or boil over a range of temperatures. Candle wax is an example. It is not a single, pure substance, and some of the substances in it melt at lower temperatures than others.
Dissolving things in water changes the boiling point and freezing point of the water. For example, salty water boils at a higher temperature than pure water and freezes at a lower temperature.
Not all substances melt or boil when they are heated. Some burn, and others decompose (break down) into simpler substances before they have a chance to change state.

ACTIVITY 9.2

Breaking free

Figure 10.20

Create a comic strip to tell the story of what happens to the particle when it gets what it wants (energy) and changes from a solid to a liquid and then to a gas.
Your comic strip should be:
  • engaging (use pictures, colour and humour)
  • scientifically correct (check back through the chapter so far; use the correct scientific words and explain what is happening).

PEER ASSESSMENT

Swap comic strips with another group. Study their work and give them feedback. Include:
  • three ways it works well
  • one suggestion for how it could be even better; this could be something they need to make clearer, or something they could add.

Questions

9 Look at Figure 10.17. Sketch the graph and add notes explaining what is happening during each section.
10 Use Figure 10.19b to determine the melting point of stearic acid.
11 Use the information in Table 9.2 to explain why air doesn’t have a fixed melting or boiling point. (Hint: air is roughly 80 % 80 % 80%80 \% nitrogen and 20 % 20 % 20%20 \% oxygen).

Evaporation

Figure 10.21: In warm weather puddles evaporate quickly.
A liquid can change state without boiling. After it rains, the puddles dry up even though the temperature is much lower than 100 C 100 C 100^(@)C100^{\circ} \mathrm{C}. The water from the puddles has evaporated. The liquid water has become a gas called water vapour in the air. This is the process of evaporation. We can think of a vapour as a gas at a temperature below its boiling point.
A liquid evaporates more quickly as its temperature approaches its boiling point. That is why puddles disappear faster on a hot day than a cold day.
Figure 10.22: Fast-moving particles leave the surface of a liquid. This is how the liquid evaporates.
How can we use the kinetic molecular model of matter to explain evaporation? Imagine a beaker of water. The water will gradually evaporate. Figure 10.22 shows the particles that make up the water. The particles of the water are moving around, and some are moving faster than others. Some may be moving fast enough to escape from the surface of the water. They become particles of water vapour in the air. In this way, all of the water particles may eventually escape from the beaker, and the water will have evaporated.
If the temperature of the liquid is higher, more of its particles will have enough energy to escape. This means the liquid will evaporate more quickly. The hottest particles are most likely to escape as they have most energy. When they escape, the average energy of the remaining particles is less, so the liquid cools down.

Cooling by evaporation

If you get wet, perhaps in the rain or after swimming, you will notice that you can quickly get cold. The water on your body is evaporating, and this cools you down. Why does evaporation make things cooler?
Look again at Figure 10.22. The particles that are escaping from the water are the fastest-moving ones. They are the particles with the most kinetic energy. This means that the particles that remain are those with less energy. Now the particles of the liquid have less energy (on average) and so the temperature of the water decreases. The water cools down.

Comparing evaporation and boiling

Evaporation and boiling both involve a liquid turning into a gas. Evaporation is different from boiling.
  • Boiling only happens at the boiling point of the substance. Evaporation occurs at all temperatures.
  • For a liquid to boil, it has to be heated - the kinetic energy of its particles must be increased. Evaporation happens when the most energetic particles escape, so evaporation takes energy from the substance.
  • Boiling happens throughout the liquid. Evaporation only happens at the surface.
  • A boiling liquid bubbles. A liquid can evaporate without bubbles.

Speeding up evaporation

We can use the kinetic model to explain some ways of speeding up evaporation. Increasing the temperature
Figure 10.23: Increasing the temperature increases the rate of evaporation.
Increasing the temperature of the liquid means the particles on average have more kinetic energy. More of the particles will have enough energy to escape. This means the liquid will evaporate more quickly.
Increasing the surface area
Figure 10.24: Increasing the surface area increases the rate of evaporation.
In Figure 10.24, the liquid has a greater surface area more of the particles are close to the surface, and so they can escape more easily. This means the liquid evaporates more quickly.
Blowing air across the surface
Figure 10.25: A draught blowing across the surface increases the rate of evaporation.
A draught is moving air. When particles escape from the water, they are blown away so that they cannot fall back in to the water. This helps the liquid evaporate quickly.
As a liquid evaporates, the remaining liquid cools. This means that thermal energy will flow to the liquid from any objects in contact with it. When we get hot we sweat. The sweat evaporates, causing thermal energy to flow from the skin. This helps us cool down.
Figure 10.26: Sweating helps us regulate our body temperature in hot conditions.
Fridges use the cooling effect of evaporation (see Figure 10.27). A liquid is compressed then squirted through a narrow hole so its pressure is reduced and it evaporates. This draws thermal energy from inside the fridge into the liquid. The liquid is then pumped out of the fridge to the pipes on the back of the fridge where it is compressed and condenses, releasing the thermal energy to the surroundings.
Figure 10.27: The refrigerant liquid absorbs thermal energy from the fridge as it evaporates.

Questions

12 Copy and complete the following sentences.
a qquad\qquad is the change from a liquid to a gas at a temperature below the boiling point.
b Evaporation causes a liquid to cool because the qquad\qquad moving, hotter particles are the most likely to escape. This means the particles left behind are the slower moving, qquad\qquad ones.
13 Explain in terms of movement and position of particles what happens to an ice cube as it is heated and melts.
14 Tungsten melts at a much higher temperature than iron. What can you say
about the forces between the tungsten atoms, compared to the forces between the iron atoms?
15 A solid material is heated but its temperature does not rise.
a What is happening to the solid?
b What happens to the energy that is being supplied to the material?
16 Use the kinetic theory to explain why a wet towel will dry much faster if hung outside on a warm, windy day than if left folded up in a bag.
17 Explain how covering a bottle of milk with a damp cloth will help to cool the milk.

PROJECT

Hot stuff

Figure 10.28: Heating gold causes it to melt. Liquid gold can be poured into moulds, where it will cool and solidify.

Option 1: Changing state

Create a video or article for a revision guide to explain changes of state. Show objects changing state, such as an ice cube melting or a puddle of water evaporating (you could use time lapse photography). You should show what is happening to the particles. You may want to use marbles or drawings to represent the particles. You must include the following key words:
  • melting
  • boiling
  • condensing
  • solidifying
  • internal energy
  • temperature.
Write three multiple choice questions for the students to test their understanding to go with your video.

Option 2: Specific heat capacity questions

Specific heat capacity is important in engineering and in the choice of material for different products. Investigate the applications of specific heat capacity in answering questions such as:
  • Which is the best metal for a pan?
  • What makes a good coolant for the cooling system for an engine?
  • Why do the different specific heat capacities of sand and water create sea breezes?
  • Which types of food store most thermal energy and so stay hot the longest?
Research one or more of these questions. Present your results as a website entry for a site which answers scientific questions posed by science students. Write a question to test the students’ understanding. Include a calculation and answer.

PEER ASSESSMENT

Present your work to another group. Ask them to answer your questions and comment on what they have learnt.

SUMMARY

Solids, liquids and gases expand when their temperatures rise.
Solids, liquids and gases contract when their temperatures drop.
Solids expand least because their particles are held together by strong attractive forces.
Gases expand most as their particles are free to move.
Expansion can cause problems such as the bending of train tracks, but can also be useful, for example in thermometers.
Melting is the change from a solid to a liquid without any change in temperature.
Boiling is the change from a liquid to a gas without a change in temperature.
Supplying heat energy increases the internal energy of a substance and its temperature rise.
The amount of energy needed to raise the temperature of 1 Kg of a substance by 1 C 1 C 1^(@)C1^{\circ} \mathrm{C} is called the specific heat capacity of the substance.
The energy needed to heat a mass m m mm of a substance by Δ θ Δ θ Delta theta\Delta \theta degrees can be calculated using the equation c = Δ E m Δ θ c = Δ E m Δ θ c=(Delta E)/(m Delta theta)c=\frac{\Delta E}{m \Delta \theta}.
During a change of state, energy is supplied or given out but the temperature does not change.
When a liquid evaporates the most energetic particles escape from the surface, so the liquid cools.
Evaporation occurs at temperatures below the boiling point of the liquid. It happens faster at higher temperatures, with a larger surface area or if there is a draught across the surface.

EXAM-STYLE QUESTIONS

1 Which one of the following statements is true?
A The firework has the highest temperature and it contains the most internal energy.
B The bath has the highest temperature and contains the most internal energy.
C The firework has the highest temperature but contains the least internal energy.
D The bath has the lowest temperature and contains the least internal energy.
2 Which statement about the difference between evaporation and boiling is true?
A Evaporation happens at any temperature.
B Evaporation happens throughout a liquid.
C Evaporation happens at a specific temperature.
D Evaporation and boiling are not different.
3 What is thermal expansion?
A An increase in density due to an increase in temperature.
B A decrease in density due to a decrease in temperature.
C A decrease in density due to an increase in temperature.
D A decrease in volume due to an increase in temperature.
4 A jeweller is making a silver ring. She heats the silver until it melts, and then continues to heat it for another minute after it has melted. She then pours the molten (melted) silver into a mould and leaves it to cool down.

a Describe what happens to the arrangement and movement of the silver atoms as the silver is heated from room temperature and poured into the mould.
b The graph shows how the temperature of the silver varies as it is heated.
Use the graph to find the melting point of silver.
c State how long it takes for the silver to melt.
[Total: 5]
5 The diagram shows the circuit for a fire alarm using a bimetallic strip.

a Brass expands more than iron. Which metal should be at the top of the strip?
b Describe what happens as the temperature rises in case of a fire.
[Total: 4]
6 A cook spills water. Some falls on the floor, some on the hot stove top. The temperature of the stove top is about 80 C 80 C 80^(@)C80^{\circ} \mathrm{C}. He goes to get a cloth to wipe up. When he returns, he sees the stove is dry but the floor is still wet.
a Name the process by which water becomes a gas at a temperature below its boiling point.
b why energy must be supplied to a liquid to turn it into a gas. In your answer, refer to the particles of the liquid and the forces between them.
c Explain why the floor will dry more quickly if the cook turns on an electric fan.
7 Huda follows the cooking instructions for a tin of baked beans.
  • Adjust times according to your particular oven.
Huda’s microwave oven has a power of 650 W . She heats the beans for the recommended time and checks the temperature. The beans have heated from 20 C 20 C 20^(@)C20^{\circ} \mathrm{C} to 90 C 90 C 90^(@)C90^{\circ} \mathrm{C}.
a Use the equation energy = = == power × × xx\times time to calculate the energy supplied.
b the specific heat capacity of the beans. Give your answer to two significant figures.
c She notices that the dish she heated the beans in has also become hot and realises this will affect her calculation. Explain whether her result will be too high or too low.
d Huda also measures the specific heat capacity of tinned spaghetti. She calculates it as 3100 J / kg C 3100 J / kg C 3100J//kg^(@)C3100 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C}. If she heats both the beans and spaghetti to the same temperature, which will cool down quicker? Explain your answer.

SELF-EVALUATION CHECKLIST

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
I can
See
Topic...
See Topic...| See | | :--- | | Topic... |
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
Almost
there
Almost there| Almost | | :--- | | there |
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |
Describe everyday examples of
thermal expansion.
Describe everyday examples of thermal expansion.| Describe everyday examples of | | :--- | | thermal expansion. |
10.1
State which expands most, solid,
liquid or gas, and explain this in terms
of the particles.
State which expands most, solid, liquid or gas, and explain this in terms of the particles.| State which expands most, solid, | | :--- | | liquid or gas, and explain this in terms | | of the particles. |
10.1
State the effect of raising an objects,
temperature on its internal energy.
State the effect of raising an objects, temperature on its internal energy.| State the effect of raising an objects, | | :--- | | temperature on its internal energy. |
10.2
Recall and use the equation c = Δ E m Δ θ c = Δ E m Δ θ c=(Delta E)/(m Delta theta)c=\frac{\Delta E}{m \Delta \theta}
Recall and use the equation c=(Delta E)/(m Delta theta)| Recall and use the equation $c=\frac{\Delta E}{m \Delta \theta}$ | | :--- |
10.2
Describe experiments to measure the
specific heat capacity of solids and
liquids.
Describe experiments to measure the specific heat capacity of solids and liquids.| Describe experiments to measure the | | :--- | | specific heat capacity of solids and | | liquids. |
10.2
Describe changes of state as a change
in internal energy without a change in
temperature.
Describe changes of state as a change in internal energy without a change in temperature.| Describe changes of state as a change | | :--- | | in internal energy without a change in | | temperature. |
10.3
Know why evaporation causes
cooling.
Know why evaporation causes cooling.| Know why evaporation causes | | :--- | | cooling. |
10.3
Explain the factors that increase the
rate of evaporation.
Explain the factors that increase the rate of evaporation.| Explain the factors that increase the | | :--- | | rate of evaporation. |
10.3
Explain how an evaporating liquid
causes objects in contact with it to
cool.
Explain how an evaporating liquid causes objects in contact with it to cool.| Explain how an evaporating liquid | | :--- | | causes objects in contact with it to | | cool. |
10.3
I can "See Topic..." "Needs more work" "Almost there" "Confident to move on" "Describe everyday examples of thermal expansion." 10.1 "State which expands most, solid, liquid or gas, and explain this in terms of the particles." 10.1 "State the effect of raising an objects, temperature on its internal energy." 10.2 "Recall and use the equation c=(Delta E)/(m Delta theta)" 10.2 "Describe experiments to measure the specific heat capacity of solids and liquids." 10.2 "Describe changes of state as a change in internal energy without a change in temperature." 10.3 "Know why evaporation causes cooling." 10.3 "Explain the factors that increase the rate of evaporation." 10.3 "Explain how an evaporating liquid causes objects in contact with it to cool." 10.3 | I can | See <br> Topic... | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :--- | :--- | :--- | :--- | | Describe everyday examples of <br> thermal expansion. | 10.1 | | | | | State which expands most, solid, <br> liquid or gas, and explain this in terms <br> of the particles. | 10.1 | | | | | State the effect of raising an objects, <br> temperature on its internal energy. | 10.2 | | | | | Recall and use the equation $c=\frac{\Delta E}{m \Delta \theta}$ | 10.2 | | | | | Describe experiments to measure the <br> specific heat capacity of solids and <br> liquids. | 10.2 | | | | | Describe changes of state as a change <br> in internal energy without a change in <br> temperature. | 10.3 | | | | | Know why evaporation causes <br> cooling. | 10.3 | | | | | Explain the factors that increase the <br> rate of evaporation. | 10.3 | | | | | Explain how an evaporating liquid <br> causes objects in contact with it to <br> cool. | 10.3 | | | |

> Chapter 11

Thermal energy transfers

In this chapter you will:

  • carry out experiments to demonstrate conduction, convection and radiation
    > > >> explain why some materials conduct and others do not
  • describe and explain convection currents
  • explain thermal energy radiation
    > > >> investigate the differences between good and bad emitters of radiation
  • research applications and consequences of thermal energy transfer.

GETTING STARTED

Figure 11.1 shows items that are used to transfer thermal energy or prevent thermal energy transfer. Work in a group to decide what each item is for and how it works. Your teacher will stop you and pick a member of the group to report back on one of the photos. If they can do this they score a point for the team. If they use any of the following terms correctly, your teacher will award bonus points.
Figure 11.1: A selection of items that transfer thermal energy, or prevent thermal energy transfer.
conduction convection radiation conductor insulator
convection
current
convection current| convection | | :---: | | current |
infrared
radiation
infrared radiation| infrared | | :---: | | radiation |
conduction convection radiation conductor insulator "convection current" "infrared radiation"| conduction | convection | radiation | conductor | insulator | convection <br> current | infrared <br> radiation | | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
Rules:
  • All members of the group need to be able to answer.
  • You cannot write down your ideas.
  • You can write a maximum of three words for each picture as a prompt.

WHAT CAN WE LEARN FROM A REINDEER'S NOSE?

Figure 11.2: Reindeer are superbly adapted to prevent thermal energy losses.
Reindeer live in some of the most difficult conditions on the planet. They need to cope with temperatures dropping to 50 C 50 C -50^(@)C-50^{\circ} \mathrm{C} and below.
Reindeer are adapted so they lose very little thermal energy from their bodies. They have two layers of fur: soft fur near the body, then a layer of guard hairs which are hollow tubes containing air. Air is an excellent insulator. The hairs are densely packed, insulating the body so well reindeer can sleep on snow without melting it.
The reindeer’s nasal passages serve as a complex thermal energy exchange system. They have a complex system of tubes which allow warm air being breathed out to pass over cold air being breathed in. Thermal energy is transferred to the incoming air so that the air leaving a reindeer’s body is cold, and the reindeer keeps the thermal energy. The warm air is cooled by about 21 C 21 C 21^(@)C21^{\circ} \mathrm{C} before leaving the reindeer’s nose.
The reindeer are so well insulated that they are in danger of overheating when running from predators such as wolves.
They lose thermal energy from their large tongues. Also they circulate more blood to their legs, which are less insulated so can give out the thermal energy.
It is important that a reindeer does not lose feeling in its nose while finding food in the snow. To keep the nose warm and sensitive, it has extra blood vessels. This means that when a reindeer is photographed with a temperature-sensitive camera, the nose seems to glow.
Figure 11.3: A reindeer’s nose has a complex thermal exchange system. Reindeer lose thermal energy through their large tongues.
Reindeer have other adaptations that make them suited to life in the Arctic. They need to stick together in blizzards, because they cannot see each other. Their feet make a clicking sound as tendons move across bones. This means they can hear each other. They also have a gland in the leg that leaves a scent on the ground, which helps them locate each other.
Reindeer’s eyes change in winter. In summer, they have a golden reflective layer at the back of the eye, like a cat’s eye. In winter, this layer turns dark blue, which means less light escapes and the eyes are more sensitive.
In this chapter you will study the ways in which thermal energy is transferred and how we can use this to our advantage.

Discussion questions

1 Human nasal passages also transfer thermal energy from exhaled air. Try breathing onto your hand, first from your mouth, then from your nose. You will notice the air from your nose is cooler. Discuss how we use this to help us cool down or to keep warm.
2 The reindeer’s nasal passages can be referred to as a heat energy exchange system. They exchange the thermal energy from the warm hair breathed out to the cold air breathed in. Thermal energy exchangers can reduce the amount of fossil fuels we use, helping
reduce the amount of energy we use and cutting costs. Discuss any other uses of thermal energy exchangers. These may be examples you have seen, or situations where you think they could be useful.

11.1 Conduction

As we discussed in Chapter 6, thermal energy transfers from a hotter place to a colder place, that is, from a higher temperature to a lower temperature. Thermal energy requires a temperature difference if it is to be transferred. In this chapter, we look at the various ways in which thermal energy is transferred. We start with thermal conduction.
Lying on the table are two spoons: one is metal, the other is wooden (Figure 11.4). You pick up the metal spoon - it feels cold. You pick up the wooden spoon - it feels warm. In fact, both are at the same temperature (room temperature) as a thermometer would prove to you.
Figure 11.4a: Touching a metal spoon. b: Touching a wooden spoon.
How can this be? What you are detecting is the fact that metal is a good conductor of thermal energy, and wood is a poor conductor of thermal energy.
When your finger touches a metal object, thermal energy is conducted out of your finger and into the metal. Because metal is a good thermal conductor, thermal energy spreads rapidly through the metal. Thermal energy continues to escape from your finger, leaving it colder than before. The temperature-sensitive nerves in your finger tip tell your brain that your finger is cold. So, you think you are touching something cold.
When you touch a wooden object, thermal energy conducts into the area that your finger is in direct contact with. However, because wood is a good thermal insulator, the thermal energy travels no further. Your finger loses no more thermal energy and remains warm. The message from the nerves in your finger tip is that your finger is warm. So, you think you are touching something warm.
Note that the nerves in your finger tell you how hot your finger is, not how hot the object that you are touching is! This is similar to our discussion of thermometers in Chapter 10. A thermometer in water indicates its own temperature, and we have to assume that the temperature of the water is the same as this.

EXPERIMENTAL SKILLS 11.1

Demonstrating conduction

In this series of experiments, you will investigate how materials conduct, and which are the best conductors. This information is vital for engineers and designers whose work involves planning how to keep temperatures safe and comfortable.
Safety: These experiments will involve heating materials to high temperatures. Plan carefully so that you can leave hot materials to cool safely. Wear safety goggles while using the Bunsen burner.
Carry out a risk assessment and make sure your teacher has checked it before beginning any practical work.

Getting started

Look at the apparatus in Figure 11.5. Explain how attaching paper clips using petroleum jelly will allow you to observe conduction.
Figure 11.5: Experimental set-up for part 1.
Part 1: How is thermal energy conducted along ametal bar?
You will need:
  • Bunsen burner
  • heatproof mat
  • clamp and stand
  • copper rod
  • paper clips
  • petroleum jelly.

Method

1 Use small blobs of petroleum jelly to attach paper clips along the copper rod, as shown in Figure 11.5.
2 Secure the rod in the clamp and heat the other end.
3 Watch carefully what happens to the paper clips.

Questions

1 Describe what happened to the paper clips.
2 What does this tell you about how thermal energy is conducted along the rod?
Part 2: Which metal is the best conductor?

You will need:

  • Bunsen burner
  • heatproof mat
  • tripod
  • rods made from different metals
  • paper clips
  • timer
  • petroleum jelly.

Method

1 Use small blobs of petroleum jelly to attach a paper clip to the end of each metal rod.
2 Place the rods on the tripod, as shown in Figure 11.6, and heat the ends.
3 Use a table like this to record the time taken for each paper clip to fall.
Figure 11.6: Experimental set-up for part 2.
Metal Time taken for clip to fall / seconds
Metal Time taken for clip to fall / seconds | Metal | Time taken for clip to fall / seconds | | :--- | :--- | | | | | | |

Questions

1 Write the metals in order from the best conductor to the worst conductor.
2 Do you think this was a fair test? Explain why or why not. Describe any possible sources of experimental error.

Part 3: Is water a good conductor of thermal energy?

You will need:

  • Bunsen burner
  • heatproof mat
  • boiling tube
  • small piece of wire gauze
  • test tube holder
  • cold water
  • ice cube.

Method

1 Fill three-quarters of the boiling tube with water. Add a small ice cube and then a small piece of wire gauze to hold the ice in place, as shown in Figure 11.7.
Figure 11.7: Experimental set-up for part 3.
2 Look at the boiling tube. Touch it to check the temperature.
3 Hold the tube with the test tube holder and gently heat the water at the top of the tube.
4 Watch carefully to see what happens to the water at the top of the tube and to the ice.

Questions

1 Describe your observations.
2 Explain why this experiment shows that water is a poor conductor of thermal energy.
Part 4: What materials make good insulators?
You will need:
  • 4 equal sized beakers with cardboard lids
  • 4 the rmometers, pushed through thecardboard lids
  • timer
  • a range of insulating materials
  • electric kettle.

Method

Figure 11.8: Experimental set-up for part 4.
1 Wrap three of the beakers in different insulating materials. Leave the fourth beaker without insulation.
2 Design a table to record your results.
3 Boil the kettle and carefully pour equal amounts of water into each of the beakers.
4 Measure the temperature of the water in each beaker every minute for ten minutes.

Questions

1 Why was the beaker with no insulation included in the experiment?
2 Draw a graph of your results. Plot temperature on the y y yy-axis against time on the x x xx-axis. Plot all four sets of results on the same axis.
3 Which insulator kept the water the hottest?
4 List the insulators in order from best to worst.

REFLECTION

Think about the experiments you have done and write a short sentence to sum up what you have learnt in each. What was the most powerful learning moment for you in these experiments? Think about why this was and how it helped your understanding.
Table 11.1 compares conductors and insulators. In general, metals are good conductors of thermal energy and non-metals are poor conductors. Air and water are very poor conductors of thermal energy.
Table 11.1: Comparing conductors of thermal energy, from the best conductors to the worst. A bad conductor is a good insulator. Almost all good conductors are metals; polymers (plastics) are at the bottom of the list. Glass wool is an excellent insulator because it is mostly air.

Explaining conduction in metals and non-metals

Both metals and non-metals conduct thermal energy. Metals are generally much better conductors than non-metals. We need different explanations of conduction for these two types of material.
We will start with non-metals. Imagine a long glass rod (Figure 11.9). One end is being heated, the other end is cold. This makes a temperature difference between the two ends, and so thermal energy flows along the rod. What is going on inside the rod?
Figure 11.9: Conduction of thermal energy in non-metals. A glass rod is heated at one end. Thermal energy travels from the hot end to the cold end.
Look at the particles inside the rod in Figure 11.9. At the hot end of the rod, the atoms are vibrating much more than they are at the cold end. As the atoms vibrate, they collide with their neighbours. This process results in each atom sharing its energy with its neighbouring atoms. Atoms with a lot of energy end up with less, and those with a little end up with more. The collisions gradually transfer energy from the atoms at the hot end to those at the cold end. Energy is steadily transferred down the rod, from the hot end to the cold end.
This is how poor conductors (such as glass, ice and plastic) conduct thermal energy. It is not a very efficient method of thermal energy transfer.
Metals are good conductors for another reason. Many of the electrons in metallic conductors are free to move (they are delocalised). These are the particles which carry electric current. They also carry thermal energy as they get hot and move through the metal. Figure 11.10 shows a copper rod, heated in the same way as the glass rod. Notice the free electrons, which carry thermal energy through the metal.
Figure 11.10: Conduction of thermal energy in metals. Metals have free electrons which carry thermal energy, making metals good conductors.
Liquids can also conduct thermal energy, because the particles of which they are made are in close contact with one another. However, as the particles are free to move, vibrations are not passed on as easily as in a solid. The particles in gases are very spread out, making gases very poor conductors of thermal energy.

Questions

1 Copy these sentences. Choose the correct word from each set of brackets.
Conduction happens mostly in {solids / liquids / gases}. Thermal energy flows from the {hotter / cooler} parts of an object to the {hotter / cooler} parts. A material which does not conduct thermal energy well is called { a { a {a\{\mathrm{a} conductor / an insulator / a resistor}. An example is {copper / polystyrene / gold}.
2 Explain why a wooden spoon is better than a metal one to stir a saucepan of hot soup.
3 Use the information in Table 11.1 to explain why walking on a marble floor in bare feet would feel colder than walking on a wooden floor.
4 Explain why two thin layers of clothing are often warmer than one thick layer.
5 Explain why:
a copper is a better conductor than wood
b wood is a better conductor than air.

11.2 Convection

As we have seen, liquids and gases are not usually good conductors of thermal energy. Thermal energy transfer in liquids and gases is mainly by another method known as convection.
‘Hot air rises’ is a popular saying - one of the things many adults remember from their science lessons. Figure 11.11 shows this science being put to use. When air is heated, it expands and so its density decreases. The warm air is less dense than its surroundings, so it floats upwards. For the balloon to rise, the density of the hot air, the balloon itself, plus the basket and its occupants, must altogether have a density less than that of the surrounding cold air.
Figure 11.11: The heater is heating the air going into the balloon. This makes the air expand so it becomes less dense and makes the balloon float up.
The opposite effect can be seen in Figure 11.12. Cold air sinks. This photo is taken with a technique which shows the flow of air. You can see cold air sinking down below the frozen pizza.
Figure 11.12: Cold air sinks below any object which is colder than its surroundings.
The rising of hot air is just one example of convection. Hot air can rise because air is a fluid, and convection can be observed in any fluid any (liquid or gas). Convection doesn’t happen in solids as the particles are in fixed positions so they cannot flow.
Figure 11.13 shows how a convection current can be observed in water. Above the flame, water is heated and expands. Now its density is less than that of the surrounding water, and it floats upwards. The purple dye shows how it moves. Colder water, which is more dense, flows in to replace it.
Figure 11.13: Because water is clear and colourless, it can be difficult to see how the water moves to form a convection current. Crystals of potassium manganate(VII) act as a purple dye to show the movement of the water.
A convection current is a movement of a fluid that carries energy from a warmer place to a cooler one. This highlights an important difference between convection and conduction.
  • In convection, energy is transferred through a material from a warmer place to a cooler place by the movement of the material itself.
  • In conduction, energy is transferred through a material from a warmer place to a cooler place without the material itself moving.

EXPERIMENTAL SKILLS 11.2

Demonstrating convection

Convection in air and water is difficult to see as air and water are transparent and colourless. These experiments will let you observe convection currents in water and air.
You will need:
  • Bunsen burner
  • heatproof mat
  • tripod
  • beaker
  • tweezers
  • thin card
  • scissors
  • thin thread
  • potassium manganate(VII) crystals.
Safety: Wear eye protection when using a Bunsen burner. Wear protective gloves when handling potassium manganate(VII). Do not hold the card spiral near a flame.

Getting started

What is the purpose of the potassium manganate(VII) crystal in this experiment?

Method part 1: Convection in a liquid

1 Fill three-quarters of a beaker with water.
2 When the water is settled, use tweezers to place two or three small crystals of potassium manganate(VII) on the bottom of the beaker, at one side. (See Figure 11.14.)
Figure 11.14: Set-up for experiment.
3 Use a Bunsen burner to heat the water gently, just below the crystals.
4 Observe what happens to the colour of the potassium manganate(VII) as it moves. This shows how the water is flowing.
Method part 2: Convection in air
Figure 11.15: Cutting a spiral and using it to demonstrate convection.
1 Cut a spiral from thin card.
2 Attach a thin thread to the centre of the spiral.
3 Hold the thread above a heat source (such as a radiator) and observe
how it moves.

Questions

1 Draw the beaker of water and arrows to show the movement of the purple colour when you heated it.
2 The experiment shown in Figure 11.16 uses smoke from burning paper to show convection currents in air. Explain why the smoke moves down one chimney and up the other.
Figure 11.16: Experiment with candle and burning paper.

Convection currents at work

Convection currents help to share energy between warm and cold places. If you are sitting in a room with an electric heater, thermal energy will be moving around the room from the heater as a result of convection currents, which rise from the heater. You are likely to be the source of convection currents yourself, since your body is usually warmer than your surroundings (see Figure 11.21).
Many biting insects make use of this effect. For example, bed bugs crawl across the bedroom ceiling. They can detect a sleeping person below by finding the warmest spot on the ceiling. Then they drop straight down on to the sleeper. This is a lot easier than crawling about on top of the bedding.
Figure 11.17: Convection currents rise above the warm objects in a room.
Cold objects also produce convection currents. You may have noticed cold water sinking below an ice cube in a drink. In a refrigerator, the freezing surface is usually positioned at the top and the back, so that cold air will sink to the bottom. Warm air rises to replace the cold air as it sinks. The warm air is then cooled. (see Figure 11.18).
Figure 11.18: In a refrigerator, cold air sinks from the freezing compartment. If the freezer was at the bottom, cold air would remain there, and the food at the top would not be cooled.

Explaining convection

We have already seen that convection results from the expansion of a fluid when it is heated. Expansion means an increase in volume while mass stays constant. This means that density decreases. A less dense material is lighter and so is pushed upwards by the surrounding denser material.
The particles in the hotter fluid have more kinetic energy so they move around faster. As they flow from place to place, they take this energy with them.
Convection is the main method of thermal energy transfer in fluids. Thermal energy can be conducted through a liquid but this is generally a slow process compared with convection.

Questions

6 Copy Figure 11.19. Add arrows labelled ‘cold dense water’ and ‘hot, less dense water’ to show the convection current in the pan.

heat
Figure 11.19: A pan of water.
7 The ice cube in Figure 11.20 floats at the top of the water. Draw and label a diagram to show how the ice cube creates a convection current which cools all of the water.
Figure 11.20: Ice in glass of water.
8 An inventor makes an electric kettle with the heating element at the top. Explain why it will not work.
9 a Draw diagrams to show the difference in the arrangement of particles in a hot gas and a cold gas.
b Use your diagrams to explain why hot gases rise. Use the words ‘expand’ and ‘density’ in your answer.
10 Explain why convection does not happen in solids.

11.3 Radiation

At night, when it is dark, you can see much further than during the day. In the daytime, the most distant object you are likely to be able to see is the Sun, about 150 million kilometres away. At night, you can see much further, to the distant stars. The most distant object visible to the naked eye is the Andromeda galaxy, about 20 million million million kilometres away.
The light that reaches us from the Sun and other stars travels to us through space in the form of electromagnetic radiation. This radiation travels as electromagnetic waves. It travels over vast distances, following a straight line through empty space. Radiation is the only form of thermal energy transfer which does not involve the movement of particles. Thermal radiation does not need a medium to travel through. It can travel through a vacuum. As well as light, the Earth receives other forms of electromagnetic radiation from the Sun, including infrared and ultraviolet radiation. (You will learn much more about electromagnetic radiation in Chapter 15.)
All objects emit infrared radiation. The hotter an object, the more infrared radiation it gives out. You can use this idea to help you to do a bit of detective work. Imagine you arrive at a crime scene. The suspect says he has just arrived by car. You suspect he has been at the scene for an hour, giving him time to commit the crime. Holding your hand over the engine compartment will quickly tell you if the engine is radiating thermal energy.
Our skin detects the infrared radiation produced by a hot object. Nerve cells just below the surface respond to thermal energy. You notice this if you are outdoors on a sunny day.
To summarise, infrared radiation:
  • is produced by warm or hot objects
  • is a form of electromagnetic radiation
  • travels through empty space (and through air) in the form of waves
  • travels in straight lines
  • warms the object that absorbs it
  • is invisible to the naked eye
  • can be detected by nerve cells in the skin.
Figure 11.21 shows another way of detecting infrared radiation: by using a temperature-sensitive camera. The photograph of a woman sitting at a desk is taken with a camera which detects infrared radiation instead of light. It is very sensitive to slight differences in temperature.
Figure 11.21: Using an infrared-sensitive camera. Slight variations in temperature show up as different colours. The scale shows how the colour relates to temperature.

Questions

11 Which statement about infrared radiation is true?
A Infrared radiation travels slower than light.
B Infrared radiation cannot be reflected.
C Infrared radiation can travel through a vacuum.
D Infrared radiation is transferred by the movement of particles.
12 Explain why thermal energy from the Sun can only reach us by radiation, not conduction or convection.
13 What evidence is there in this infrared photograph (Figure 11.22) to suggest that the car has only just broken down?
Figure 11.22: Infrared photograph of a car.

ACTIVITY 9.1

Thermal imaging

Figure 11.23: Thermal image of a fingerprint.
A science museum is preparing a display about infrared photography. You are a researcher and have been given the following brief.
Investigate one of the uses of this technology and prepare an information board about it. Your board should be one side of A4 paper. It must include at least one eye-catching picture taken with an infrared camera, and a maximum of 150 words describing the science and how it is used.
The board should be interesting to the general public and also detailed enough to be relevant to high school physics students.
Uses of infrared technology include:
  • medicine
  • military
  • detection of drug farms
  • astronomy
  • wildlife photography
  • forensic science
  • studying thermal energy loss from buildings.

PEER ASSESSMENT

Display each group’s work from Activity 9.1 around the classroom. Take a class vote on which board best meets the brief. Discuss what makes it a good piece of work. Adapt your board to add some of the features you identified.

Good absorbers, good emitters

On a hot, sunny day, car drivers may park their cars with a sunshield behind the windscreen (Figure 11.24). Such a sunshield is usually shiny, because this reflects light and infrared radation from the Sun. This stops the car getting uncomfortably hot. The black plastic parts of the car (such as the steering wheel and dashboard) are very good absorbers of infrared radaition, and they can become too hot to touch.
Figure 11.24: A sunshield reflects unwanted radiation, which would otherwise make the car very hot.
It is the surface that determines whether an object absorbs or reflects infrared radiation. A surface that is a good reflector is a poor absorber. On a hot day, you may have noticed how the black surface of a tarred (metalled) road emits
thermal energy. Black surfaces readily absorb infrared radiation. They are also good emitters.
You may also have noticed that you stay cooler on a hot day if you wear light coloured clothes. Light surfaces do not absorb much thermal radiation. Instead, they reflect the radiation. The house in Figure 11.25 is painted white to reflect infrared radiation.
Figure 11.25: This house is built from bottles filled with sand. It is painted white to reduce absorption of infrared radiation.
  • Shiny or white surfaces are the best reflectors (the worst absorbers).
  • Matte black surfaces are the best absorbers (the worst reflectors).
  • Matte black surfaces are the best emitters.

Questions

14 Explain why the worker in Figure 11.26 is wearing a shiny suit.
Figure 11.26: A worker supervising the flow of hot molten metal.
15 Which will stay hot longer: tea in a shiny silver teapot or tea in dark brown one? Explain your answer.

Factors affecting infrared radiation

All objects emit radiation and absorb radiation from their surroundings. The hotter an object is, the more radiation it emits each second, or the more power it radiates. Any object which is hotter than its surroundings radiates more energy per second than it absorbs and so will cool down. An object which is cooler than its surroundings absorbs more energy per second than it radiates until it reaches the temperature of its surroundings. An object with a temperature that remains constant absorbs thermal energy at the same rate as it emits thermal energy. An object with a large surface area emits thermal energy at faster rate.
Figure 11.27 shows three beakers of water. The room temperature is 20 C 20 C 20^(@)C20^{\circ} \mathrm{C}. All three beakers radiate and absorb energy, but the amount or radiation and absorption depends on the temperature.
Investigating the emission and absorption of infrared radiation
In this investigation you will use cans of water with different surfaces to investigate which emits most infrared radiation and which absorbs most infrared radiation.
The amount of infrared radiation emitted also depends on the surface area and temperature of the objects so these must be controlled in your experiment.

You will need:

  • 1 shiny silver can and 1 dull black can
  • 2 lids with holes for thermometers
  • 2 thermometers or electronic temperature probes
  • timer
  • kettle
  • board to isolate the cans from each other
  • Bunsen burner or electrical heater.
Safety: Take care when using hot water. Do not touch or move the cans while they are full of very hot water. Wear eye protection when using a Bunsen burner.

Getting started

Look at Figure 11.28. Use what you know about thermal energy transfers to explain why the cans must be fitted with lids, and why they should stand on a wooden or plastic surface.
Figure 11.28: Set-up for the experiment.
Experimen A: Which surface radiates better, black or shiny? Method
1 Set up the experiment as shown in Figure 11.28.
2 Fill the two cans with equal volumes of hot water.
3 Use thermometers or electronic temperature probes to measure the temperature of the water in each can every minute for ten minutes.

Questions

1 What features of the experimental design ensure that this is a fair test?
2 Plot a graph of temperature on the y y yy-axis against time on the x x xx-axis. Plot both sets of results on the same graph.
3 What can you conclude from your results?
Experiment B: Which surface absorbs thermal energy better, black or shiny?
Method
1 Set up the experiment as shown in Figure 11.29; use the same cans as in experiment A.
Figure 11.29: Set-up for the experiment.
2 Fill the cans with equal volumes of cold water.
3 Place the cans at equal distances from the Bunsen flame or electric heater.
4 Use thermometers or electronic temperature probes to measure the temperature of the water in each can every minute for ten minutes.

Questions

1 Plot a similar graph to experiment A.
2 What can you conclude from your results?

11.4 Consequences of thermal energy transfer

In this section, we will see how we can use ideas about thermal energy transfers to understand a lot of different situations. Remember that:
  • Thermal energy travels from a hotter place to a colder place. It is the temperature difference that makes it flow.
  • Conduction is the main way in which energy can pass through a solid. Energy travels through the solid but the solid itself cannot move.
  • Convection is the main way in which energy is transferred in a fluid. Warm fluid moves around, carrying energy with it.
  • Radiation is the only way in which thermal energy can travel through empty space. Infrared radiation can also pass through some transparent materials such as air.
Hot objects have a lot of internal energy. As we have seen, energy tends to escape from a hot object, spreading to its cooler surroundings by conduction, convection and radiation. This can be a great problem. We may use a lot of energy (and money) to heat our homes during cold weather, and the energy simply escapes. We eat food to supply the energy we need to keep our bodies warm, but energy escapes from us at a rate of about 100 watts ( 100 W = 100 J / s 100 W = 100 J / s 100W=100J//s100 \mathrm{~W}=100 \mathrm{~J} / \mathrm{s} ).
To keep energy in something that is hotter than its surroundings, we need to insulate it. Knowing about conduction, convection and radiation can help us to design effective insulation.
Remember that all three mechanisms of energy transfer (conduction, convection and radiation) may be involved when an object warms up or cools down.

Home insulation

A well-insulated house can avoid a lot of energy wastage during cold weather. Insulation can also help to prevent the house from becoming uncomfortably hot during warm weather. Figure 11.30 shows where thermal energy is lost from a house, and some ways to reduce thermal energy losses. More details are listed in Table 11.2.
Figure 11.30: Insulating a house reduces thermal energy loss and heating costs.
Method Why it works
thick curtains,
draught excluders
thick curtains, draught excluders| thick curtains, | | :--- | | draught excluders |
stops convection
currents, and so
prevents thermal
energy transfer
stops convection currents, and so prevents thermal energy transfer| stops convection | | :--- | | currents, and so | | prevents thermal | | energy transfer |
loft and underfloor
insulating
materials
loft and underfloor insulating materials| loft and underfloor | | :--- | | insulating | | materials |
prevents
conduction of
thermal energy
through floors and
ceilings
prevents conduction of thermal energy through floors and ceilings| prevents | | :--- | | conduction of | | thermal energy | | through floors and | | ceilings |
double and triple
glazing of
windows
double and triple glazing of windows| double and triple | | :--- | | glazing of | | windows |
vacuum between
glass panes cuts
out losses or gains
by conduction and
convection
vacuum between glass panes cuts out losses or gains by conduction and convection| vacuum between | | :--- | | glass panes cuts | | out losses or gains | | by conduction and | | convection |
Method Why it works "thick curtains, draught excluders" "stops convection currents, and so prevents thermal energy transfer" "loft and underfloor insulating materials" "prevents conduction of thermal energy through floors and ceilings" "double and triple glazing of windows" "vacuum between glass panes cuts out losses or gains by conduction and convection" | Method | | Why it works | | | :--- | :--- | :--- | :--- | | | thick curtains, <br> draught excluders | stops convection <br> currents, and so <br> prevents thermal <br> energy transfer | | | | | loft and underfloor <br> insulating <br> materials | prevents <br> conduction of <br> thermal energy <br> through floors and <br> ceilings | | | double and triple <br> glazing of <br> windows | vacuum between <br> glass panes cuts <br> out losses or gains <br> by conduction and <br> convection | |
| | | :--- | :--- | :--- | :--- |
Table 11.2: Methods of retaining energy in a house in a cold climate, and of keeping a house cool in a hot climate.
Double-glazed windows usually have a vacuum between the two panes of glass. This means that energy can only escape by radiation, since conduction and convection both require a material. Modern houses are often built with cavity walls, with an air gap between the two layers of bricks. It is impossible to have a vacuum in the cavity, and convection currents can transfer energy across the gap (see Figure 11.31a). Filling the cavity with foam means that a small amount of energy is lost by conduction, although the foam is a very poor conductor. However, this does stop convection currents from flowing (Figure 11.31b), so there is an overall benefit.
Figure 11.31a: A cavity wall reduces thermal energy loss by conduction because air is a good insulator. However, a convection current can transfer energy from the inner wall to the outer wall. b: Filling the cavity with foam or mineral (glass or rock) wool prevents convection currents from forming.

Keeping cool

Vacuum flasks are used to keep hot drinks hot. They can also be used to keep cold drinks cold. Giant vacuum flasks are used to store liquid nitrogen and helium at very low temperatures, ready for use in such applications as body scanners in hospitals. They also have medical uses, such as for storing frozen embryos for IVF treatment.
Figure 11.32 shows the construction of a vacuum flask. Glass is generally used, because glass is a good insulator. However, some flasks are made of steel for added strength. Air is removed from the gap between the double walls, creating a vacuum. This reduces losses by conduction and convection because both of them need a material to travel through. The silver coating on the glass reduces losses by radiation by reflecting any infrared radiation. The stopper is made of plastic and it prevents losses by convection and evaporation.
Figure 11.32: A vacuum flask is cleverly designed to keep hot things hot by reducing thermal energy losses. It also keeps cold things cold. Although we might say ‘it stops the cold getting out’, it is more correct to say that it prevents thermal energy from getting in. The first such flask was designed by James Dewar, a Scottish physicist, in the 1870s. He needed flasks to store liquefied air and other gases at temperatures as low as 200 C 200 C -200^(@)C-200^{\circ} \mathrm{C}. Soon after, people realised that a flask like this was also useful for taking hot or cold drinks on a picnic.
Figure 11.33: This couple are hoping to have a baby by IVF. The flask they are holding contains their frozen embryos.

ACTIVITY 11.2

Marketing a vacuum flask

Imagine you have just invented the vacuum flask. You want to go into business producing and selling flasks but need finance. Prepare a presentation for potential investors. You will need to explain the technical details of how it works. You must use the words ‘conduction’, ‘convection’ and ‘radiation’. The investors want to know you are an expert and understand the science. You should also explain why your product will be popular with customers.
Your presentation can include an information leaflet, a video or an advertising poster.
A car engine burns fuel and so gets very hot. The cooling system (Figure 11.34a) transfers some of this thermal energy to the surroundings so the engine does not
overheat. This system uses many of the things you have learnt:
Figure 11.34a: Transfer of thermal energy in a car cooling system. b: The black metal radiator cooler fins in a car.
  • Specific heat capacity: water flows around the block to absorb thermal energy. Water is a good choice as it has a very high specific heat capacity.
  • Convection: as the water is heated, a convection current flows in the direction shown by the arrows. The pump is used to speed up this flow.
  • Conduction: the radiator has metal fins (Figure 11.34b) so the thermal energy is conducted to all parts of the radiator.
  • Radiation: the fins have a large surface area and are black to increase the rate of thermal energy radiation.
Wood burns on a bonfire to produce thermal energy. The heating effect you experience when sitting by a fire is almost entirely due to radiation. Air is a bad conductor and thermal energy transfer due to convection will heat the air above the fire.
Figure 11.35: The older child will feel hotter than the younger child as his black T-shirt absorbs infrared radiation whereas the white T-shirt reflects the infrared radiation.

Thermal energy transfer, climate and weather

Radiation from the Sun is essential for life on Earth. The Sun’s radiation warms the Earth. The warm Earth emits some infrared radiation. Gases in the Earth’s atmosphere, such as carbon dioxide, absorb some of this thermal energy and this warms our atmosphere. This is the greenhouse effect (Figure 11.36) and without it life on Earth would be impossible. However, the amount of greenhouse gases in the atmosphere is increasing, trapping more thermal energy. This means that Earth and its atmosphere are absorbing more infrared radiation than they emit. This is the cause of global warming.
Figure 11.36: For the Earth to maintain a constant temperature, infrared radiation must be emitted at the same rate as it is absorbed. Upsetting this balance is causing global warming.
Convection currents explain the origins of winds and ocean currents, which are two of the major factors that control climate patterns around the world. For example, warm air rises above the Equator, and colder air sinks in subtropical areas. This creates the pattern of trade winds that are experienced in the tropics.
Ocean currents (Figure 11.37) help to spread thermal energy from equatorial regions to cooler parts of the Earth’s surface. Warm water at the surface of the sea flows towards the poles. In polar regions, colder water sinks and flows back towards the Equator. Provided this pattern remains constant, this helps to make temperate regions of the world more habitable. However, there is evidence that the pattern of ocean currents is changing, perhaps as a consequence of global warming.
Figure 11.37: Convection currents in the oceans flow in predictable directions. The red lines show warm surface water and the blue lines show the flow of the deeper, colder water. The warm currents affect the climate. For example, the British Isles are warmed by the Gulf steam (a flow of warm water from the Gulf of Mexico).

Questions

16 In a rolling mill, iron is heated to make it malleable and it is then passed through rollers to produce thin sheets of the metal. Explain how the following become hot in this process:
a the rollers which press the metal
b the face of a worker
c the air in the building.
17 The coat in Figure 11.38 is designed for a cold climate.
Figure 11.38:
Describe the features of the coat which prevent thermal energy loss by:
a conduction
b convection
c radiation.
18 Figure 11.39 shows a solar water heater. Cold water flows through the pipes and is heated by the Sun.
Figure 11.39: A solar water heater.
Suggest reasons why:
a the inside of the panel is painted black
b the back of the panel is insulated
c the cold water enters at the bottom of the panel, and leaves at the top.

PROJECT

Be the teacher

A student has submitted their answers to a homework question. As you will see, they have not understood the topic. Your task is to:
  • Mark their work and identify any mistakes. Is there anything that shows they have been listening in class? Write brief encouraging feedback notes so the student will be willing to try again. Discuss with your group where the student is going wrong and what you might do to help them avoid making this mistake again.
  • Write a model answer. Each member of the group should do this individually, then as a group, share your ideas and make sure you have the best possible answer.
  • Plan a five minute revision session in which you explain the main points the student has missed. You can do this using presentation software, a short video, an illustrated talk or you could demonstrate some experiments (check with your teacher). Include study hints on how to remember key points as well as the factual information. Make your presentation as lively and relevant as you can.
  • Deliver this revision session to another group of students.
  • Prepare a follow up worksheet, with answers, to check that the students have understood your presentation.

Homework questions

1 The photograph shows a reusable insulated coffee cup.
Explain how the cup reduces thermal energy loss by conduction, convection and radiation.
2 A student heats some ice. She records the temperature and plots a graph.
Describe what is happening at each stage of the graph and identify any important temperatures.

Student's answers

1 The cup keeps the coffee hot because it is made of metal and metal is a good insulator so no thermal energy transfers through the metal walls. The silver sides stop it cooling by radiation because any thermal
energy from the Sun will be reflected away and will not affect the coffee. Silver surfaces reflect infrared radiation.
There is no thermal energy loss due to convection because the cup is solid. This means its particles are fixed in place and so convection currents can’t happen.
2 A-B the ice is melting; B-C she has stopped heating the ice; C-D the water is getting hotter; D-E she has stopped heating it again; E-F the water is getting really hot. There are no numbers on the graph so I can’t identify any temperatures.

PEER ASSESSMENT

Once your peers have completed your worksheet, read their answers and give them feedback. Comment on how well they use scientific terms, and how clear their explanations are.

REFLECTION

What problems did you encounter while working on this project?
Did playing the part of the teacher tell you anything about the way you learn?

SUMMARY

Metals are good thermal conductors. Most non-metals are good insulators.
Metals are good thermal conductors because they have free electrons.
Hot fluids are less dense than cold fluids. This causes convection currents.
Infrared radiation transfers thermal energy using electromagnetic waves.
Infrared radiation does not require a medium (it can travel through a vacuum).
Shiny, white surfaces reflect infrared radiation. They are poor emitters and absorbers of infrared radiation.
Dull, black surfaces are good absorbers and emitters of infrared radiation, but poor reflectors.
The amount of infrared radiation emitted also depends on the surface area and temperature of the object.

EXAM-STYLE QUESTIONS

1 Which word describes a material that does not let thermal energy pass through it?
A conductor
B vacuum
C resistor
D insulator
2 Which states of matter can convention happen in?
A solids and liquids
B liquids and gases
C gases and solids
D solids, liquids and gases
3 Hot fluids rise. This can be explained because, as the fluid gets hotter, there is a decrease in the fluid’s:
A mass
B temperature
C density
D volume
4 This diagram shows an electric water heater.

a The copper wall is hot to touch. Name the process by which thermal energy from the water passes through the wall.
b Describe how a heater at the bottom of the tank heats all the water in the tank.
c The hot walls transfer thermal energy to the surroundings. Suggest a way this thermal energy loss could be reduced.
[Total: 5]
5 a Suggest a question which can be answered using the apparatus in the diagram.

b Explain which rod you would expect the drawing pin to fall from first.
c Describe two features of the experiment which make it a fair test.
[2]
[Total: 5]
6 The diagram shows a Leslie cube. It is a metal box and each side has a diff erent surface.
Success groups (S S 4 6 N ) S 4 6 N {:S_(4)^(6)N)\left.S_{4}^{6} N\right)
Saraya uses a Leslie cube to investigate infrared radiation. She fills the cube with boiling water so all sides are at the same temperature. She uses the infrared detector to investigate the radiation from each surface.
a Why is it important to keep the detector the same distance from each side?
b Match the surfaces to the temperatures. The first one has been done for you.
shiny black 64.5 C 64.5 C 64.5^(@)C64.5^{\circ} \mathrm{C}
matte black 71.2 C 71.2 C 71.2^(@)C71.2^{\circ} \mathrm{C}
shiny silver 60.4 C 60.4 C 60.4^(@)C60.4^{\circ} \mathrm{C}
matte white 65.3 C 65.3 C 65.3^(@)C65.3^{\circ} \mathrm{C}
shiny black 64.5^(@)C matte black 71.2^(@)C shiny silver 60.4^(@)C matte white 65.3^(@)C| shiny black | $64.5^{\circ} \mathrm{C}$ | | :--- | ---: | | matte black | $71.2^{\circ} \mathrm{C}$ | | shiny silver | $60.4^{\circ} \mathrm{C}$ | | matte white | $65.3^{\circ} \mathrm{C}$ |
c Saraya’s infrared detector gave the temperature to the nearest 0.1 C 0.1 C 0.1^(@)C0.1^{\circ} \mathrm{C}.
Zain repeats the experiment using a different infrared detector which gives the temperature to the nearest degree.
Suggest how Zain’s conclusion will differ from Saraya’s.
7 Some double glazed windows have a plastic frame and two panes of glass with a vacuum between them.
a Explain why plastic is used for the frame.
b Name the types of thermal energy transfer which are stopped by the vacuum.
c Double glazing reduces thermal energy losses from a house. Describe two other ways to reduce thermal energy losses and state the type of thermal energy transfer which each reduces.
8 Two beakers are filled with water at 20 C 20 C 20^(@)C20^{\circ} \mathrm{C}. Beaker A is placed outside in the snow at 5 C 5 C -5^(@)C-5^{\circ} \mathrm{C}. Beaker B is placed in a room at 25 C 25 C 25^(@)C25^{\circ} \mathrm{C}. The temperature of each beaker is taken half an hour later.
a State what you would expect to happen to the temperatures of the two beakers.
b Explain your answer in terms of the energy radiated or absorbed by the water in each beaker.
c A third beaker, also filled with water at 20 C 20 C 20^(@)C20^{\circ} \mathrm{C} remains at this temperature. State and explain what you can deduce from this about where the third beaker has been placed…

SELF-EVALUATION CHECKLIST

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
I can
See
Topic...
See Topic...| See | | :--- | | Topic... |
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
Almost
there
Almost there| Almost | | :--- | | there |
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |
Describe experiments to show the
properties of conductors and
insulators.
Describe experiments to show the properties of conductors and insulators.| Describe experiments to show the | | :--- | | properties of conductors and | | insulators. |
11.1
Explain thermal conduction in metallic
conductors in terms of free electrons.
Explain thermal conduction in metallic conductors in terms of free electrons.| Explain thermal conduction in metallic | | :--- | | conductors in terms of free electrons. |
11.1
State that thermal energy is transferred
by convection in fluids.
State that thermal energy is transferred by convection in fluids.| State that thermal energy is transferred | | :--- | | by convection in fluids. |
11.2
Explain how changes in density cause
convection currents.
Explain how changes in density cause convection currents.| Explain how changes in density cause | | :--- | | convection currents. |
11.2
Describe experiments to demonstrate
convection.
Describe experiments to demonstrate convection.| Describe experiments to demonstrate | | :--- | | convection. |
11.2
Describe experiments to demonstrate
convection.
Describe experiments to demonstrate convection.| Describe experiments to demonstrate | | :--- | | convection. |
11.2
State that infrared radiation is a part of
the electromagnetic spectrum and can
pass through a vacuum.
State that infrared radiation is a part of the electromagnetic spectrum and can pass through a vacuum.| State that infrared radiation is a part of | | :--- | | the electromagnetic spectrum and can | | pass through a vacuum. |
11.3
Describe how the surface of an object
affects how it emits, reflects or
absorbs infrared radiation.
Describe how the surface of an object affects how it emits, reflects or absorbs infrared radiation.| Describe how the surface of an object | | :--- | | affects how it emits, reflects or | | absorbs infrared radiation. |
11.3
Describe experiments to show which
surfaces absorb and emit most infrared
radiation.
Describe experiments to show which surfaces absorb and emit most infrared radiation.| Describe experiments to show which | | :--- | | surfaces absorb and emit most infrared | | radiation. |
11.3
Describe and explain some practical
applications of thermal energy
transfer.
Describe and explain some practical applications of thermal energy transfer.| Describe and explain some practical | | :--- | | applications of thermal energy | | transfer. |
11.4
I can "See Topic..." "Needs more work" "Almost there" "Confident to move on" "Describe experiments to show the properties of conductors and insulators." 11.1 "Explain thermal conduction in metallic conductors in terms of free electrons." 11.1 "State that thermal energy is transferred by convection in fluids." 11.2 "Explain how changes in density cause convection currents." 11.2 "Describe experiments to demonstrate convection." 11.2 "Describe experiments to demonstrate convection." 11.2 "State that infrared radiation is a part of the electromagnetic spectrum and can pass through a vacuum." 11.3 "Describe how the surface of an object affects how it emits, reflects or absorbs infrared radiation." 11.3 "Describe experiments to show which surfaces absorb and emit most infrared radiation." 11.3 "Describe and explain some practical applications of thermal energy transfer." 11.4 | I can | See <br> Topic... | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :--- | :--- | :--- | :--- | | Describe experiments to show the <br> properties of conductors and <br> insulators. | 11.1 | | | | | Explain thermal conduction in metallic <br> conductors in terms of free electrons. | 11.1 | | | | | State that thermal energy is transferred <br> by convection in fluids. | 11.2 | | | | | Explain how changes in density cause <br> convection currents. | 11.2 | | | | | Describe experiments to demonstrate <br> convection. | 11.2 | | | | | Describe experiments to demonstrate <br> convection. | 11.2 | | | | | State that infrared radiation is a part of <br> the electromagnetic spectrum and can <br> pass through a vacuum. | 11.3 | | | | | Describe how the surface of an object <br> affects how it emits, reflects or <br> absorbs infrared radiation. | 11.3 | | | | | Describe experiments to show which <br> surfaces absorb and emit most infrared <br> radiation. | 11.3 | | | | | Describe and explain some practical <br> applications of thermal energy <br> transfer. | 11.4 | | | |

Chapter 12

Sound

In this chapter you will:

  • describe how sounds are produced and how they travel
  • measure the speed of sound
  • describe how the amplitude and frequency of a sound wave are linked to its loudness and pitch
  • state the range of human hearing
  • define the term ‘ultrasound’ and describe some of its applications.

Getting started

Draw a table like this. Your table should fill a piece of A4 paper.
How are sounds
made?
How are sounds made?| How are sounds | | :--- | | made? |
How does sound travel? How fast is sound?
"How are sounds made?" How does sound travel? How fast is sound? | How are sounds <br> made? | How does sound travel? | How fast is sound? | | :--- | :--- | :--- | | | | |
How do we detect
sound?
How do we detect sound?| How do we detect | | :--- | | sound? |
How do sounds differ
from each other?
How do sounds differ from each other?| How do sounds differ | | :--- | | from each other? |
Are there sounds we
cannot hear?
Are there sounds we cannot hear?| Are there sounds we | | :--- | | cannot hear? |
"How do we detect sound?" "How do sounds differ from each other?" "Are there sounds we cannot hear?"| How do we detect <br> sound? | How do sounds differ <br> from each other? | Are there sounds we <br> cannot hear? | | :--- | :--- | :--- |
Write or draw something in each cell of the table. Make it as detailed as you can.
Pair up with another student and compare your tables. Make any additions or changes you want to based on your discussion.
Join with another pair and share your ideas. Again, make any changes.

THE SOUND OF SILENCE

Figure 12.1: Outer space is a vacuum. This means sounds cannot travel through it. Astronauts communicate by using radio waves.
We are surrounded by sounds. Natural sounds include birds, animals and the wind. Other sounds include cars, computers and musical instruments. Sounds wake us, sounds alert us that our food is cooked, our plane is ready to board or that a car is approaching. It is hard for most of us to imagine a world without sound. How would we communicate our emotions? How would a baby attract attention when it is hungry? How different would our forms of entertainment be?
Our world is increasingly noisy as we add more machines and technology.
Silence is not something city dwellers hear often. Psychologists believe noise is a major cause of stress. Acoustic engineers study ways to make a better sound environment. Mobile phone noise is alleviated by using vibration rather than sound to inform us of a call.
It is important for our wellbeing to reduce unnecessary noise, but sound reduction is not always entirely positive. Silent electric cars raise fears of more accidents as people are unaware of their approach.
In this chapter we will look at how sounds are created, how they travel and how we hear sounds. We will also look at why sounds can be so varied. Why are some sounds high or low pitched? Why are some sounds loud and others quiet?

Discussion questions

1 Evelyn Glennie (Figure 12.2) is one of the world’s top percussionists despite being profoundly deaf. Can you work out how she experiences sound? (Hint: she plays barefoot.)
Figure 12.2: Some people are born or become deaf, or they are only able to hear a limited range of sound. They learn how to navigate the world using other senses.
2 Discuss with a partner all the ways in which you have used sound today. What adaptations could you make to get through your daily
activities if you could not hear sounds?

12.1 Making sounds

All sounds are caused by something vibrating. The vibrations are not always visible because they may be too small or too fast. Vibrating sources cause the air around them to vibrate. These vibrations are passed through the air to our ears where they cause the eardrum to vibrate and we hear sound.
Figure 12.3: The water particles on this speaker are vibrating as the speaker makes a sound. The vibrations of the speaker and the water will change if the pitch or loudness of the sound is changed.
Figures 12.4-12.7 link sounds to the vibrating sources which cause them.
Figure 12.4: Hitting the tuning fork causes the prongs to vibrate.
Figure 12.5: Hitting the gong with a hammer causes it to vibrate.
Figure 12.6: The cicada has ribbed membranes at the base of its abdomen which vibrate causing a particularly loud sound.
Figure 12.7: Vocal folds in the human throat vibrate to create speech.

Musical sounds

Musical instruments are engineered to vibrate in ways that make a range of interesting and beautiful sounds.
Figure 12.8: Djembe drums are made in a range of sizes to produce different notes. The skin vibrates when hit and produces sound waves.
String instruments are plucked or bowed to cause them to vibrate. The length of the string can be changed (usually by holding the string down with a finger) and this changes the note produced. The body of the instrument and the air inside it also vibrate and this gives the instrument its distinctive sound. This is why an oud and a violin can play the same note but sound very different.
Figure 12.9: The musician’s left hand varies the length of the strings whilst his right plucks the strings to cause vibrations.
Wind instruments are blown to cause the column of air inside to vibrate. Players cover and uncover holes to change the length of the air column. This changes the pitch of the note.
Figure 12.10: The system of holes on a flute allows a range of notes to be played.
An orchestra produces vibrations by a complex combination of hitting, plucking, bowing and blowing. These vibrations pass through the air to the audience, causing their eardrums to vibrate, so they hear the sound.
If the sound is very loud, such as at a rock concert, the audience may feel vibrations throughout their bodies and through the floor.

Questions

1 What are all sounds caused by?
2 A drum, a flute and a violin all play a note. For each instrument state what vibrates to create the sound.
3 Describe how a drummer hitting a drum leads to a listener hearing sound.

12.2 How does sound travel?

Sound is a series of vibrations passing through air or another material. The source of the sound vibrates and this makes the air particles around it vibrate back and forward in the direction the sound is travelling. These vibrations make a sound wave. This type of wave is called a longitudinal wave.
You will learn more about waves in Chapter 14.
The movement of the air particles can be demonstrated using a slinky spring, as shown in Figure 12.11.
Figure 12.11: Moving the end of the spring back and forwards causes the coils of the spring to be squashed together and then stretched out. The squashed up area moves along the spring but the spring remains in its original place.

More about sound waves

When a tuning fork is hit, it the prongs start to vibrate. Figure 12.12 shows how this creates a sound wave. As the right-hand prong moves forward, it squashes the air particles together. This makes a dense region in the air called a compression. As it moves back, the air particles are more spread out, making a less dense region called a rarefaction. Air does not move from the tuning fork to the listener - the vibrations pass through the air by the patterns of compressions and rarefactions caused by the air molecules being pushed back and forth.
Figure 12.12: The tuning fork’s prongs move back and forth creating compressions and rarefactions in the air.
Figure 12.13:The number on the tuning fork tells us that the fork vibrates 440 times a second.

What can sound travel through?

Sound waves are vibrations caused by particles moving back and forth. In a vacuum there are no particles, so it is impossible for sound to travel.
Sound needs a medium (material) to travel through. This can be shown using a bell in a glass jar. Figure 12.14 shows an experiment to demonstrate this.
Figure 12.14: When the battery is connected, the bell can be seen and heard. Vibrations from the bell pass through the air in the jar, through the glass and then through the air to your ear. When the pump removes the air from the jar, the bell can still be seen vibrating, but cannot be heard.
Sound can travel through solids. You may be able to hear sounds from outside the classroom as you read this. Sound vibrations can travel through the walls. Sounds can also pass through liquids. Many sea animals such as dolphins and whales use sound to communicate with each other and to navigate.
Figure 12.15: This is a solar flare, where a huge amount of matter is emitted by the Sun. An explosion like this on Earth would create a deafening sound.
The Sun is very active and yet we hear no sound from it. This is because there are no particles to carry the huge disturbances caused by explosions and solar flares. We can see the Sun because, unlike sound waves, light waves can travel through a vacuum.
Another major difference between sound and light waves is their speed. Light travels at 300000000 m / s 300000000 m / s 300000000m//s300000000 \mathrm{~m} / \mathrm{s}, about a million times faster than sound. This means we see the lightning almost as it happens but hear the sound later. To calculate how many kilometres away the storm is, measure the time between the lightning and the thunder and divide by three. This works because sound travels 1 km in about 3 seconds.

Questions

4 Explain why sound cannot be heard in a vacuum.
5 Describe and explain what is heard when the vacuum pump in figure 12.14 is switched on.
6 A boy sees lightning and hears the thunderclap 9 seconds later. Calculate how far away is the storm is.

12.3 The speed of sound

Sound travels at between 330 m / s 330 m / s 330m//s330 \mathrm{~m} / \mathrm{s} and 350 m / s 350 m / s 350m//s350 \mathrm{~m} / \mathrm{s} in air. The speed changes slightly depending on the temperature and humidity of the air. This is much slower than light, but still so fast that we are usually not aware of the time it takes for sounds to reach us, unless the distance it travels is large.
An echo is a reflected sound wave. If you bang two wooden blocks together in the classroom you will hear only one bang. The sound will reflect from the walls, but the echo will be so close to the original bang that you will not hear it as a separate sound. Banging the blocks outside will mean the sound has further to travel so you may hear an echo. This can be used to measure the speed of sound.

EXPERIMENTAL SKILLS 12.1

Measuring the speed of sound in air

You can calculate the speed of sound in air by measuring the time taken for an echo to be heard.
You will need:
  • 2 wooden blocks
  • stopwatch
  • long tape measure or trundle wheel.
Safety: You will be creating very loud sounds. Avoid doing this near anyone’s ears as loud sounds can damage the ear.

Getting started

Why is it necessary to stand a long distance from the wall?
Why will it be difficult to measure the time from hitting the blocks together to hearing the echo?

Method

Figure 12.16: Measuring the speed of sound in air.
1 Stand a measured distance (ideally at least 50 metres) from a large flat wall. Bang the blocks together and listen for the echo.
2 Now try to bang the blocks in an even rhythm so that each clap coincides with the echo of the previous bang. This will mean you don’t hear the echo separately from the next bang. This may take some practice.
3 Your partner should then measure the time for 20 of your bangs.
4 Record the time taken for 20 bangs and the distance from the wall.
5 Calculate the total distance travelled by the sound. Each bang involves the sound travelling to the wall and back, so the total distance is 20 × 2 20 × 2 20 xx220 \times 2 × × xx\times distance from wall.
6 Calculate the speed of sound in air. Use the equation:
speed = distance travelled time taken =  distance travelled   time taken  =(" distance travelled ")/(" time taken ")=\frac{\text { distance travelled }}{\text { time taken }}

Questions

1 What is the actual speed of sound in air? How does the value you calculated compare to this?
2 Which measurement (distance or time) do you think was least

REFLECTION

Discuss the experiment with a partner. Suggest ways to improve your experiment to increase the accuracy of this measurement.
A more precise value of the speed of sound can be obtained using an electronic timer and microphones. Figure 12.17 shows this method. The wooden blocks and the two microphones are arranged in a straight line.
Figure 12.17: Another method for measuring the speed of sound.
When the students bangs the two blocks of wood together, it creates a sudden, loud sound. The sound reaches microphone 1 and a pulse of electric current is sent to the timer. The timer starts timing. A fraction of a second later the sound reaches microphone 2 .
A second pulse of current is sent to the timer and stops it. The timer now
indicates the time it took for the sound to travel from microphone 1 to microphone 2. If the distance between the two microphones is measured, the speed of sound can be calculated using the equation:
speed = distance time =  distance   time  =(" distance ")/(" time ")=\frac{\text { distance }}{\text { time }}

WORKED EXAMPLE 12.1

A man blows a whistle and hears the echo from a rock face after 3.6 seconds. How far away from the rock is he? Assume speed of sound in air = 340 m / s = 340 m / s =340m//s=340 \mathrm{~m} / \mathrm{s}.
Step 1: Calculate the distance travelled by the sound.
speed = distance time so, distance = speed × time = 340 m / s × 3.6 s = 1244 m  speed  =  distance   time   so, distance  =  speed  ×  time  = 340 m / s × 3.6 s = 1244 m {:[" speed "=(" distance ")/(" time ")],[{:[" so, distance "=" speed "xx" time "],[=340m//sxx3.6s],[=1244m]:}]:}\begin{aligned} & \text { speed }=\frac{\text { distance }}{\text { time }} \\ & \begin{aligned} \text { so, distance } & =\text { speed } \times \text { time } \\ & =340 \mathrm{~m} / \mathrm{s} \times 3.6 \mathrm{~s} \\ & =1244 \mathrm{~m} \end{aligned} \end{aligned}
Step 2: Halve this distance. (The distance you have already calculated is the total distance travelled by the sound, to the rock face and back. The distance to the rock face is half of this.)
1224 2 = 612 m 1224 2 = 612 m (1224)/(2)=612m\frac{1224}{2}=612 \mathrm{~m}

Answer

The rock face is 612 metres away from the man.
Sound travels at different speeds in different materials. Figure 12.18 shows the speed in different materials. This can be explained by considering the spacing of particles. Particles are closer together in solids than in liquids, so the vibrations can be passed on more easily in solids. This means that the sound wave travels faster in a solid.
Figure 12.18: The speed of sound in different materials.

Questions

7 A spectator at a cricket match sees the batsman hit the ball, then 1.2 seconds later he hears the strike. How far away is the spectator? The speed of sound in air is 330 m / s 330 m / s 330m//s330 \mathrm{~m} / \mathrm{s}.
8 Sound travels at 1500 m / s 1500 m / s 1500m//s1500 \mathrm{~m} / \mathrm{s} in fresh water and at 1530 m / s 1530 m / s 1530m//s1530 \mathrm{~m} / \mathrm{s} in salt water. Explain the difference in speeds.
9 Explain why the method shown in Figure 12.17 is more accurate than the echo method when measuring the speed of sound.

12.4 Seeing and hearing sounds

Seeing sounds

A cathode ray oscilloscope and microphone can be used to represent sounds on a display screen (Figure 12.19). The microphone picks up the sound and converts it to an electrical signal. The oscilloscope converts this to a line which represents the vibrations that make up the sound wave.
Figure 12.19: As the student blows, he creates vibrations in the air. The vibrations are detected by the microphone and converted to electrical signals, which are displayed on the oscilloscope.
The vibration from a musical instrument is complicated because it is produced by vibrations of the air and the instrument itself. A signal generator can be used to produce a pure sound wave. Pure notes are easier to measure, but not so musical.
The oscilloscope trace that represents a pure note is a simple curve as shown in Figure 12.20a. When representing a musical note from a particular musical instrument, the pattern is more complicated. Figures 12.20 b and c show this.
Figure 12.20: The three representations of sound waves shown here are all the same note. Each wave has a repeating pattern. The design of the instrument adds extra vibrations called overtones which give each instrument its distinctive sound. All three waves have four repeats, meaning that they all are the same note.
The oscilloscope can be used to observe two important things about the wave (Figure 12.21).
Figure 12.21: The frequency of the note can be calculated from the oscilloscope trace. This oscilloscope is set at 20 ms / 20 ms / 20ms//20 \mathrm{~ms} / div. This means each division on the grid represents 20 ms ( 0.02 s ) 20 ms ( 0.02 s ) 20ms(0.02s)20 \mathrm{~ms}(0.02 \mathrm{~s}). The time for one wave (marked as T T TT ) is two divisions or 0.04 seconds. One wave takes 0.04 seconds, so the number of waves each second is 1 ÷ 0.04 = 25 Hz 1 ÷ 0.04 = 25 Hz 1-:0.04=25Hz1 \div 0.04=25 \mathrm{~Hz}.
  • The amplitude is the furthest distance the particles move from their undisturbed position. This is shown by the height or depth of the oscilloscope trace.
  • The frequency is the number of vibrations each second. The more waves on the screen, the higher the frequency. Frequency is measured in hertz.
    One hertz means one wave per second.
    Connecting a signal generator and a speaker to an oscilloscope allows you to both see a representation of the sound wave and hear the sound.
Increasing the loudness of the sound produces taller waves - they have a bigger amplitude. Increasing the frequency means more waves will be seen on the screen - they have a higher frequency, and the sound heard will be higher pitched. Remember:
  • high frequency means high pitch
  • large amplitude means loud sound.

Hearing sounds

Young humans can hear sounds from 20 Hz up to 20000 Hz . As we grow older, the sensory cells in the ear which detect vibrations deteriorate. This means that the range of sounds which can be heard decreases with age. These cells can also be damaged by repeated exposure to very loud noise.
Sounds which have a higher frequency than 20000 Hz are too high pitched to be heard by the human ear. These sounds are known as ulttrasound.
Many animals can hear high pitched sounds that we cannot. Many animals such as dolphins communicate using ultrasound. Whistles creating ultrasound can be used to train animals (Figure 12.22).
Figure 12.22: This whistle was invented by a scientist called Francis Galton in around 1900 to help him investigate human hearing. Similar whistles are used in animal training.

ACTIVITY 12.1

An annoying noise
The mosquito sound alarm is a device intended to stop young people congregating in areas where they are not wanted. The device emits a high pitched pulsing sound which young people can hear but older people cannot. It is not harmful but is annoying to those who can hear it.
Prepare a leaflet to inform young people about the device and the science behind it. Use the words ‘frequency’ and ‘ultrasound’ in your explanation.

PEER ASSESSMENT

Swap leaflets with another student. Complete a grid like this to give them feedback:
Question ddots\ddots ddots\ddots Comments
Does the leaflet
explain what the
frequency of a sound
is?
Does the leaflet explain what the frequency of a sound is?| Does the leaflet | | :--- | | explain what the | | frequency of a sound | | is? |
Does it explain the
difference in hearing
range between
teenagers and adults?
Does it explain the difference in hearing range between teenagers and adults?| Does it explain the | | :--- | | difference in hearing | | range between | | teenagers and adults? |
Does it explain why
the mosquito device
will stop young
people hanging
around where they
are not wanted?
Does it explain why the mosquito device will stop young people hanging around where they are not wanted?| Does it explain why | | :--- | | the mosquito device | | will stop young | | people hanging | | around where they | | are not wanted? |
Question ddots ddots Comments "Does the leaflet explain what the frequency of a sound is?" "Does it explain the difference in hearing range between teenagers and adults?" "Does it explain why the mosquito device will stop young people hanging around where they are not wanted?" | Question | $\ddots$ | $\ddots$ | Comments | | :--- | :--- | :--- | :--- | | Does the leaflet <br> explain what the <br> frequency of a sound <br> is? | | | | | Does it explain the <br> difference in hearing <br> range between <br> teenagers and adults? | | | | | Does it explain why <br> the mosquito device <br> will stop young <br> people hanging <br> around where they <br> are not wanted? | | | |

Applications of ultrasound

Sonar

Sonar is a method used to measure the depth of water or to locate an underwater object. Figure 12.23 shows how this works.
Figure 12.23: Using sonar to measure depth. What effect might the fish have?
A pulse of ultrasound is sent down from a boat and reflects from the seabed. The time taken for the reflected pulse to be received is measured. This is used, with the speed of sound in water to calculate the depth of the water.

WORKED EXAMPLE 12.2

A ship sends out an ultrasound pulse and receives the echo after 3 seconds. The speed of sound in water is 1500 m / s 1500 m / s 1500m//s1500 \mathrm{~m} / \mathrm{s}. Calculate the depth of the water.
Step 1: Calculate the distance travelled by the pulse:
distance = speed × time = 1500 m / s × 3 s = 4500 m  distance  =  speed  ×  time  = 1500 m / s × 3 s = 4500 m {:[" distance "=" speed "xx" time "=1500m//sxx3s],[=4500m]:}\begin{aligned} \text { distance }=\text { speed } \times \text { time } & =1500 \mathrm{~m} / \mathrm{s} \times 3 \mathrm{~s} \\ & =4500 \mathrm{~m} \end{aligned}
Step 2: Halve this to get the depth. Remember the pulse goes down and back up again.
depth = 4500 m 2 = 2250 m depth = 4500 m 2 = 2250 m depth=(4500(m))/(2)=2250m\operatorname{depth}=\frac{4500 \mathrm{~m}}{2}=2250 \mathrm{~m}

Answer

The depth of the water is 2250 m .

Material testing

Ultrasound can be used to detect flaws inside materials. A small crack in a metal girder could cause a building to collapse. Figure 12.24 shows ultrasound being passed through uncracked (A) and cracked (B) metal. The original and reflected pulses are shown on oscilloscope trace A. Oscilloscope trace B has an extra peak. This indicates to the engineers that some ultrasound is being reflected from a crack or flaw inside the metal.
Figure 12.24: The exact location of the flaw can be calculated from the speed of sound in the metal and the time for the reflected pulse to be detected.
Ultrasound is also used in medicine (Figure 12.25). Ultrasonic waves are partially reflected from boundaries between different materials, such as the chambers of a patient’s heart, or a fetus. Computer analysis of the reflected waves produces an image.
Figure 12.25: The ultrasound image shows the doctor and the mother how this fetus is developing.

Questions

10 State the range of sound a young person can typically hear.
11 Describe what happens to this range as the person gets older. What else can have this effect on hearing?
12 Figure 12.26 shows the traces produced on an oscilloscope by three different sounds.
Figure 12.26: Three sounds produced on an oscilloscope.
a Which sound is quietest? Explain your answer.
b Which two sounds have the same pitch? Explain your answer.
13 A ship positioned above a shoal of fish (Figure 12.27) sends out an
ultrasound pulse and receives two reflected pulses, one after 0.2 seconds and the other after 0.5 seconds. The speed of sound in water is 1500 m / s 1500 m / s 1500m//s1500 \mathrm{~m} / \mathrm{s}.
Figure 12.27: Using ultrasound to find depth.
a Calculate the depth of the sea.
b Calculate the depth of the shoal of fish.
c Explain why the first reflected pulse lasts for longer than the second.

PROJECT

How do musical instruments produce a range of notes?
Your task is to investigate how different instruments create a range of sounds.

Wind instruments

Wind instruments produce sounds by making columns of air vibrate. Putting different amounts of air into test tubes creates columns of air above the water. The air can be made to vibrate by tapping the glass or by blowing over the top of the tube.
Investigate the effect of changing the length of the air column. How does it affect the pitch of the note produced?

Plucked instruments

Plucking a stretched string creates a sound. Use a stringed instrument or a homemade elastic band guitar to investigate the relationship between the length of the string and the pitch of the sound.

Optional

Extend your investigation to answer one of the following:
  • What other methods are used to produce different notes in musical instruments?
  • Is the note produced by a string affected by other factors such as the material it is made from, its thickness or the tension (how tightly it is stretched)?
  • How is the frequency of a note from a string related to length? You can measure the frequency of the sound directly or by using a microphone and oscilloscope. You can draw a graph of frequency against length.
    Present your results to the class. This could be as a poster, a talk, a presentation, or playing a recognisable tune on an improvised instrument. You should include evidence such as diagrams, tables or graphs.
Figure 12.28: A musical instrument that is plucked.

SUMMARY

All sounds are caused by vibrating sources.
Sound waves are longitudinal waves.
Sound waves consist of a series of compressions and rarefactions.
Sound waves need a medium to travel through.
Sound travels at between 330 m / s 330 m / s 330m//s330 \mathrm{~m} / \mathrm{s} and 350 m / s 350 m / s 350m//s350 \mathrm{~m} / \mathrm{s} in air.
Sound travels fastest in solids and slowest in gases.
An echo is a reflected sound.
The greater the amplitude of a sound, the louder the sound.
The greater the frequency of a sound, the higher its pitch.
Humans can hear sounds between 20 Hz and 20000 Hz .
Ultrasound can be used in medical scans, testing materials and depth calculation.

EXAM-STYLE QUESTIONS

1 Which arrow shows the amplitude of the sound wave?
2 A student stands 60 metres from a wall. He makes a sound and hears the echo 0.3 seconds later.
Which is the correct calculation to help him find the speed of sound?
A 60 ÷ 0.3 60 ÷ 0.3 60-:0.360 \div 0.3
B 120 ÷ 0.3 120 ÷ 0.3 120-:0.3120 \div 0.3
C 120 ÷ 0.6 120 ÷ 0.6 quad120-:0.6\quad 120 \div 0.6
D 60 × 0.3 60 × 0.3 60 xx0.360 \times 0.3
3 What is the typical range of sound a human ear can detect?
A 200 20000 Hz 200 20000 Hz quad200-20000Hz\quad 200-20000 \mathrm{~Hz}
B 20 2000 Hz 20 2000 Hz quad20-2000Hz\quad 20-2000 \mathrm{~Hz}
C 200 2000 Hz 200 2000 Hz quad200-2000Hz\quad 200-2000 \mathrm{~Hz}
D 20 20000 Hz 20 20000 Hz 20-20000Hz20-20000 \mathrm{~Hz}
4 a What is meant by the term ‘frequency of a sound’?
[1]
b What units is frequency measured in?
[1]
c What name is given to sounds with a frequency of more than 20000 Hz ?
d The diagram shows the traces produced on an oscilloscope by two different sounds. Describe two ways in which sound B B BB is different from sound A A AA.

[Total: 5]
5 A drummer is playing a solo.
a Describe how the sound produced travels through the air.
b What type of wave is a sound wave?
c Describe an experiment to show that sound needs a medium to travel through.
6 A ship uses echo sounding to find the depth of the water. It sends out an ultrasound pulse and detects the echo after 3 seconds.

a Calculate the depth of the water. Assume that the speed of sound in water is 1500 m / s 1500 m / s 1500m//s1500 \mathrm{~m} / \mathrm{s}.
b The ship’s siren is heard by a boy windsurfing nearby. He is 400 metres away from the boat. How long will the sound take to reach him. Assume that the speed of sound in air is 340 m / s 340 m / s 340m//s340 \mathrm{~m} / \mathrm{s}.
c The sound of the siren also travels through the metal body of the ship. Which of these could be the speed of sound in the steel?

SELF-EVALUATION CHECKLIST

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
I can
See
Topic...
See Topic...| See | | :--- | | Topic... |
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
Amost
there
Amost there| Amost | | :--- | | there |
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |
Describe how sounds are produced. 12.1
Describe sound waves. 12.1
Use the terms 'compression' and
'rarefaction'.
Use the terms 'compression' and 'rarefaction'.| Use the terms 'compression' and | | :--- | | 'rarefaction'. |
12.2
State that sound needs a medium. 12.2
State the speed of sound in air. 12.3
Describe a method to measure the
speed of sound in air.
Describe a method to measure the speed of sound in air.| Describe a method to measure the | | :--- | | speed of sound in air. |
12.3
Compare the speed of sound in solids,
liquids and gases.
Compare the speed of sound in solids, liquids and gases.| Compare the speed of sound in solids, | | :--- | | liquids and gases. |
12.3
Describe how the amplitude of a sound
affects its loudness.
Describe how the amplitude of a sound affects its loudness.| Describe how the amplitude of a sound | | :--- | | affects its loudness. |
12.4
Describe how the frequency of a sound
affects its pitch.
Describe how the frequency of a sound affects its pitch.| Describe how the frequency of a sound | | :--- | | affects its pitch. |
12.4
Explain how an echo is made. 12.3
Define the term 'ultrasound'. 12.4
State the range of sounds which
humans can hear.
State the range of sounds which humans can hear.| State the range of sounds which | | :--- | | humans can hear. |
12.4
Describe medical and engineering
applications of ultrasound.
Describe medical and engineering applications of ultrasound.| Describe medical and engineering | | :--- | | applications of ultrasound. |
12.4
I can "See Topic..." "Needs more work" "Amost there" "Confident to move on" Describe how sounds are produced. 12.1 Describe sound waves. 12.1 "Use the terms 'compression' and 'rarefaction'." 12.2 State that sound needs a medium. 12.2 State the speed of sound in air. 12.3 "Describe a method to measure the speed of sound in air." 12.3 "Compare the speed of sound in solids, liquids and gases." 12.3 "Describe how the amplitude of a sound affects its loudness." 12.4 "Describe how the frequency of a sound affects its pitch." 12.4 Explain how an echo is made. 12.3 Define the term 'ultrasound'. 12.4 "State the range of sounds which humans can hear." 12.4 "Describe medical and engineering applications of ultrasound." 12.4 | I can | See <br> Topic... | Needs <br> more <br> work | Amost <br> there | Confident <br> to move <br> on | | :--- | :---: | :--- | :--- | :--- | | Describe how sounds are produced. | 12.1 | | | | | Describe sound waves. | 12.1 | | | | | Use the terms 'compression' and <br> 'rarefaction'. | 12.2 | | | | | State that sound needs a medium. | 12.2 | | | | | State the speed of sound in air. | 12.3 | | | | | Describe a method to measure the <br> speed of sound in air. | 12.3 | | | | | Compare the speed of sound in solids, <br> liquids and gases. | 12.3 | | | | | Describe how the amplitude of a sound <br> affects its loudness. | 12.4 | | | | | Describe how the frequency of a sound <br> affects its pitch. | 12.4 | | | | | Explain how an echo is made. | 12.3 | | | | | Define the term 'ultrasound'. | 12.4 | | | | | State the range of sounds which <br> humans can hear. | 12.4 | | | | | Describe medical and engineering <br> applications of ultrasound. | 12.4 | | | |

> Chapter 13 Light

IN THIS CHAPTER YOU WILL:

  • use the law of reflection of light to explain how an image is formed in a plane mirror
  • construct ray diagrams for reflection
  • investigate the refraction of light
  • draw ray diagrams to show how lenses form images
  • describe the difference between real and virtual images
  • describe total internal reflection and how it is used
  • describe how the visible spectrum is formed.

GETTING STARTED

The statements opposite are all true, but there are some exceptions to each statement. Discuss each statement and identify when light behaves differently. Rewrite each statement to include these exceptions.
  • Light travels in straight lines.
  • Light passes straight through glass.
  • Light always travels at the same speed.

THE WORLD IN A DROP OF RAIN

This raindrop on a blade of glass shows an image of its surroundings. The light rays causing this have travelled about 150 million kilometres from the Sun to the Earth. Some rays striking the grass are absorbed and supply energy for photosynthesis, while others are reflected, allowing you to see the grass. Those rays passing through the water drop are refracted (bent) to form the image that you see.
Notice that the Sun is at the bottom and the trees appear to grow downwards. The image is upside down. Your eyes contain lenses which produce images in a very similar way to the water drop. Light from the Sun reflects from different objects and enters your eyes. The lens forms an upside-down image, like the water drop. Your brain knows the image is upside down so automatically turns it upright. It has had years of practice, so you are not aware of this process.
The image in Figure 13.1 has been captured using a camera which also contains a lens to produce an image.
Figure 13.1: The light is bent by the raindrop, forming an upside down image of the scene.
Light travels at 300 million metres per second, so it takes eight minutes and 20 seconds to travel from the Sun to the Earth. This means that if the Sun stopped producing light, we would not know for eight minutes and 20 seconds.
In this chapter you will learn more about the processes of reflection and refraction, which can help us explain how light behaves.
When Apollo astronauts visited the Moon, they left behind reflectors on its surface. These are used to measure the distance from the Earth to the Moon. A laser beam is directed from an observatory on Earth so that it reflects back from these reflectors left on the lunar surface. The time taken by the light to travel there and back is measured and, because the speed of light is known, the distance can be calculated.
A similar method can be used with sound waves to measure the depth of water. Sound is reflected from the sea bed and the time taken is measured. Knowing the speed of sound in water allows the depth to be calculated.

Discussion questions

1 How are these two methods similar?
2 What are the differences?
3 Why is light suitable for one and sound for the other?

13.1 Reflection of light

Light usually travels in straight lines. It changes direction if it hits a shiny surface. This change in direction at a shiny surface such as a mirror is called reflection. We will look at reflection in this section.
You can see that light travels in a straight line using a ray box, as shown in Figure 13.2. A light bulb produces light, which spreads out in all directions. A ray box produces a broad beam. By placing a narrow slit in the path of the beam, you can see a single narrow beam or ray of light. The ray shines across a piece of paper. You can record its position by making dots along its length. Laying a ruler along the dots shows that they lie in a straight line.
Figure 13.2a: A ray box produces a broad beam of light, which can be narrowed down using a metal plate with a slit in it. b: Marking the line of the ray with
crosses allows you to record its position.
You may see demonstrations using a different source of light, a laser. A laser (Figure 13.3a) has the great advantage that all of the light it produces comes out in a narrow beam. All of the energy is concentrated in this beam, rather than spreading out in all directions (as with a light bulb). The total amount of energy coming from the laser is probably much less than that from a bulb, but it is much more concentrated. That is why it is dangerous if a laser beam gets into your eye.
Figure 13.3a: Laser beams are very intense. It is important to protect your eyes when using a laser. b: Scarring (just left of centre) to the retina of a child after looking at a laser pointer.

Looking in the mirror

Most of us look in a mirror, at least once a day, to check on our appearance (Figure 13.4). It is important to us to know how people see us.
Figure 13.4: Psychologists use mirrors to test the intelligence of animals. Do they recognise that they are looking at themselves? Chimpanzees’ reactions show they clearly do recognise their images. Other animals, such as cats and dogs, do not - they may even try to attack their own reflection.
Archaeologists have found bronze mirrors more than 2000 years old, so the desire to see ourselves clearly has been around for a long time.
Modern mirrors give a very clear image. When you look in a mirror, rays of light from your face reflect off the shiny surface and back to your eyes. You seem to see an image of yourself behind the mirror. To understand why this is, we need to use the law of reflection of light.
When a ray of light reflects off a mirror or other reflecting surface, it follows a path as shown in Figure 13.5. The ray bounces off, rather like a ball bouncing off a wall. The two rays are known as the incident ray and the reflected ray.
Figure 13.5: This ray diagram shows the reflection of light. The normal is drawn at 90 90 90^(@)90^{\circ} to the mirror. Then the angles are measured between the rays and the normal. The angle of incidence and the angle of reflection are equal: i = r i = r i=ri=r.
The angle of incidence, i i ii, and the angle of reflection, r r rr, are found to be equal to each other. This is the law of reflection:
angle of incidence = angle of reflection
i = r i = r i=ri=r
Angles of incidence and reflection are always measured between the ray and the normal to the surface. The angles between the rays and the mirror are also equal, but it would be hard to measure them if the mirror was curved. The law of reflection also works for curved mirrors.

ACTIVITY 13.1

Investigating the law of reflection
The student in Figure 13.6 is investigating the law of reflection.
Figure 13.6: Investigating the law of reflection.
Write step-by-step instructions for his investigation. Include a list of the apparatus needed, and the measurements that he should take.

The image in a plane mirror

Why do we see such a clear image when we look in a plane mirror? And why does it appear to be behind the mirror?
Figure 13.7a shows how an observer sees an image of a candle in a plane mirror. Light rays from the flame are reflected by the mirror. Some of them enter the observer’s eye. The observer has to look forward and slightly to the left to see the image of the candle.
Figure 13.7a: Looking in the mirror, the observer sees an image of the candle. The image appears to be behind the mirror. b: The ray diagram shows how the
image is formed. Rays from the candle flame are reflected according to the law of reflection.
Figure 13.7b shows how the image is formed. The solid lines show the light from the candle reflecting from the mirror. The girl’s brain assumes that the rays have travelled in straight lines from a point behind the mirror, shown by the dashed lines. (Our brains assume that light travels in straight lines, even though we know that light is reflected by mirrors.) In reality, no light is coming from behind the mirror. The dashed lines appear to be coming from a point behind the mirror, at the same distance behind the mirror as the candle is in front of it. You can see this from the symmetry of the diagram.
The image looks as though it is the same size as the candle. Also, it is (of course) a mirror image - it is left-right reversed, or laterally inverted. You will know this if you have seen writing reflected in a mirror. If you could place the object and its image side-by-side, you would see that they are mirror images of each other, in the same way that your left and right hands are mirror images of each other.
The image of the candle in the mirror is not a real image. A real image is an image that can be formed on a screen. If you place a piece of paper at the position of the image, you will not see a picture of the candle on it, because no rays of light from the candle reach that spot. That is why we drew dashed lines, to show where the rays appear to be coming from. We say that it is a virtual image.
To summarise, when an object is reflected in a plane mirror, its image is:
  • the same size as the object
  • the same distance behind the mirror as the object is in front of it
  • laterally inverted
  • virtual.

Questions

1 a Why is the word ambulance written in reverse on the front of this vehicle?
Figure 13.8: An ambulance.
b Write the word POLICE in the same way.
2 A student investigated the law of reflection. She increased her angle of incidence by 20 20 20^(@)20^{\circ} each time.
Angle of incidence Angle of reflection
20 20 20 20^(@)20^{\circ}
40 40 40^(@)40^{\circ} 39 39 39^(@)39^{\circ}
60 60 60^(@)60^{\circ} 30 30 30^(@)30^{\circ}
Angle of incidence Angle of reflection 20 20^(@) 40^(@) 39^(@) 60^(@) 30^(@)| Angle of incidence | Angle of reflection | | :---: | :---: | | 20 | $20^{\circ}$ | | $40^{\circ}$ | $39^{\circ}$ | | $60^{\circ}$ | $30^{\circ}$ |
a Which angle of reflection did she measure incorrectly?
b Suggest what she may have done wrong.
3 Draw a diagram to show a ray hitting a mirror with an angle of incidence of 40 40 40^(@)40^{\circ}. Draw the reflected ray. Label both rays, the normal and the angles of incidence and reflection.
4 What angle must ray hit a mirror at for the direction of the ray to be turned through 90 90 90^(@)90^{\circ} ? Draw a diagram to illustrate your answer.

Ray diagrams

Light rays follow strict rules. By following the same rules we can construct detailed ray diagrams. These show where an image is formed and what type of image it is. Worked Example 13.1 shows the steps in constructing a ray diagram to show the formation of an image in a plane mirror.

WORKED EXAMPLE 13.1

A small lamp is placed 5 cm in front of a plane mirror. Draw an accurate scale diagram, and use it to show that the image of the lamp is 5 cm behind the mirror.
Step 1: Draw a line to represent the mirror. Indicate its reflecting surface by drawing short lines on the back of the line. Mark the position of the object (the lamp) with a cross and label it O.
Figure 13.9a:
The steps needed to draw the ray diagram are listed below and shown in
Figures 13.9a-d. (It helps to work on squared paper or graph paper.)
Step 2: Draw two rays from the object to the mirror. Draw in the normal lines where they strike the mirror.
Figure 13.9b:
Step 3: Using a protractor, measure the angle of incidence for each ray. Mark the equal angle of reflection. Draw in the reflected rays.
Figure 13.9c:
Step 4: Extend the reflected rays back behind the mirror. Draw this using dotted lines. The point where they cross is where the image is formed. Label this point I.
Figure 13.9d:

Answer

From the diagram for Step 4, it is clear that the image is 5 cm from the mirror, directly opposite the object. The line joining O to I is perpendicular to the mirror.

Questions

5 Figure 13.10 shows an object placed 6 cm from a plane mirror.
Figure 13.10: Object and mirror.
a Copy Figure 13.10 and use the law of reflection to find the path of the rays after they hit the mirror.
b Trace these rays back to find the position of the image.
c Measure the perpendicular distance from the image to the mirror.
6 Figure 13.11 shows how a periscope uses mirrors to allow the user to see beyond obstacles.
Figure 13.11: Periscope.
a Copy and complete the diagram in Figure 13.11 to show how the light reaches the observer.
b The image the observer sees is not laterally inverted. Explain why.

13.2 Refraction of light

If you look down at the bottom of a swimming pool, you may see patterns of shadowy ripples as in Figure 13.12 The surface of the water is not flat. There are often small ripples on the water, and these cause the rays of sunlight to change direction. Where the pattern is darker, rays of light have been bent away, producing a sort of shadow. This bending of rays of light when they travel from one medium to another is called refraction.
Figure 13.12: The ripples seen on the bottom of the pool are caused by water bending the light.
There are many effects caused by the refraction of light. Some examples are the sparkling of diamonds, the way the lens in your eye produces an image of the world around you, and the twinkling of the stars in the night sky. The image of a bent straw in liquid (Figure 13.13) is another consequence of refraction.
Figure 13.13: The straw is partly immersed in the drink. Because of the refraction of the light coming from the part of the straw that is underwater, the straw appears bent.
Refraction occurs when a ray of light travels from one medium into another. The ray of light may change direction. Refraction happens at the boundary between the two materials. The ray approaching the boundary is called the incident ray and the ray leaving the boundary is called the refracted ray. The angle of incidence, i i ii, and angle of refraction, r r rr, are measured to the normal drawn at the point where the ray hits the boundary (see Figure 13.14).
Figure 13.14: Refraction of a light ray entering glass.

EXPERIMENTAL SKILLS 13.1

Investigating refraction

You can investigate refraction by shining a ray of light into a glass or Perspex ® ® ^(®){ }^{\circledR} block and tracing the path of the ray.

You will need:

  • ray box
  • power pack
  • rectangular glass or Perspex ® ® ^(®){ }^{\circledR} block
  • glass or Perspex ® ® ^(®){ }^{\circledR} of different shapes
  • plain paper
  • optical pins.
Safety: You will be working in a darkened area, so keep your work area
tidy to avoid trip hazards.

Getting started

How will you mark the path of the rays going in and out of the block?
How will you determine the path of the ray inside the glass block?

Method

1 Place the block on paper and draw round it.
2 Shine a ray into the block as shown in Figure 13.15. Look carefully at the ray inside the glass. You will see it travels in a straight line. It only bends at the surface.
Figure 13.15: Ray hitting the glass block.
3 Mark the ray going into the block and the ray coming out of the block with crosses or optical pins.
4 Remove the block and draw the ray going in to the block and the ray coming out.
5 Join these two rays to show the path of the light through the glass.
6 Draw the normal at the points where the ray enters and leaves the block.
7 Measure angles i i ii and r r rr for the ray entering the block and record them in a table.
Angle of incide1ce in air, i i ii Angle of refraction in glass, r r rr
Angle of incide1ce in air, i Angle of refraction in glass, r | Angle of incide1ce in air, $i$ | Angle of refraction in glass, $r$ | | :--- | :--- | | | | | | | | | |
8 Repeat this for rays entering the block at different angles.
9 If you have blocks of different materials, repeat the experiment to find out which material bends light the most.
10 Investigate what happens if a ray of light hits the glass block at 90 90 90^(@)90^{\circ}.
11 Investigate the refraction of light as it passes through different shaped blocks, such as those in Figure 13.16.
Figure 13.16: Ray hitting the glass block.
12 Record your observations in sketches. Note any unexpected effects you observe - these will be explained later in this chapter.

Questions

1 Copy and complete these sentences:
When a ray goes from air to glass it bends qquad\qquad the normal.
For a ray going from air to glass, the angle of qquad\qquad is smaller than the angle of qquad\qquad .
When a ray goes from glass to air it bends qquad\qquad the normal.
2 Describe what happens when a ray hits the block at 90 90 90^(@)90^{\circ}.

Changing direction

Figure 13.17 shows light passing through a rectangular block. Notice that the light only bends at the point where it enters or leaves the block, so it is the change of material that causes the bending.
Figure 13.17: Demonstrating the refraction of a ray of light when it passes through a rectangular block of glass or Perspex®. The ray bends as it enters the block. As it leaves, it bends back to its original direction.
From Figure 13.17, you can see that the direction in which the ray bends depends on whether it is entering or leaving the glass.
  • The ray bends towards the normal when entering the glass.
  • The ray bends away from the normal when leaving the glass.
One consequence of this is that, when a ray passes through a parallel-sided block of glass or Perspex ® ® ^(®){ }^{\circledR}, it returns to its original direction of travel, although it is shifted to one side.
When we look at the world through a window, we are looking through a parallelsided sheet of glass. Although the rays of light are shifted slightly as they pass through the glass, we do not see a distorted image because they all reach us
travelling in their original direction.
A ray of light may strike a surface at an angle of incidence of 0 0 0^(@)0^{\circ}, as shown in Figure 13.18. In this case, it does not bend - it simply passes straight through and carries on in the same direction. Usually we say that refraction is the bending of light when it passes from one medium to another. However, we should bear in mind that, when the light is perpendicular to the boundary between the two materials, there is no bending.
Figure 13.18: When a ray hits a boundary at 90 90 90^(@)90^{\circ} it is not refracted.

Explaining refraction

Light is refracted because it travels at different speeds in different materials. Light travels fastest in a vacuum (empty space) and almost as fast in air. It travels more slowly in glass, water and other transparent substances.
Figure 13.19 shows how a change of speed can cause a change of direction. Imagine a truck is driving along a road across the desert. The driver is careless and allows the wheels on the left to drift off the road onto the sand. Here, they spin around, so that the left-hand side of the truck moves more slowly. The righthand side is still on the road, and keeps moving quickly, so that the truck starts
to turn to the left.
Figure 13.19: To explain why a change in speed explains the bending caused by refraction, picture a truck’s wheels slipping off the road into the sand. The truck turns to the side because it cannot move so quickly through the sand.
The boundary between the two materials is the edge of the road. The normal is at right angles to the road. The truck has turned to the left, so its direction has moved towards the normal.
This matches what we have seen with light. Light slows down when it goes from air to glass, and it bends towards the normal (Figure 13.14).

Questions

7 Describe what happens to a ray of light that passes from:
a air to glass
b glass to air.
8 Why are the angle of incidence and the angle of refraction always measured between the ray and the normal?
9 Draw a diagram to show a light ray passing from glass to air. Mark the incident and refracted rays, the normal and the angles of incidence and refraction.
10 A swimming pool is lit from the bottom.
Use Figure 13.20 to explain why the swimming pool appears to be shallower than it is.
Figure 13.20: Light in a swimming pool.

Refractive index

Light travels very fast. As far as we know, nothing can travel any faster than light. The speed of light as it travels though empty space is exactly 299792458 m / s m / s m//s\mathrm{m} / \mathrm{s}. This is usually rounded to 300000000 m / s 300000000 m / s 300000000m//s300000000 \mathrm{~m} / \mathrm{s} or 3 × 10 8 m / s 3 × 10 8 m / s 3xx10^(8)m//s3 \times 10^{8} \mathrm{~m} / \mathrm{s}.
When a ray of light passes from air into glass, it slows down and bends towards the normal. This happens when light passes from one transparent medium to another, or when light passes between different regions, such as from hot water to cold water.
The refractive index of a material is a measure of how much the light slows, or how much it is bent. If the speed of light is halved when it enters a material, the refractive index is 2 , and so on. The refractive index is the ratio of the speeds of light in two different media or different regions.
Material Speed of light / m / s / m / s //m//s/ \mathrm{m} / \mathrm{s} speed in vacuum speed in material  speed in vacuum   speed in material  (" speed in vacuum ")/(" speed in material ")\frac{\text { speed in vacuum }}{\text { speed in material }}
vacuum 2.998 × 10 8 2.998 × 10 8 2.998 xx10^(8)2.998 \times 10^{8} 1 exactly
air 2.997 × 10 8 2.997 × 10 8 2.997 xx10^(8)2.997 \times 10^{8} 1.0003
water 2.308 × 10 8 2.308 × 10 8 2.308 xx10^(8)2.308 \times 10^{8} 1.33
Perspex ® ® ^(®){ }^{\circledR} 2.000 × 10 8 2.000 × 10 8 2.000 xx10^(8)2.000 \times 10^{8} 1.5
glass ( 1.800 2.000 ) × 10 8 ( 1.800 2.000 ) × 10 8 (1.800-2.000)xx10^(8)(1.800-2.000) \times 10^{8} 1.5 1.7 1.5 1.7 1.5-1.71.5-1.7
Material Speed of light //m//s (" speed in vacuum ")/(" speed in material ") vacuum 2.998 xx10^(8) 1 exactly air 2.997 xx10^(8) 1.0003 water 2.308 xx10^(8) 1.33 Perspex ^(®) 2.000 xx10^(8) 1.5 glass (1.800-2.000)xx10^(8) 1.5-1.7| Material | Speed of light $/ \mathrm{m} / \mathrm{s}$ | $\frac{\text { speed in vacuum }}{\text { speed in material }}$ | | :--- | :--- | :---: | | vacuum | $2.998 \times 10^{8}$ | 1 exactly | | air | $2.997 \times 10^{8}$ | 1.0003 | | water | $2.308 \times 10^{8}$ | 1.33 | | Perspex ${ }^{\circledR}$ | $2.000 \times 10^{8}$ | 1.5 | | glass | $(1.800-2.000) \times 10^{8}$ | $1.5-1.7$ |
diamond 1.250 × 10 8 1.250 × 10 8 1.250 xx10^(8)1.250 \times 10^{8} 2.4
diamond 1.250 xx10^(8) 2.4| diamond | $1.250 \times 10^{8}$ | 2.4 | | :--- | :--- | :--- |
Table 13.1:The speed of light in some transparent materials. Note that the values are only approximate.

Calculating refractive index

ACTIVITY 13.2

Interpreting data

You can use the measurements you took in Experimental skills 13.1 to investigate the relationship between the angles of incidence and refraction.
Angle of incidence, i i ii
Angle of
refraction, r r rr
Angle of refraction, r| Angle of | | :---: | | refraction, $r$ |
sin i sin i sin i\sin i sin r sin r sin r\sin r sin i sin r sin i sin r (sin i)/(sin r)\frac{\sin i}{\sin r}
Angle of incidence, i "Angle of refraction, r" sin i sin r (sin i)/(sin r) | Angle of incidence, $i$ | Angle of <br> refraction, $r$ | $\sin i$ | $\sin r$ | $\frac{\sin i}{\sin r}$ | | :---: | :---: | :---: | :---: | :---: | | | | | | |
Use a calculator or tables to find the sine of each angle, and divide sin i i ii by sin r sin r sin r\sin r.
If you investigated different blocks, draw and complete a table for each.

Optional

1 What can you conclude about the value of:
sin i sin r ? sin i sin r ? (sin i)/(sin r?)\frac{\sin i}{\sin r ?}
The value of sin i sin r sin i sin r (sin i)/(sin r)\frac{\sin i}{\sin r} is always the same for a particular material. This value is the refractive index of the material. This value is the refractive index of the material. The refractive index of glass is about 1.5 and the refractive index of water is about 1.33.
Refractive index ( n n nn ) can be calculated using the equation:
n = sin i sin r n = sin i sin r n=(sin i)/(sin r)n=\frac{\sin i}{\sin r}

KEY EQUATION

Refractive index:

n = sin i sin r n = sin i sin r n=(sin i)/(sin r)n=\frac{\sin i}{\sin r}

WORKED EXAMPLE 13.2

A ray of light hits the surface of water at an angle of incidence of 30 30 30^(@)30^{\circ}. It is refracted at an angle of 22 22 22^(@)22^{\circ}. Find the refractive index, n n nn, of water.
Step 1: Write down what you know and what you want to find out.
i = 30 r = 22 n = ? i = 30 r = 22 n = ? {:[i=30^(@)],[r=22^(@)],[n=?]:}\begin{aligned} & i=30^{\circ} \\ & r=22^{\circ} \\ & n=? \end{aligned}
Step 2: Write down the equation and substitute these values.
n = sin i sin r = sin 30 sin 22 n = sin i sin r = sin 30 sin 22 n=(sin i)/(sin r)=(sin 30)/(sin 22)n=\frac{\sin i}{\sin r}=\frac{\sin 30}{\sin 22}
Step 3: Use a calculator or tables to find the sines and calculate n n nn
n = 0.5 0.375 = 1.33 n = 0.5 0.375 = 1.33 {:[n=(0.5)/(0.375)],[=1.33]:}\begin{aligned} n & =\frac{0.5}{0.375} \\ & =1.33 \end{aligned}
Answer
n = 1.33 n = 1.33 n=1.33n=1.33

Questions

11 A ray of light enters a block of glass at an angle of incidence of 40 40 40^(@)40^{\circ}. The angle of refraction in the glass is 25 25 25^(@)25^{\circ}. Calculate the refractive index of the glass.
12 a Define ‘refractive index’.
b Explain why refractive index does not have a unit.
13 a Describe what happens to the speed of light as it passes from air to glass.
b Use your answer to part a to explain why light entering glass at an angle is refracted.
c Explain in terms of the speed of light why a ray entering the glass along the normal is not refracted.
14 Figure 13.21 shows a ray of light passing from air into an unknown material (X). The ray is deflected by 19 19 19^(@)19^{\circ}.
Figure 13.21: Ray of light passing into air.
a Determine the angles of incidence and refraction.
b Calculate the refractive index of material X. Give your answer to two significant figures.
15 a Refraction can happen when light passes from one transparent solid to another. Use the information in Table 13.1 to deduce what will happen to a ray of light passing from diamond to glass. Sketch a diagram to show what will happen.
16 A ray of light enters glass with a refractive index of 1.52 at an angle of incidence of 60 60 60^(@)60^{\circ}.
a Calculate the angle of refraction.
b Calculate the speed of light in the glass.

13.3 Total internal reflection

When you investigated refraction you may have noticed that not all the light is refracted. Some is reflected back from the surface, as shown in Figure 13.17. You can also see that as the light emerges from the glass, some light is reflected back inside the glass. This is called internal reflection.
Like all reflected light, these reflected rays obey the law of reflection:
angle of incidence = = == angle of reflection
Rays reflected back inside materials such as glass can be very useful. Figure 13.22 shows internal reflection. A ray of light is incident on a semicircular glass block. The ray enters the curved side of the block at right angles and so passes straight through to the midpoint of the straight side.
Figure 13.22: Using a ray box to investigate reflection when a ray of light strikes a glass block. The ray enters the block without bending, because it is directed along the radius of the block.
What happens next depends on the angle of incidence of the ray at the midpoint. Figure 13.23 shows the possibilities.
In Figure 13.23a, the angle of incidence is small, so most of the light emerges
from the block. There is a faint reflected ray inside the glass block. The refracted ray bends away from the normal.
In Figure 13.23b, the angle of incidence has increased, so more light is reflected inside the block. The refracted ray bends even further away from the normal.
In Figure 13.23c, the refracted ray emerges along and parallel to the surface of the block for a particular angle of incidence. This angle is called the critical angle. Most of the light is reflected inside the block.
In Figure 13.23d, the angle of incidence is even greater and all of the light is reflected inside the block. No refracted ray emerges from the block.
Figure 13.23: How a ray of light is reflected or refracted inside a glass block depends on the angle of incidence.
We have been looking at how light is reflected inside a glass block. We have seen that, if the angle of incidence is greater than the critical angle, the light is
entirely reflected inside the glass. This is known as total internal reflection (TIR):
  • total, because 100 % 100 % 100%100 \% of the light is reflected
  • internal, because it happens inside the glass
  • reflection, because the ray is entirely reflected.
For total internal reflection to happen, the angle of incidence of the ray must be greater than the critical angle. The critical angle depends on the material being used. For glass, it is about 42 42 42^(@)42^{\circ}, depending on the type of glass. For water, the critical angle is greater, about 49 49 49^(@)49^{\circ}. For diamond, the critical angle is small, about 25 25 25^(@)25^{\circ}. Rays of light that enter a diamond are very likely to be totally internally reflected, so they bounce around inside, eventually emerging from one of the diamond’s cut faces. That explains why diamonds are such sparkly jewels.

EXPERIMENTAL SKILLS 13.2

Total internal reflection and the critical angle
You will shine a ray of light into a block and investigate when and how it is reflected.

You will need:

  • ray box with a slit
  • power pack
  • glass or Perspex ® ® ^(®){ }^{\circledR} semi-circular block
  • plain paper
  • ruler and protractor.
Safety: As you will be working in a darkened area, you should keep your work area tidy to avoid trip hazards.

Getting started

How will you record the path of the rays of light that you observe as they pass through the block?

Method

1 Place the semi-circular block on the paper and draw around it. Mark the exact centre of the straight side with a normal line.
2 Shine a ray of light into the block along the curved side so that it hits the midpoint of the straight side. Notice that the ray enters the block at right angles.

raybox
Figure 13.24: Shining the ray of light onto the glass block.
3 Observe the path of the ray. You should see two rays. One is refracted as it emerges from the block and the other is reflected back into the block.
4 Move the ray box to increase the angle of incidence until the refracted ray emerges along the side of the block. The angle of incidence at which this happens is called the critical angle ©.
5 Mark the rays with crosses. Remove the block and draw the ray as it passes through the block.
Figure 13.25: Critical angle.
6 Measure the critical angle, c c cc.
7 Draw round the block again, and shine a ray into the block at an angle greater than the critical angle. The ray will be totally internally reflected.
Figure 13.26: Total internal reflection.
8 Mark the rays, remove the block and measure angles i i ii and r r rr.

Questions

1 Why is the ray not bent when it enters the block?
2 What was the critical angle for your block?
3 What conclusion can you draw from your measurements of angles i and r r rr for the totally internally reflected ray?

Questions

13 Explain the meaning of the words ‘total’ and ‘internal’ in the expression ‘total internal reflection’.
18 The critical angle for water is 49 49 49^(@)49^{\circ}. If a ray of light strikes the upper surface of a pond at an angle of incidence of 45 45 45^(@)45^{\circ}, will it be totally internally reflected? Explain your answer.
19 Look at Figure 13.27.
Figure 13.27
a Name angles x , y x , y x,yx, y and z z zz.
b Write down any relationships you know between the angles.
c Describe what will happen when angle x x xx is increased.

Critical angle and refractive index

As we have seen, the critical angle depends on the material through which a ray is travelling. The greater the refractive index of the material, the smaller the critical angle (see the example of diamond in Worked Example 13.3). We can use the equation n = sin i sin r n = sin i sin r n=(sin i)/(sin r)n=\frac{\sin i}{\sin r} to see how critical angle c c cc and refractive index n n nn are related. To do this we need to be aware than the refractive index for glass relates to light passing from air to glass, not from glass to air. During your reflection and refraction experiments you always marked the rays with arrows. This is because all light rays are reversible. When you shine light from the opposite
direction, you will get the same lines, only the arrows tell you which way the ray was travelling.
Figure 13.28 shows what happens when we reverse the rays.
Figure 13.28: Reversing the rays.
You can see from Figure 13.28 that the angle of incidence is 90 90 90^(@)90^{\circ}, and the angle of refraction as the ray goes into glass is the critical angle. Substituting in the equation gives:
n = sin i sin r = sin 90 sin c n = sin i sin r = sin 90 sin c n=(sin i)/(sin r)=(sin 90^(@))/(sin c)n=\frac{\sin i}{\sin r}=\frac{\sin 90^{\circ}}{\sin c}
Since sin 90 = 1 sin 90 = 1 sin 90^(@)=1\sin 90^{\circ}=1
n = 1 sin c n = 1 sin c n=(1)/(sin c)n=\frac{1}{\sin c}

KEY EQUATION

critical angle:
n = 1 sin c n = 1 sin c n=(1)/(sin c)n=\frac{1}{\sin c}
Note that you need to know this equation, but you do not need to remember how it is derived.
Note also that the letter c c c\mathbf{c} is used both for critical angle and the speed of light.
Read questions carefully and make sure you know which meaning is being used.

WORKED EXAMPLE 13.3

Find the critical angle, c c cc, for diamond. Assume that refractive index n = 2.40 n = 2.40 n=2.40n=2.40.
Step 1: quad\quad Substitute the value of n n nn in the equation:
sin c = 1 n = 1 2.40 = 0.417 sin c = 1 n = 1 2.40 = 0.417 sin c=(1)/(n)=(1)/(2.40)=0.417\sin c=\frac{1}{n}=\frac{1}{2.40}=0.417
Step 2: Rearrange to make c c cc the subject:
c = sin 1 0.417 c = sin 1 0.417 c=sin^(-1)0.417c=\sin ^{-1} 0.417
Step 3: Use a calculator (remember to check your calculator is set to degrees not radians) to find sin 1 00.417 sin 1 00.417 sin^(-1)00.417\sin ^{-1} 00.417.

Answer

c = 24.6 c = 24.6 c=24.6^(@)c=24.6^{\circ}

Optical fibres

A revolution in telecommunications has been made possible by the invention of fibre optics. Telephone messages and other electronic signals such as internet computer messages or cable television signals are passed along fine glass fibres in the form of flashing laser light, which is a digital signal. Figure 13.29a shows how fine these fibres can be. Each of these fibres is capable of carrying thousands of telephone calls simultaneously.
Figure 13.29: The use of fibre optics has greatly increased the capacity and speed of the world’s telecommunications networks. Without this technology, cable television and the Internet would not be possible. a: Each of these very
fine fibres of high-purity glass can carry many telephone messages simultaneously. b: Light travels along a fibre by total internal reflection.
Inside a fibre, light travels along by total internal reflection (see Figure 13.29b). It bounces along inside the fibre because, each time it strikes the inside of the fibre, its angle of incidence is greater than the critical angle. This means no light is lost as it is reflected. The fibre can follow a curved path and the light bounces along inside it, following the curve. For signals to travel over long distances, the glass used must be of a very high purity, so that it does not absorb the light.
Optical fibres are also used in medicine. An endoscope is a device that can be used by doctors to see inside a patient’s body, for example, to see inside the stomach (Figure 13.30). One bundle of fibres carries light down into the body, while another bundle carries an image back up to the user. The endoscope may also have a small probe or cutting tool built in, so that minor operations can be performed without the need for major surgery.
Figure 13.30: Passing a fibre optic cable down the oesophagus to the stomach allows the doctor to see inside the patient without the need for major surgery.
processing issues. Using optical fibres means children can play freely with lights with no risk of electrocution or burning.

Questions

20 List three uses of total internal reflection.
21 a The critical angle for a material is 38 38 38^(@)38^{\circ}. Calculate its refractive index.
b Calculate the critical angle for diamond. Use data from Table 13.1.
22 Sketch a diagram to show how a ray of light can travel along a curved glass fibre. Indicate the points where total internal reflection occurs.
23 Figure 13.31 shows a bicycle reflector. It reflects light by total internal reflection.
Figure 13.31: How light is reflected in a bicycle reflector.
a Why is light not refracted when it enters the plastic?
b What is the angle of incidence at the diagonal surface?
c Copy and complete the diagram in Figure 13.31 to show the path of the rays from the car.
d What can you deduce about the critical angle of the plastic?

13.4 Lenses

Lenses are all around us, for example in spectacles and cameras. Lenses are particularly important to scientists in instruments including microscopes and telescopes. Figure 13.32 shows two uses of lenses which helped to revolutionise science.
Figure 13.32a: More than 400 years ago, Galileo ground his own glass lenses to make telescopes like this. What he saw through them suggested the Earth was not the centre of the universe. b: A few years later, Van Leeuwenhoek used a simple single lens microscope to observe bacteria from his teeth, providing a clue as to how infectious diseases are spread.

Converging and diverging lenses

Lenses can be divided into two types, according to their effect on light (Figure 13.33):
Figure 13.33: Converging lenses are fattest in the middle. Diverging lenses are thinnest in the middle. They are given these names because of their effect on parallel rays of light. Usually, we simply draw the cross-section of the lens.
  • converging lenses are fatter in the middle than at the edges
  • diverging lenses are thinner in the middle than at the edges.
You have probably used a magnifying glass to look at small objects. This is a converging lens. ‘Converging’ means bringing together. You may even have used a magnifying glass to focus the rays of the Sun onto a piece of paper, to set fire to it. (More than a thousand years ago, an Arab scientist described how people used lenses for starting fires.)
Figure 13.34a shows how a converging lens focuses the parallel rays of the Sun. On one side of the lens, the rays are parallel to the principal axis of the lens. After they pass through the lens, they converge on a single point: the principal focus (or focal point). After they have passed through the principal focus, they spread out again. A converging lens can be used to produce a beam of parallel rays. A source of light, such as a small light bulb, is placed at the principal focus. As they pass through the lens, the rays are bent so that they become a parallel beam (Figure 13.34b). This diagram is the same as Figure 13.34a, but in reverse.
Figure 13.34: The effect of a converging lens on rays of light. a: A converging lens makes parallel rays converge at the principal focus. b: Rays from the principal focus of a converging lens are turned into a parallel beam of light.
A converging lens is so-called because it makes parallel rays of light converge. The principal focus is the point where the rays are concentrated together, and where a piece of paper needs to be placed if it is to be burned. The distance from the centre of the lens to the principal focus is called the focal length of the lens. The fatter the lens, the closer the principal focus is to the lens. A fat lens has a shorter focal length than a thin lens. Figure 13.35 shows this.
Figure 13.35a: A fatter lens is more powerful so bends the light more. This gives a shorter focal length. b b b\mathbf{b} : A thinner lens is not as powerful and does not bend the light as much.
Lenses work by refracting light. When a ray strikes the surface of the lens, it is refracted towards the normal. When it leaves the glass of the lens, it bends away
from the normal. The clever thing about the shape of a converging lens is that it bends all rays just enough for them to meet at the principal focus.

Forming a real image

When the Sun’s rays are focused onto a piece of paper, a tiny image of the Sun is created. It is easier to see how a converging lens makes an image if you focus an image of a light bulb or a distant window onto a piece of white paper. The paper acts as a screen to catch the image. (Be careful if you try this yourself - you could set fire to the paper or burn yourself. And remember: never look directly at the Sun.)
Figure 13.36 shows an experiment in which an image of a light bulb (the object) is formed by a converging lens. The image in the raindrop in Figure 13.1 was formed in the same way.
Figure 13.36: Forming a real image of a light bulb using a converging lens. The image is upside down on the screen at the back right.
Images can be described in terms of their size (enlarged, the same size, or diminished), which way up they are (inverted or upright), and where they are formed.
The image of the lightbulb in Figure 13.36 is:
  • inverted
  • diminished
  • nearer to the lens than the object
  • real.
We say that the image is real, because light really does fall on the screen to make the image. If light only appeared to be coming from the image, we would say that the image was virtual. The size of the image depends on how fat or thin the lens is.

Drawing ray diagrams for lenses

We can explain how this real image is formed using a ray diagram. These ray diagrams are drawn to scale and show the path of two particular rays. The steps needed to draw an accurate ray diagram are shown in Worked Example 13.4. Remember that it helps to work on squared paper or graph paper when drawing a ray diagram.

WORKED EXAMPLE 13.4

Draw a ray diagram to find the image formed of a 3 cm tall object placed 12 cm from a converging lens which has a focal length of 5 cm .
Step 1: Draw the lens (a simple outline shape will do) with a horizontal axis through the middle of it.
Figure 13.37a
Step 2: Mark the positions of the principal focus (F) on either side, at 5 cm from the centre of the lens. Mark the position of the object, O, along with a 3 cm arrow standing on the axis. Place the arrow 12 cm from the lens.

Figure 13.37b

Step 3: Draw ray 1, a straight line from the top of the object arrow which passes undeflected through the middle of the lens.
Figure 13.37c
Step 4: Draw ray 2 from the top of the object arrow parallel to the principal axis. As it passes through the lens, it is refracted through the principal focus. To make things easier when we draw ray diagrams, we only show rays bending once, at the centre of the lens.
Figure 13.37d
Look for the point where the two rays cross. This is the position of the top of the image (I).
With an accurately drawn ray diagram, you can see that the image is inverted, diminished and real.
So, to construct a ray diagram like this, draw two rays starting from the top of the object:
  • ray 1: unrefracted through the centre of the lens
  • ray 2: parallel to the axis and then refracted through the principal focus.

ACTIVITY 13.3

Ray diagrams

Imagine you work for a company which makes cameras and projectors.
You have been asked to produce short, technical guides explaining the use of lenses in these products.
Drawing ray diagrams lets us predict what an image will look like. A lens can make different images depending on its position. Draw the following ray diagrams:
  • A lens with a focal length of 3 cm with the object 8 cm from the lens.
  • The same lens but with the object 5 cm from the lens.
Describe the image formed in each case. Which is for a camera, and which for a film projector?
Draw diagrams to investigate the difference the thickness of a lens makes
to the image. Remember, a fat lens has a short focal length, a thin lens has a longer focal length.

PEER ASSESSMENT

Swap ray diagrams with a partner and check if they are correctly drawn.
Give them a smiley face, straight face or sad face for each of the following features:
  • axis drawn horizontally through the middle of the lens
  • focal point, F, marked on each side of the lens
  • ray from the top of the object passing though the centre of the lens without being deflected
  • ray from the top of the object parallel to the axis striking the lens then passing through the principal focus
  • rays drawn as straight lines using a ruler
  • correct arrows on both rays
  • image clearly marked, including which way up it is.
Discuss what, if anything, was missing and how you can both improve your diagrams.

Questions

24 Copy and complete these sentences:
A converging lens refracts parallel rays of light to a point called the
qquad\qquad .
The distance from this point to the lens is called the qquad\qquad .
The fatter the lens the qquad\qquad this distance will be.
25 Copy and complete the ray diagrams in Figure 13.38 to show what happens when the rays hit the lenses.

b
C
Figure 13.38: Ray diagrams.

Magnifying glasses

A magnifying glass is a converging lens. You hold it close to a small object and peer through it to see a magnified image. Figure 13.39 shows how a magnifying glass can help to magnify print for someone with poor eyesight.
Figure 13.39: A converging lens produces a magnified image making small print easier to read.
The object viewed by a magnifying glass is closer to the lens than the principal focus. We can draw a ray diagram using the same two rays as in Figure 13.40.
  • Ray 1 is unrefracted as it passes through the centre of the lens.
  • Ray 2 starts off parallel to the axis and is refracted by the lens so that it passes through the principal focus.
Rays 1 and 2 do not cross over each other. They are diverging (spreading apart) after they have passed through the lens. However, by extending the rays backwards, as shown by the dashed lines, we can see that they both appear to be coming from a point behind the object. This is the position of the image (I). We draw dashed lines because light does not actually travel along these parts of the rays. We cannot catch the image on a screen, because there is no light there.
Figure 13.40: A ray diagram to show how a magnifying glass works. The object is between the lens and the focus. The image produced is virtual. To find its position, the rays have to be extrapolated back (dashed lines) to the point where they cross.
From the ray diagram (Figure 13.40), we can see that the image produced by a magnifying glass is:
  • upright
  • enlarged
  • further from the lens than the object
  • virtual.
So, if you read a page of a book using a magnifying glass, the image you are looking at is behind the page that you are reading. This image is a virtual image.

Using lenses to correct eyesight problems

Our eyes contain converging lenses which form an image on the retina at the back of the eye. The lenses in our eyes are flexible and muscles can change the shape and strength of the lens. This allows us to focus on objects at different distances. Figure 13.41a shows an eye forming an image of a distant object. Figure 13.41b shows the eye focusing on a much closer object. The light from the closer object is diverging so the lens needs to be thicker and stronger to form an image on the retina.
Figure 13.41a: Parallel light from a distant object is focused by a weak lens. b: Diverging light from a close object needs a stronger lens.
Some eyes are unable to change their strength enough to focus on either close or distant objects. An extra lens, worn as spectacles or contact lenses, can work with the eye lens to let it focus as needed.

Short sight

A person with short sight can see close up objects clearly, but cannot form a clear image of distant objects. The image is formed in front of the retina. This is usually because the eyeball is slightly too long so that the rays meet in front of the retina. To correct this, a diverging lens is used to make the rays from the distant object diverge. The eye lens is then able to form a focused image. Figure 13.42 shows the problem and how it is corrected.
Figure 13.42a: With short sight, the image is formed before the retina. b: Using a diverging lens to help the lens in the eye to form an image on the retina.
Figure 13.43: Without glasses, distant objects appear blurred to a short-sighted person. The diverging lens in the glasses lets the person focus on distant objects.

Long sight

A long-sighted person has the opposite problem. A long-sighted eye can focus on distant objects but not close objects. This can be because the eyeball is too short, or the lens cannot become strong enough so the rays from a close object cannot be converged enough to a form an image on the retina. A converging lens causes the rays to converge, allowing the eye lens to form a focused image of close objects, as shown in Figure 13.44.
Figure 13.44a: With long sight, the image is formed behind the retina. b: A converging lens works with the lens in the eye to form an image on the retina.

Questions

26 Look at Figure 13.40. How can you tell from the diagram that the image formed by the magnifying glass is a virtual image?
27 a A converging lens has focal length 3 cm . An object, 2 cm tall, is placed 5 cm from the centre of the lens, on the principal axis. Draw an accurate ray diagram to represent this.
b Use your diagram to determine the distance of the virtual image formed from the lens, and the height of the image.
28 Figure 13.45 shows an eye looking at a distant object. The person sees a blurred image.
Figure 13.45: Eye looking at a distant object.
a Name the eyesight problem this person has.
b Describe how the shape of the eye causes this problem.
c Draw a diagram to show how a lens can be used to help this person to see a focused image of the distant object.

13.5 Dispersion of light

When white light passes through glass, it refracts as it enters and leaves the glass, and can be split into a spectrum of colours. Figure 13.46a shows light being refracted as it passes through a glass prism. You can see that the colours merge into one another, and they are not all of equal widths in the spectrum. A rainbow is a naturally occurring spectrum. White light from the Sun is split up into a spectrum of colours as it enters and leaves droplets of water in the air. It is also reflected back to the viewer by total internal reflection, which is why you must have the Sun behind you to observe a rainbow.
Figure 13.46a: White light is dispersed by a prism. b: Raindrops take the place of the prism to form a rainbow.
This splitting up of white light into a spectrum is known as dispersion. Isaac Newton set out to explain how it happens. It had been suggested that light is coloured by passing it through a prism. Newton showed that this was the wrong idea by arranging for the spectrum to be passed back through another prism. The colours recombined to form white light again. He concluded that white light is a mixture of all the different colours of the spectrum. Newton described the visible light spectrum as being made up of seven colours - red, orange, yellow, green, blue, indigo and violet. You may wonder why indigo and violet are separate rather than just being purple. This is because, in the 17th century, seven was considered to be a mystical number. So, Newton wanted seven colours not just
six. The colours can be remembered as a name, Roy G Biv.
Figure 13.47: The first prism refracts the light causing it to disperse. The second prism refracts the light back, causing the colours to recombine.
So, what happens in a prism to produce a spectrum? As the white light enters the prism, it slows down. We say that it is refracted and, as we have seen, its direction changes. Dispersion occurs because each colour is refracted by a different amount (Figure 13.47). Violet light slows down the most, and so it is refracted the most. Red light is least affected.
Light of a single colour is not dispersed by a prism. It is refracted so that it changes direction, but it is not split up into a spectrum. You can see this in Figure 13.48. This is because it is light of a single colour. This light is described as monochromatic (mono = = == one, chromatic = = == coloured). Monochromatic light is light of a single frequency.
Figure 13.48: Monochromatic light from a laser is not dispersed by a prism as the light is all one frequency.

Questions

29 Copy and complete these sentences.
When light enters glass it slows down, causing it to change direction. This is called qquad\qquad .
Red light changes direction qquad\qquad than violet light.
The light is split into different colours. This is called qquad\qquad .
30 Draw a diagram to show how white light can be dispersed into a spectrum using a glass prism. Label the red and violet light.
31 List the colours of the visible spectrum in order starting with the colour which is refracted the least.

REFLECTION

When we study light, we use very similar words and diagrams for reflection and refraction. Have you found this difficult? How have you learnt the rules for ray diagrams? How successful have you been?

PROJECT

Light fantastic

In this chapter you have seen lots of optical effects caused by the reflection and refraction of light.
Some are useful, some are a nuisance. Your task is to investigate, illustrate and explain one of these effects. Your results should be presented as a poster and should include:
  • a photo - preferably one you have taken - of the effect
  • a ray diagram to show what is happening
  • a description of the scientific laws and principles which explain the effect.
You can use an effect you have seen in class, such as the formation of an image in a plane mirror, or you can apply what you have learnt to a new situation. For example:
  • How does the turtle in Figure 13.49 see itself? (With a waterproof camera you could take a similar photo of yourself in a swimming pool.)
  • Why do diamonds sparkle so much?
  • Why do lenses create upside down images?
  • How can interior designers use mirrors to make a space seem bigger?
  • How does the eye form an image (and how can glasses help solve vision problems)?
  • How does a periscope work?
Figure 13.49: Reflection at the surface of the water allows this turtle to see itself.

SUMMARY

The law of reflection: angle of incidence = angle of reflection.
The image in a plane mirror is upright, as far behind the mirror as the object is in front and swapped round left to right.
The image in a plane mirror is virtual.
Refraction is the bending of light as it goes from one substance to another.
Refraction is caused by light travelling at different speeds in different materials.
When light passes from air to glass it bends towards the normal. When it passes from glass to air it bends away from the normal.
When light travelling through glass hits a boundary with air, some light passes from glass to air, some light is internally reflected back into the glass.
When the angle of incidence in glass is equal to, or greater than, the critical angle, all the light is reflected back into the glass. This is total internal reflection.
The refractive index is a measure of how much light is slowed, or bent, by a material.
Refractive index can be calculated using the equation n = sin i sin r n = sin i sin r n=(sin i)/(sin r)n=\frac{\sin i}{\sin r}
Refractive index = 1 critical angle = 1  critical angle  =(1)/(" critical angle ")=\frac{1}{\text { critical angle }}.
Optical fibres can transmit information rapidly and efficiently using total internal reflection. This is useful in telecommunications and medicine.
Converging lenses bend parallel rays together so they meet at a point called the principal focus.
Drawing a ray parallel to the axis, and a ray which strikes the centre of a lens allows us to draw a ray diagram and find the type of image formed.
A magnifying glass produces a virtual image.
Our eyes use a flexible convex lens to form images.
A short-sighted eye has a lens which is too powerful. This can be corrected using a diverging lens.
A long-sighted eye has a lens which is too weak. This can be corrected using a converging lens.
White light can be dispersed by passing it through a glass prism. This creates the visible spectrum.

EXAM-STYLE QUESTIONS

1 The diagram shows a ray of light striking a plane mirror.
Which row gives the correct labels for the diagram?
X Y Z
A normal incident ray angle of refraction
B angle of incidence incident ray normal
C reflected ray normal angle of reflection
D normal incident ray reflected ray
X Y Z A normal incident ray angle of refraction B angle of incidence incident ray normal C reflected ray normal angle of reflection D normal incident ray reflected ray | | X | Y | Z | | :--- | :--- | :--- | :--- | | A | normal | incident ray | angle of refraction | | B | angle of incidence | incident ray | normal | | C | reflected ray | normal | angle of reflection | | D | normal | incident ray | reflected ray | | | | | |
2 Which one of the following statements about the image formed in a plane mirror is true?
A The image can be formed on a screen.
B The image is diminished.
C The image is upright.
D The image is on the surface of the mirror.
3 A ray of light passes from air to glass at an angle of 30 30 30^(@)30^{\circ}. Which statement correctly describes what happens?
A The ray bends towards the normal.
B The ray passes through the glass without bending.
C The ray is totally internally reflected.
D The ray bends away from the normal.
4 This diagram shows light being refracted by a converging lens.

a What name is given to the distance marked X?
b This diagram shows one ray of light passing through the lens. The lens can be used to form an image of an object.
Copy the diagram and draw another ray to complete it. Draw an arrow to represent the image.
c Choose two of these words to describe the image in b b b\mathbf{b}. upright inverted diminished enlarged
5 This diagram shows a ray of light entering a glass prism. The refractive index of the glass is 42 42 42^(@)42^{\circ}.

a Explain why the ray is not refracted as it enters the glass.
b Copy and complete the diagram to show what happens to the ray.
[1]
c What name is given to the effect you have drawn?
d If the prism was removed, what other piece of apparatus could produce the same effect on the light ray?
6 The diagram shows a ray of light striking the inner surface of a glass block. The critical angle for the glass is 42 42 42^(@)42^{\circ}.

a Copy the diagram. Without calculating, continue the ray from the point where it strikes the surface until it leaves the block.
b What can you say about the speed of light in the glass block?
c Calculate the refractive index of the glass. Write down the equation you use.
d Calculate the angle of incidence at which the ray enters the glass block.
7 a Construct a ray diagram to show how a magnifying glass with a focal length of 6 cm produces an image of an object placed 4 cm from the lens.
b The image formed is virtual. Explain what this means.

SELF-EVALUATION CHECKLIST

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
I can
See
Topic...
See Topic...| See | | :--- | | Topic... |
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
Almost
there
Almost there| Almost | | :--- | | there |
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |
Describe how an image is formed in
a plane mirror.
Describe how an image is formed in a plane mirror.| Describe how an image is formed in | | :--- | | a plane mirror. |
13.1
Recall and use the equation: angle of
incidence = angle of reflection.
Recall and use the equation: angle of incidence = angle of reflection.| Recall and use the equation: angle of | | :--- | | incidence = angle of reflection. |
13.1
Draw a ray diagram to show the
formation of a virtual image in a
plane mirror.
Draw a ray diagram to show the formation of a virtual image in a plane mirror.| Draw a ray diagram to show the | | :--- | | formation of a virtual image in a | | plane mirror. |
13.1
Describe an experiment to
investigate how light is refracted
when it passes through a glass block.
Describe an experiment to investigate how light is refracted when it passes through a glass block.| Describe an experiment to | | :--- | | investigate how light is refracted | | when it passes through a glass block. |
13.2
Draw a labelled diagram of light
passing through a glass block,
labelling the normal and the angles
of incidence and refraction.
Draw a labelled diagram of light passing through a glass block, labelling the normal and the angles of incidence and refraction.| Draw a labelled diagram of light | | :--- | | passing through a glass block, | | labelling the normal and the angles | | of incidence and refraction. |
13.2
Describe internal reflection and
recall that if the angle of incidence is
greater than the critical angle, total
internal reflection occurs.
Describe internal reflection and recall that if the angle of incidence is greater than the critical angle, total internal reflection occurs.| Describe internal reflection and | | :--- | | recall that if the angle of incidence is | | greater than the critical angle, total | | internal reflection occurs. |
13.3
Recall and use the equations:
n = s i n n = s i n n=sinn ~=~ s i n ~
sin r r rr
Recall and use the equations: n=sin sin r| Recall and use the equations: | | :--- | | $n ~=~ s i n ~$ | | sin $r$ |
1 13.2,
n = 1 critical angle n = 1  critical angle  n=(1)/(" critical angle ")n=\frac{1}{\text { critical angle }}
n=(1)/(" critical angle ")| $n=\frac{1}{\text { critical angle }}$ | | :--- |
13.3
Describe and explain how total
internal reflection is used in
telecommunications and medicine.
Describe and explain how total internal reflection is used in telecommunications and medicine.| Describe and explain how total | | :--- | | internal reflection is used in | | telecommunications and medicine. |
13.3
I can "See Topic..." "Needs more work" "Almost there" "Confident to move on" "Describe how an image is formed in a plane mirror." 13.1 "Recall and use the equation: angle of incidence = angle of reflection." 13.1 "Draw a ray diagram to show the formation of a virtual image in a plane mirror." 13.1 "Describe an experiment to investigate how light is refracted when it passes through a glass block." 13.2 "Draw a labelled diagram of light passing through a glass block, labelling the normal and the angles of incidence and refraction." 13.2 "Describe internal reflection and recall that if the angle of incidence is greater than the critical angle, total internal reflection occurs." 13.3 "Recall and use the equations: n=sin sin r" 1 13.2, "n=(1)/(" critical angle ")" 13.3 "Describe and explain how total internal reflection is used in telecommunications and medicine." 13.3 | I can | See <br> Topic... | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :---: | :---: | :---: | :--- | | Describe how an image is formed in <br> a plane mirror. | 13.1 | | | | | Recall and use the equation: angle of <br> incidence = angle of reflection. | 13.1 | | | | | Draw a ray diagram to show the <br> formation of a virtual image in a <br> plane mirror. | 13.1 | | | | | Describe an experiment to <br> investigate how light is refracted <br> when it passes through a glass block. | 13.2 | | | | | Draw a labelled diagram of light <br> passing through a glass block, <br> labelling the normal and the angles <br> of incidence and refraction. | 13.2 | | | | | Describe internal reflection and <br> recall that if the angle of incidence is <br> greater than the critical angle, total <br> internal reflection occurs. | 13.3 | | | | | Recall and use the equations: <br> $n ~=~ s i n ~$ <br> sin $r$ | 1 | 13.2, | | | | $n=\frac{1}{\text { critical angle }}$ | 13.3 | | | | | Describe and explain how total <br> internal reflection is used in <br> telecommunications and medicine. | 13.3 | | | |
Draw a fully labelled ray diagram to
show the formation of a real image
by a converging lens.
Draw a fully labelled ray diagram to show the formation of a real image by a converging lens.| Draw a fully labelled ray diagram to | | :--- | | show the formation of a real image | | by a converging lens. |
13.4
Describe the nature of the image
formed.
Describe the nature of the image formed.| Describe the nature of the image | | :--- | | formed. |
13.4
Draw a fully labelled ray diagram to
show the formation of a virtual
image by a converging lens and
explain how this is used in a
magnifying glass.
Draw a fully labelled ray diagram to show the formation of a virtual image by a converging lens and explain how this is used in a magnifying glass.| Draw a fully labelled ray diagram to | | :--- | | show the formation of a virtual | | image by a converging lens and | | explain how this is used in a | | magnifying glass. |
13.4
Draw diagrams to show the
correction of long and short-
sightedness using lenses.
Draw diagrams to show the correction of long and short- sightedness using lenses.| Draw diagrams to show the | | :--- | | correction of long and short- | | sightedness using lenses. |
13.4
Name the seven colours of the
visible spectrum in the correct order.
Name the seven colours of the visible spectrum in the correct order.| Name the seven colours of the | | :--- | | visible spectrum in the correct order. |
13.5
"Draw a fully labelled ray diagram to show the formation of a real image by a converging lens." 13.4 "Describe the nature of the image formed." 13.4 "Draw a fully labelled ray diagram to show the formation of a virtual image by a converging lens and explain how this is used in a magnifying glass." 13.4 "Draw diagrams to show the correction of long and short- sightedness using lenses." 13.4 "Name the seven colours of the visible spectrum in the correct order." 13.5 | Draw a fully labelled ray diagram to <br> show the formation of a real image <br> by a converging lens. | 13.4 | | | | | :--- | :---: | :--- | :--- | :--- | | Describe the nature of the image <br> formed. | 13.4 | | | | | Draw a fully labelled ray diagram to <br> show the formation of a virtual <br> image by a converging lens and <br> explain how this is used in a <br> magnifying glass. | 13.4 | | | | | Draw diagrams to show the <br> correction of long and short- <br> sightedness using lenses. | 13.4 | | | | | Name the seven colours of the <br> visible spectrum in the correct order. | 13.5 | | | |

> Chapter 14
Properties of waves

IN THIS CHAPTER YOU WILL:

  • describe a wave in terms of speed, amplitude, frequency and wavelength
  • identify differences between transverse waves and longitudinal waves
  • calculate wave speed
  • describe reflection and refraction of waves
  • describe diffraction of waves.

GETTING STARTED

In a group, discuss all the different types of wave you can think of.
What do these waves have in common? What do waves do?

MAKING WAVES

Figure 14.1: Einstein’s last prediction was confirmed after 100 years.
Einstein hypothesized that the universe is made of a fabric that he called the space-time continuum. He predicted that interactions between massive objects would create ripples in space-time in much the same way that throwing a stone into a pond causes ripples in the water.
The ripples are very hard to detect as they are tiny. The space-time continuum is very stiff and only really huge events cause any measurable ripples. Lasers are used to pick up tiny variations in space-time that could be caused by a passing gravitational wave. In 2016, scientists working at the Laser Interferometer Gravitational-Wave observatory (LIGO) detected a ripple in space-time caused by two black holes spiralling towards each other and merging into one.
The discovery of gravitational waves is important for two reasons:
  • It provides further evidence to support Einstein’s theory of relativity.
  • Gravitational waves travel huge distances without the signal being
    affected. This could allow astronomers to investigate areas of the universe which we cannot access using light. This could provide information about the Big Bang, and the existence of dark matter.
    The idea of a wave is a very useful model in physics. Physicists talk about light waves, sound waves, electromagnetic waves, and so on. It is easy to see how waves behave in water and springs. It is not obvious that light, gravity and sound are waves in the way we think of waves in the sea. In this chapter, we will investigate waves in water and springs, and see how these waves can act as a good model for both light and sound.

Discussion questions

1 Does the discovery of gravitational waves prove that Einstein was right?
2 A member of the LIGO team commented that the discovery of gravitational waves means we are at the start of the era of gravitational wave astronomy. Gravitational waves will provide us with information that will help us to understand the universe. Discuss other ways in which humans have used waves to gather information about the earth and beyond.

14.1 Describing waves

Physicists use waves as a model to explain the behaviour of light, sound and electromagnetic radiation. Water waves can help us understand the behaviour of waves as they are easy to observe. Waves are what we see on the sea or a lake, but physicists have a more specialised idea of waves. We can begin to understand this model in the laboratory using a ripple tank (Figure 14.2).
A ripple tank is a shallow glass-bottomed tank containing a small amount of water. A light shining downwards through the water casts a shadow of the ripples on the floor below, showing the pattern that they make.
Figure 14.2: The ripples on the surface of the water in this ripple tank are produced by the bar, which vibrates up and down. The pattern of the ripples is seen by shining a light downwards through the water. This casts a shadow of the ripples on the screen beneath the tank.
Figure 14.3 shows two patterns of ripples, straight and circular, which are produced in different ways.
Figure 14.3: Two patterns of ripples on water. a: Straight ripples are a model for a broad beam of light. b: Circular ripples are a model for light spreading out from a lamp.
One way of making ripples on the surface of the water in a ripple tank is to have a wooden bar that just touches the surface of the water (as in Figure 14.2). The bar vibrates up and down at a steady rate. This sends equally spaced straight ripples across the surface of the water.
A spherical dipper can produce a different pattern of ripples. The dipper just touches the surface of the water. As it vibrates up and down, equally spaced circular ripples spread out across the surface of the water.
In each case, the ripples are produced by something vibrating up and down vertically, but the ripples move out horizontally. The vibrating bar or dipper pushes water molecules up and down. Each molecule drags its neighbours up and down. These then start their neighbours moving, and so on. Each molecule simply moves up and down. Energy is transferred by the wave, but the water molecules remain in the same place after the wave has passed. A wave transfers energy but not matter.
How can these patterns of ripples be a model for the behaviour of light? The straight ripples are like a beam of light, perhaps coming from the Sun. The ripples move straight across the surface of the water, just as light from the Sun travels in straight lines. The circular ripples spreading out from a vibrating dipper are like light spreading out from a lamp. (The dipper is the lamp.) In this chapter we will use waves in the ripple tank as a model to help us understand the behavior of light and sound waves.

Wavelength and amplitude

Waves are often represented by a wavy line, as shown in Figure 14.4. We have already used this idea for sound waves (in Chapter 12) and we will do so again for electromagnetic waves (in Chapter 15). This wavy line is like a downward slice though the ripples in the ripple tank. It shows the succession of crests (also called peaks) and troughs of which the ripples are made.
Figure 14.4a: A sideways view of a wave in a ripple tank.
Figure 14.4b: The same wave as a represented as a smoothly varying wavy line on a graph. This shape is known as a sine wave. If you have a graphics calculator, you can use it to display a graph of y = sin x y = sin x y=sin xy=\sin x, which will look like this graph.
The graph in Figure 14.4b shows a wave travelling from left to right. The horizontal axis ( x x xx-axis) shows the distance, x x xx, travelled horizontally by the wave. The vertical axis ( y y yy-axis) shows how far (distance y y yy ) the surface of the water has been displaced from its normal level. We can think of the x x xx-axis as the level of the surface of the water when it is undisturbed. The blue line on the graph shows how far the surface of the water has been displaced from its undisturbed level.
Two measurements are marked on the graph. These define quantities used to describe waves.
  • The wavelength λ λ lambda\lambda of a wave is the distance from one crest of the wave to the next or between any two points on the wave which are in step. Since the wavelength is a distance, it is measured in metres, m . Its symbol is λ λ lambda\lambda, the Greek letter lambda.
  • The amplitude, A A AA, of a wave is the maximum distance that the surface of the water is displaced from its undisturbed level, that is, the height of a crest (or the depth of a trough). For ripples on the surface of water, the amplitude is a distance, measured in metres, m . Its symbol is A A AA.
For ripples in a ripple tank, the wavelength might be a few millimetres and the amplitude a millimetre or two. Waves on the open sea are much bigger, with wavelengths of tens of metres, and amplitudes varying from a few centimetres up to several metres.

Frequency and period

As the bar in the ripple tank vibrates, it sends out ripples. Each up-and-down movement sends out a single ripple. The more times the bar vibrates each second, the more ripples it sends out. This is shown in the graph in Figure 14.5. Take care! This looks very similar to the previous wave graph in Figure 14.4, but here the horizontal axis shows time, t t tt, not distance, x x xx. This graph shows how the surface of the water at a particular point moves up and down as time passes.
From the representation of the wave in Figure 14.5, we can define two quantities for waves in general:
  • The frequency, f f ff, of a wave is the number of waves sent out each second.
    or ripple per second.
  • The period, T T TT, of a wave is the time taken for one complete wave to pass a point. The period is measured in seconds, s s ss.
Figure 14.5: A graph to show the period of a wave. Notice that this graph has time, t t tt, on its horizontal axis.
We have already discussed the frequency and period of a sound wave in Chapter 12. It is always important to check whether a wave graph has time, t t tt, or distance, x x xx, on its horizontal axis.
The frequency of a wave is the number of waves made each second, or the number of waves passing a point per second. The period is the number of seconds it takes for each wave to pass a point. Frequency, f f ff, and period, T T TT, are obviously related to each other. Waves with a short period have a high frequency.
frequency ( Hz ) = 1 period ( s ) f = 1 T period ( s ) = 1 frequency ( Hz ) T = 1 f  frequency  ( Hz ) = 1  period  ( s ) f = 1 T  period  ( s ) = 1  frequency  ( Hz ) T = 1 f {:[" frequency "(Hz)=(1)/(" period "(s))],[qquad{:[f=(1)/(T)],[" period "(s)=(1)/(" frequency "(Hz))],[T=(1)/(f)]:}]:}\begin{aligned} & \text { frequency }(\mathrm{Hz})=\frac{1}{\text { period }(\mathrm{s})} \\ & \qquad \begin{aligned} & f=\frac{1}{T} \\ & \text { period }(\mathrm{s})= \frac{1}{\text { frequency }(\mathrm{Hz})} \\ & T=\frac{1}{f} \end{aligned} \end{aligned}
Waves on the sea might have a period of 10 seconds. Their frequency is therefore about 0.1 Hz . A sound wave might have a frequency of 1000 Hz . Its period is therefore 1 1000 s 1 1000 s (1)/(1000(s))\frac{1}{1000 \mathrm{~s}}, which means that a wave arrives every 1 ms (one millisecond).

Wave speed

The wave speed is the rate at which the crest of a wave travels. For example, it could be the speed of the crest of a ripple travelling over the surface of the water.
Speed is measured in metres per second ( m / s m / s m//s\mathrm{m} / \mathrm{s} ).
Waves can have very different speeds. Ripples in a ripple tank travel a few centimetres per second. Sound waves travel at 330 m / s 330 m / s 330m//s330 \mathrm{~m} / \mathrm{s} through air. Light waves travel at about 300000000 m / s 300000000 m / s 300000000m//s300000000 \mathrm{~m} / \mathrm{s} through air.

Waves and energy

We can also think of the speed of a wave as the speed at which it transfers energy from place to place.
Think of the Sun. It is a source of energy. Its energy reaches us in the form of radiation - light waves and infrared waves - which travels through the vacuum of space and which is absorbed by the Earth.
Think of a loudspeaker. It vibrates and causes the air nearby to vibrate. These vibrations spread out in the air as a sound wave. When they reach our ears, our eardrums vibrate. Energy has been transferred by the sound waves to our ears.
The bigger the amplitude, the more energy the wave transfers. A large amplitude means a bright light or a loud sound.
If you have ever been knocked over by a wave in the sea, you will know that water waves also carry energy. It is important to remember that, when a wave travels from one place to another, it is not matter that is moving. The wave is moving, and it is carrying energy. It may move through matter or even through a vacuum, but the matter itself is not transferred from place to place. A wave transfers energy without transferring matter.
Earthquakes show the huge amounts of energy which waves can transfer. Vibrations passing through the Earth can cause buildings to collapse. A seismometer records the vibrations caused by an earthquake.
Figure 14.6a: Vibrations caused by shifts in the Earth’s tectonic plates can cause devastation. b: The seismometer records the amplitude and frequency of the vibrations.

Transverse and longitudinal waves

Ripples in a ripple tank are one way of looking at the behaviour of waves. You can model waves in other ways. As shown in Figure 14.7, a stretched slinky spring can be used to model waves. Fix one end of the spring and move the other end from side to side (Figure 14.7a). You will see that a wave travels along the spring. (You may also notice it reflecting from the fixed end of the spring.) You can demonstrate the same sort of wave using a stretched rope or piece of elastic. You can see the link between frequency and wavelength by changing the rate at which you move the end of the spring up and down. Increasing the frequency decreases the wavelength.
Figure 14.7: Modelling waves using a spring. a: A transverse wave on a stretched spring. It is made by moving the free end from side to side. b: A longitudinal wave on a stretched spring. It is made by pushing the free end back and forth, along the length of the spring.
A second type of wave can also be modelled with a stretched slinky spring. Instead of moving the free end from side to side, move it backwards and forwards (Figure 14.7b). A series of compressions travels along the spring. These are regions in which the segments of the spring are compressed together. In between are rarefactions, regions in which the segments of the spring are further apart. This type of wave cannot be demonstrated on a stretched rope.
The direction of propagation of a wave is the direction in which it travels. You can observe this by watching the movement of crests and troughs, or compressions and rarefactions.
The demonstrations in Figure 14.7 show the two different types of wave:
  • Transverse waves: the particles carrying the wave move from side to side, at right angles to the direction of propagation of the wave.
  • Longitudinal waves: the particles carrying the wave move back and forth, along the direction of propagation of the wave.
Figure 14.8: Primary and secondary seismic waves.
A ripple on the surface of water is an example of a transverse wave. The particles of the water move up and down as the wave travels horizontally.
In an earthquake there are two types of seismic waves: primary or P P P\mathbf{P}-waves and secondary or S-waves. P-waves are longitudinal. These waves are called primary waves as they travel faster than slower secondary waves and so are felt first. Secondary waves are transverse.
A sound wave is an example of a longitudinal wave. As a sound travels through air, the air molecules move back and forth as the wave travels. Table 14.1 lists examples of transverse and longitudinal waves.
Transverse waves Longitudinal waves
ripples on water sound
light and all other electromagnetic
waves
light and all other electromagnetic waves| light and all other electromagnetic | | :--- | | waves |
primary seismic waves (P-waves)
secondary seismic waves (S-waves)
Transverse waves Longitudinal waves ripples on water sound "light and all other electromagnetic waves" primary seismic waves (P-waves) secondary seismic waves (S-waves) | Transverse waves | Longitudinal waves | | :--- | :--- | | ripples on water | sound | | light and all other electromagnetic <br> waves | primary seismic waves (P-waves) | | secondary seismic waves (S-waves) | |
Table 14.1: Transverse and longitudinal waves.

Questions

1 Copy and complete the following sentences.
a A wave transfers qquad\qquad from place to place without transferring
qquad\qquad .
b In a qquad\qquad wave the vibrations are at right angles to the direction in which the wave travels. In a qquad\qquad wave the vibrations are back and forth along the direction of the wave.
2 The two waves in Figure 14.9 represent two light waves, X and Y .

light Y Y YY
Figure 14.9: Two light waves.
Copy these sentences, selecting the correct answers from the brackets.
a The two waves have { { {\{ the same / different } } }\} wavelengths.
b The two waves have {the same / different } } }\} amplitudes.
c Light X will be { { {\{ brighter / dimmer } } }\} than light Y .
3 Draw a wave and label the amplitude and the wavelength.
4 Give one example of a transverse wave and one example of a longitudinal
wave.
5 Deduce the wavelength of the waves in Figure 14.10.
Figure 14.10: Waves.
6 The bar of a ripple tank vibrates five times per second.
a State the frequency of the waves it makes.
b Find the period of the waves it makes.
7 A student sees a flash of lightning and, three seconds later, she hears the thunder.
a Why does she see the lightning before she hears the thunder?
b Light and sound are both types of waves.
Copy and complete the following sentences, choosing the correct words from the brackets.
i Light is a {transverse / longitudinal} wave. The vibrations are {parallel to / at right angles to} the direction of travel of the wave. These waves {can / cannot } } }\} travel through a vacuum.
ii Sound is a (transverse / longitudinal} wave. The vibrations are {parallel to / at right angles to} the direction of travel of the wave. These waves {can / cannot} travel through a vacuum.
c The speed of sound in air is 330 m / s 330 m / s 330m//s330 \mathrm{~m} / \mathrm{s}. Calculate how far away the storm is from the student.

14.2 Speed, frequency and wavelength

How fast do waves travel across the surface of the sea? If you stand on the end of a pier, you may be able to answer this question.
Figure 14.11: By timing waves and measuring their wavelength, you can find the speed of waves.
Suppose that the pier is 60 metres long, and that you notice that exactly five waves fit into this length (Figure 14.11). Using this information, you can deduce that the wavelength is:
wavelength = 60 m 5 = 12 m  wavelength  = 60 m 5 = 12 m {:[" wavelength "=(60(m))/(5)],[=12m]:}\begin{aligned} \text { wavelength } & =\frac{60 \mathrm{~m}}{5} \\ & =12 \mathrm{~m} \end{aligned}
Now you time the waves arriving. The interval between crests as they pass the end of the pier is 4.0 seconds. How fast are the waves moving? One wavelength ( 12 metres) passes in 4.0 seconds. So the speed of the waves is:
speed = 12 m 4.0 s = 3.0 m / s  speed  = 12 m 4.0 s = 3.0 m / s {:[" speed "=(12(m))/(4.0(s))],[=3.0m//s]:}\begin{aligned} \text { speed } & =\frac{12 \mathrm{~m}}{4.0 \mathrm{~s}} \\ & =3.0 \mathrm{~m} / \mathrm{s} \end{aligned}
The speed, v v vv, frequency, f f ff, and wavelength, λ λ lambda\lambda, of a wave are connected. We can write the connection in the form of an equation:
speed ( m / s ) = frequency ( Hz ) × wavelength ( m ) v = f λ  speed  ( m / s ) =  frequency  ( Hz ) ×  wavelength  ( m ) v = f λ {:[" speed "(m//s)=" frequency "(Hz)xx" wavelength "(m)],[v=f lambda]:}\begin{aligned} \text { speed }(\mathrm{m} / \mathrm{s}) & =\text { frequency }(\mathrm{Hz}) \times \text { wavelength }(\mathrm{m}) \\ v & =f \lambda \end{aligned}

KEY EQUATION

wave speed = = == frequency × × xx\times wavelength
v = f λ v = f λ v=f lambdav=f \lambda
Another way to think of this is to say that the speed is the number of waves passing per second multiplied by the length of each wave. If 100 waves pass each second ( f = 100 Hz f = 100 Hz f=100Hzf=100 \mathrm{~Hz} ), and each is 4.0 m long ( λ = 4.0 m ) ( λ = 4.0 m ) (lambda=4.0m)(\lambda=4.0 \mathrm{~m}), then 400 m of waves pass each second. The speed of the waves is 400 m / s 400 m / s 400m//s400 \mathrm{~m} / \mathrm{s}.

WORKED EXAMPLE 14.1

An FM radio station broadcasts signals of wavelength 1.5 metres and frequency 200 MHz . What is their speed?
Step 1: Write down what you know, and what you want to know
f = 200 MHz = 200000000 Hz = 2 × 10 8 Hz λ = 1.5 m v = ? f = 200 MHz = 200000000 Hz = 2 × 10 8 Hz λ = 1.5 m v = ? {:[f=200MHz=200000000Hz],[=2xx10^(8)Hz],[lambda=1.5m],[v=?]:}\begin{aligned} f & =200 \mathrm{MHz}=200000000 \mathrm{~Hz} \\ & =2 \times 10^{8} \mathrm{~Hz} \\ \lambda & =1.5 \mathrm{~m} \\ v & =? \end{aligned}
Step 2: Write down the equation for wave speed. Substitute values and calculate the answer.
v = f λ v = 2 × 10 8 Hz × 1.5 m = 3 × 10 8 m / s v = f λ v = 2 × 10 8 Hz × 1.5 m = 3 × 10 8 m / s {:[v=f lambda],[v=2xx10^(8)Hzxx1.5m],[=3xx10^(8)m//s]:}\begin{aligned} v & =f \lambda \\ v & =2 \times 10^{8} \mathrm{~Hz} \times 1.5 \mathrm{~m} \\ & =3 \times 10^{8} \mathrm{~m} / \mathrm{s} \end{aligned}

Answer

The radio waves travel through the air at 3.0 × 10 8 m / s 3.0 × 10 8 m / s 3.0 xx10^(8)m//s3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}. (You might recognise this number as the speed of light.)

WORKED EXAMPLE 14.2

The highest note on a piano has a frequency of 4186 Hz . What is the wavelength of the sound waves produced when this note is played? Assume that the speed of sound in air = 330 m / s = 330 m / s =330m//s=330 \mathrm{~m} / \mathrm{s}. Give your answer to two significant figures.
Step 1: Write down what you know, and what you want to know.
f = 4186 Hz v = 330 m / s λ = ? f = 4186 Hz v = 330 m / s λ = ? {:[f=4186Hz],[v=330m//s],[lambda=?]:}\begin{aligned} f & =4186 \mathrm{~Hz} \\ v & =330 \mathrm{~m} / \mathrm{s} \\ \lambda & =? \end{aligned}
Step 2: Write down the equation for wave speed.
Rearrange it to make wavelength λ λ lambda\lambda the subject.
v = f λ λ = v f v = f λ λ = v f {:[v=f lambda],[lambda=(v)/(f)]:}\begin{aligned} v & =f \lambda \\ \lambda & =\frac{v}{f} \end{aligned}
Step 3: Substitute values and calculate the answer.
λ = 330 4186 = 0.07166 m λ = 330 4186 = 0.07166 m {:[lambda=(330)/(4186)],[=0.07166m]:}\begin{aligned} \lambda & =\frac{330}{4186} \\ & =0.07166 \mathrm{~m} \end{aligned}
To 2 s.f. this is 0.072 m

Answer

The wavelength of the note in air is 0.072 metres.

Changing material, changing speed

When waves travel from one material into another, they usually change speed.
Light travels more slowly in glass than in air. Sound travels faster in steel than in air. When this happens, the frequency of the waves remains unchanged. This means that their wavelength must change. This is illustrated in Figure 14.12, which shows light waves travelling quickly through air. They reach some glass and slow down, and their wavelength decreases. When they leave the glass, they speed up, and their wavelength increases again.
Figure 14.12: Waves change their wavelength when their speed changes. Their frequency remains constant. Here, light waves slow down when they enter glass and speed up when they return to the air.

Questions

8 For the equation v = f λ v = f λ v=f lambdav=f \lambda, write down what each symbol represents and give their SI units.
9 Calculate the speed of water waves which have a wavelength of 3 m and a frequency of 0.5 Hz .
10 A wave in air has a frequency of 1100 Hz , an amplitude of 4 cm and a wavelength of 30 cm .
a Calculate the speed of the wave.
b Suggest what type of wave it is likely to be.
c Calculate the period of the wave.
11 Blowing across pan pipes (Figure 14.13) creates a sound wave in the pipe. Explain why the longer pipes produce lower pitched notes.
Figure 14.13: Pan pipes.
12 Which have the longer wavelength, radio waves of frequency 90000 kHz or 100 MHz ?
13 Sound waves get faster when they go from air into water. What happens to their:
a speed?
b wavelength?
c frequency?
d period?

14.3 Explaining wave phenomena

When we look at ripples on the surface of water in a ripple tank, we can begin to see why physicists say that light behaves as a wave. The ripples are much more regular and uniform than waves on the sea, so this is a good model system to look at.

Reflection of ripples

Figure 14.14: The reflection of plane waves by a flat metal barrier in a ripple
tank. a: This criss-cross pattern is observed as the reflected ripples pass through the incoming ripples. b: How the ripples are produced. c c c\mathbf{c} : The arrows show how the direction of the ripples changes when they are reflected. The angle of incidence is equal to the angle of reflection, just as in the law of reflection of light.
Figure 14.14 shows what happens when a flat barrier is placed in the ripple tank. Figure 14.14a shows the pattern of the ripples observed, and Figure 14.14b shows how the ripples are produced. Straight ripples (plane waves) are reflected when they strike the flat surface of the barrier. The barrier acts like a mirror, and the ripples bounce off it. This shows an important feature of how waves behave. They pass through each other when they overlap.
Figure 14.14c shows a drawing of the same pattern. This is an overhead view of the ripples. The blue lines represent the tops of the ripples. These lines are known as wavefronts. The separation of the wavefronts is equal to the wavelength of the ripples. Figure 14.14 c also shows lines (the red arrows) to indicate how the direction of travel of the ripples changes. This diagram may remind you of the ray diagram for the law of reflection of light (see Figure 14.5).
Looking at the red arrows, you can see that the waves obey the same law of reflection as light:
aangle of incidence = = == angle of reflection

Refraction of ripples

Refraction of light occurs when the speed of light changes. We can see the same effect for ripples in a ripple tank (Figure 14.15). A glass plate is immersed in the water, to make the water shallower in that part of the tank. There, the ripples move more slowly because they drag on the bottom of the tank (which is now actually the upper surface of the submerged glass plate).
Figure 14.15a: The refraction of plane waves by a flat glass plate in a ripple tank. A submerged glass plate makes the water shallower in the grey region. In this region, the ripples move more slowly, so that they lag behind the ripples in the deeper water.
b
Figure 14.15b: This wavefront diagram shows the same pattern of ripples as a. The rays show that the refracted ray is closer to the normal, just as when light slows down on entering glass.
In the photograph in Figure 14.15a, you can see that these ripples lag behind the faster-moving ripples in the deeper water. Their direction of travel has changed. Figure 14.15b shows the same effect, but as a wavefront diagram. On the left, the ripples are in deeper water and they move faster. They advance steadily forwards. On the right, the ripples are moving more slowly. The right-hand end of a ripple is the first part to enter the shallower water, so it has spent longest time moving at a slow speed. This means that the right-hand end of each ripple lags behind.
The rays (the red arrows) marked on Figure 14.15b show the direction in which the ripples are moving. They are always at right angles to the ripples. They emphasise how the ripples turn so their direction is closer to the normal as they slow down, just as we saw with the refraction of light in Chapter 13.

Diffraction

Figure 14.16: The waves on the sea are straight. As they pass through the gap in the rocks, they start to spread out.
An interesting thing happens to waves when they go through a gap in a barrier. Figure 14.16 shows what happens. Waves passing through the gap in the rocks spread out to fill the bay. This is an example of a phenomenon called
diffraction. Figure 14.17 shows water waves in a ripple tank being diffracted as they pass through a gap.
Figure 14.17: Ripples are diffracted as they pass through a gap in a barrier they spread into the space behind the barrier.
Sound waves are diffracted as they pass through doorways and open windows. This is why we can hear a person around the corner, even though we cannot see them. This supports the idea that sound travels as a wave.
Light waves are also diffracted when they pass through very tiny gaps. You might notice that, on a foggy night, street lamps and car headlights appear to be surrounded by a halo of light. This is because their light is diffracted by the tiny droplets of water in the air. The same effect can also sometimes be seen around the Sun during the day (see Figure 14.18).
Figure 14.18: Light from the Sun is diffracted as it passes through foggy air
(which is full of tiny droplets of water), producing a halo of light.

EXPERIMENTAL SKILLS 14.1

Studying waves in a ripple tank

In this experiment you will observe reflection and refraction of ripples in a ripple tank and describe the reflection, refraction and diffraction of water waves.
You will need:
  • ripple tank and dipper with an eccentric motor
  • power supply
  • light
  • metal or plastic barriers
  • glass sheet to adjust the depth of water.
Safety: It is easiest to observe the shadow of the waves in a darkened room. Keep the work area tidy to avoid trip hazards. Clear up any spilt water immediately.

Getting started

What effect does changing the speed of the motor have on the wavefronts produced?
If you put a small piece of cork in the water, how does it move?

Method

1 Adjust the motor so that you can clearly see the shadows of the waves moving across the tank.
2 Place a barrier in the tank at an angle so that it reflects the waves.
3 Look carefully at the reflected waves. Investigate what happens when you change the angle of the barrier.
4 Place the glass sheet in the tank so that the waves hit it straight on.
Adjust the water level so that the water above the glass is very shallow.
5 Observe the waves as they enter and leave the shallower water
6 Investigate the effect of turning the glass sheet so that the waves enter the shallow water at an angle.
7 Use two barriers to create a wide gap for the waves to pass through. Observe the effect.
8 Narrow the gap until it is about the same size as the wavelength of the waves. Observe how this affects the waves.

Questions

1 Draw a diagram to show what happens to plane waves when they strike a flat reflector placed at 45 45 45^(@)45^{\circ} to their direction of travel.
2 Draw a diagram to show what happens to plane waves when:
a they enter the shallower water straight on
b they enter shallower water at an angle.
3 Draw diagrams to show diffraction at a wide gap and at a narrow gap.
4 How can the speed of ripples in a ripple tank be changed?

SELF-ASSESSMENT

Compare your ripple tank diagrams to another student’s. Think about what makes a good diagram. Are your diagrams clear, simple and correctly labelled? Add notes to your diagrams saying what you have done well and what could be improved.

ACTIVITY 14.1

Learning the definitions of key words

You will have noticed that this topic contains a lot of key words. To explain waves you need to be familiar with these words, which describe wave characteristics and how waves behave.
Design and make a revision activity to help you remember the key definitions. You could use one of the ideas below or create your own.
  • Flash cards: have a key word on one side and its definition on the other. This way, anything you get wrong, you will see the correct answer straight away which will help for the next time.
  • Questions: write a set of short questions such as, 'What is the unit of frequency?’ on a large sheet of paper. Give each question a mark (1 for easy, 2 for difficult). With a friend, take turns answering questions. Remove any which have been answered correctly as you go. The winner is the person with most points.
  • Create self-test questions on a quiz app. This way you can test yourself regularly and quickly.

REFLECTION

Learning key words is important in physics as words have very precise meanings which may be different from the everyday use of the word. Think about how you learn these words. What activities are most helpful for you? Are you regularly checking back over key words from other chapters? Have you learnt revision techniques in other subjects which you can apply here?

Questions

14 The red lines in Figure 14.19 show sound waves created by person A.
Figure 14.19
a Copy and complete Figure 14.19 to show how the sound waves reach person B.
b What type of waves are sound waves?
15 Figure 14.20 shows four ripples, a-d, in a ripple tank. Copy and complete each diagram to show what happens to the ripples. Label each diagram with the wave phenomenon it is showing.
Figure 14.20

Diffraction, more or less

Waves are diffracted when they pass through a gap or around the edge of an obstacle. The size of the gap affects diffraction. The effect is greatest when the width of the gap is equal to the wavelength of the ripples (Figure 14.21).
Figure 14.21: Diffraction is greatest when the width of the gap is equal to the wavelength of the waves being diffracted. When the gap is much smaller than the wavelength, the waves do not pass through at all.
Compare Figure 14.22 with Figure 14.17. In Figure 14.22, the gap is wider, so it is now much bigger than the wavelength of the waves. The diffraction effect is not so pronounced. The central part of the ripple remains straight after it has passed through the gap. At the edges, the ripples are curved.
Figure 14.22: The wavelength of the waves is much smaller than the gap, so the waves are diffracted less.
Sound waves have wavelengths between about 10 mm and 10 metres. This is why they are diffracted as they pass through doorways and windows. Light waves have a much shorter wavelength - less than a millionth of a metre. This is why very small gaps are needed to see light being diffracted.
Diffraction also happens as waves pass an edge. The greater the wavelength of the waves, the greater the angle at which they are diffracted (Figure 14.23).
Figure 14.23: Increasing the wavelength of waves increases the angle of diffraction.

Questions

16 Explain why person B B BB in question 14.14 can hear person A A AA, but not see them.
17 A man who lives behind a mountain finds he can receive waves with a wavelength of 1500 metres, but not shorter waves with a wavelength of 2 metres. Explain why.
18 Draw a diagram to show how a series of parallel, straight wavefronts are altered as they pass through a gap with a width that is equal to the wavelength of the waves.

PROJECT

So, what is a wave?
You work for a television company that is making a programme about the discovery of gravitational waves. The producer has asked your team to work on a five minute introduction to the general idea of waves. You need to answer the question: ‘What is a wave?’.
The director thinks viewers may think of waves as being just water waves. She wants viewers to have a clear idea of what a wave is and how waves behave. Viewers should understand why physicists describe light and sound as waves.
You will need to show that water, light and sound waves behave in the same ways, particularly that they can be reflected, refracted and diffracted.
You should also explain some of the differences, for example, that sound waves need a medium.
Your section will introduce the programme so it needs to get the audience’s interest so they don’t reach for the remote control. (You could point out that the remote control is operated by infrared or radio waves!) The producer suggests you might show a crowd doing a Mexican wave as a way of showing energy being transferred by a wave. You could also include pictures or video clips of water waves, including seismic waves and tsunamis.
Once you have the audience interested, you will need to introduce the idea that waves transmit energy but not matter. Illustrate this by referring to waves in springs, boats bobbing on the waves at sea, or any other examples you can think of. You should then cover wave behaviour, giving as many examples as possible to show the audience how many everyday things depend on waves.
You may present your ideas as a video, or as a storyboard.
When the programme is broadcast all your names will be in the credits, along with your job title. Decide early on what each person will do. Jobs could include: researcher, picture researcher, editor, producer, sound technician, cameraperson and any other roles you think are needed.

SUMMARY

Waves transfer energy without transferring matter.
Waves can be clearly seen in water and springs.
Waves can be transverse or longitudinal.
Wave speed, frequency and wavelength are related by the equation v = f λ v = f λ v=f lambdav=f \lambda.
Waves can be reflected and refracted.
Waves can be diffracted when they pass through a gap or round an obstacle.
The wave model can be used to explain reflection, refraction and diffraction.

EXAM-STYLE QUESTIONS

1 Which describes what is transferred by a wave?
A energy and matter
B just energy
C just matter
D neither energy nor matter
2 Which line gives the correct units for the wave measurements listed?
Wavelength Frequency Wave speed
Aetres metres metres per second
B metres per second hertz metres
C metres hertz metres per second
D metres metres per second hertz
Wavelength Frequency Wave speed Aetres metres metres per second B metres per second hertz metres C metres hertz metres per second D metres metres per second hertz | | Wavelength | Frequency | Wave speed | | :--- | :--- | :--- | :--- | | | Aetres | metres | metres per second | | B | metres per second | hertz | metres | | C | metres | hertz | metres per second | | D | metres | metres per second | hertz | | | | | |
3 The diagrams show waves in a ripple tank approaching a gap between two barriers. Which wave will diffract the most?
4 The diagram shows waves going from deep water into shallower water above a Perspex ® ® ^(®){ }^{\circledR} block.
Describe what happens to each of the following when the wave enters the shallower water:
a wavelength
[1]
b wave speed
[1]
c frequency
[1]
c amplitude.
[1]
[Total: 4]
5 This diagram shows a wave in the sea.

a What is the wavelength of the wave?
b A student says the amplitude is 40 cm . Explain what mistake they have made and write down the correct amplitude.
c Later in the day the sea is calmer. The amplitude is reduced to 10 cm . On graph paper, draw the new wave.
d A bird is floating on the sea. Describe the movement of the bird as the waves pass it.
d The student observes the waves closer to the beach. The wavelength here is 0.5 metres. What can you deduce from this?
6 Radio station A broadcasts radio waves with a wavelength of 1500 metres and a frequency of 200 kHz . Station B has a frequency of 100 MHz .
a Calculate the speed at which the waves travel.
b Use your answer to part a to calculate the wavelength of the waves from station B B BB.
c Explain why a man living in a hilly area can receive the waves from station B but not the waves from station A.
[Total: 4]
7 A vibration generator is used to make a wave on a string.

a What is the name of the distance labelled a a aa ?
b The distance from the pulley to the vibration generator is 90 cm .
What is the wavelength of the wave?
c The frequency of the vibrations from the generator is 60 Hz . Calculate the speed of the waves in the string.
d The vibration of the string causes a sound wave to pass through the air. What type of wave is:
i the wave in the string?
ii the sound wave in the air?
8 A student investigates waves in a ripple tank. All the water is the same depth.
She measures the wavelength of each wave as 22 cm . The period of each wave is 0.58 s .
Calculate the speed of the wave.

SELF-EVALUATION CHECKLIST

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
I can
See
Topic...
See Topic...| See | | :--- | | Topic... |
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
Almost
there
Almost there| Almost | | :--- | | there |
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |
State what is transferred by a wave. 14.1
Recall that waves (including
electromagnetic and seismic waves),
vibrations in springs and ropes, light,
and sound are all examples of wave
motion.
Recall that waves (including electromagnetic and seismic waves), vibrations in springs and ropes, light, and sound are all examples of wave motion.| Recall that waves (including | | :--- | | electromagnetic and seismic waves), | | vibrations in springs and ropes, light, | | and sound are all examples of wave | | motion. |
14.1
Illustrate wave motion using ropes,
springs and water waves.
Illustrate wave motion using ropes, springs and water waves.| Illustrate wave motion using ropes, | | :--- | | springs and water waves. |
14.1
Define and use the terms: wavefront,
wavelength, frequency, amplitude,
wave speed, crest, trough,
compression, rarefaction.
Define and use the terms: wavefront, wavelength, frequency, amplitude, wave speed, crest, trough, compression, rarefaction.| Define and use the terms: wavefront, | | :--- | | wavelength, frequency, amplitude, | | wave speed, crest, trough, | | compression, rarefaction. |
14.1
Recall and use the equation linking
wave speed, frequency and
wavelength.
Recall and use the equation linking wave speed, frequency and wavelength.| Recall and use the equation linking | | :--- | | wave speed, frequency and | | wavelength. |
14.2
Recall the direction of vibrations in
transverse and longitudinal waves.
Recall the direction of vibrations in transverse and longitudinal waves.| Recall the direction of vibrations in | | :--- | | transverse and longitudinal waves. |
14.1
Give examples of transverse and
longitudinal waves.
Give examples of transverse and longitudinal waves.| Give examples of transverse and | | :--- | | longitudinal waves. |
14.1
Draw diagrams to show reflection,
refraction and diffraction of waves.
Draw diagrams to show reflection, refraction and diffraction of waves.| Draw diagrams to show reflection, | | :--- | | refraction and diffraction of waves. |
14.3
Describe how the wavelength of a
wave affects diffraction.
Describe how the wavelength of a wave affects diffraction.| Describe how the wavelength of a | | :--- | | wave affects diffraction. |
14.3
I can "See Topic..." "Needs more work" "Almost there" "Confident to move on" State what is transferred by a wave. 14.1 "Recall that waves (including electromagnetic and seismic waves), vibrations in springs and ropes, light, and sound are all examples of wave motion." 14.1 "Illustrate wave motion using ropes, springs and water waves." 14.1 "Define and use the terms: wavefront, wavelength, frequency, amplitude, wave speed, crest, trough, compression, rarefaction." 14.1 "Recall and use the equation linking wave speed, frequency and wavelength." 14.2 "Recall the direction of vibrations in transverse and longitudinal waves." 14.1 "Give examples of transverse and longitudinal waves." 14.1 "Draw diagrams to show reflection, refraction and diffraction of waves." 14.3 "Describe how the wavelength of a wave affects diffraction." 14.3 | I can | See <br> Topic... | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :--- | :--- | :--- | :--- | | State what is transferred by a wave. | 14.1 | | | | | Recall that waves (including <br> electromagnetic and seismic waves), <br> vibrations in springs and ropes, light, <br> and sound are all examples of wave <br> motion. | 14.1 | | | | | Illustrate wave motion using ropes, <br> springs and water waves. | 14.1 | | | | | Define and use the terms: wavefront, <br> wavelength, frequency, amplitude, <br> wave speed, crest, trough, <br> compression, rarefaction. | 14.1 | | | | | Recall and use the equation linking <br> wave speed, frequency and <br> wavelength. | 14.2 | | | | | Recall the direction of vibrations in <br> transverse and longitudinal waves. | 14.1 | | | | | Give examples of transverse and <br> longitudinal waves. | 14.1 | | | | | Draw diagrams to show reflection, <br> refraction and diffraction of waves. | 14.3 | | | | | Describe how the wavelength of a <br> wave affects diffraction. | 14.3 | | | |

Chapter 15

The electromagnetic spectrum

IN THIS CHAPTER YOU WILL:

  • describe the main features of the electromagnetic spectrum
  • describe some uses of different electromagnetic waves
  • consider how electromagnetic waves can be used safely.

GETTING STARTED

Think about what you have learnt so far about waves.
Write down five bullet points which sum up what you know about waves including light and sound.
Pair up and agree five points from your combined lists. You can change or combine them.
Join with another pair and write out five points that the group agrees on.
Display these points to the class.

RADIATION - FRIEND OR FOE?

Figure 15.1a: This artist’s impression shows how high energy gamma radiation can be finely targeted to kill tumour cells. b: Jocelyn Joyce Burnell was one of the first astrophysicists to observe light from pulsars. Pulsars emit pulses of light at very regular intervals, making them enormously accurate clocks. Pulsars can be used to measure astronomical distances very accurately.
The light waves we see are part of a bigger family of waves called the electromagnetic spectrum. These range from gamma rays (seen killing a cancer cell in Figure 15.1a) to the radio waves used in radio astronomy by astronomers such as Jocelyn Joyce Burnell (Figure 15.1b). In this chapter you will learn about these waves: what they have in common and the differences that make them useful in a huge variety of ways.
All these forms of radiation have useful applications, but many can also cause harm.
Ultraviolet radiation from the Sun can cause skin cancer. However, in trying to avoid skin cancer by staying indoors and covering skin whilst
outside, we risk vitamin D deficiency. Our bodies create vitamin D when exposed to sunlight. Lack of it can cause a range of health problems, including rickets.
Gamma radiation released in nuclear disasters has been shown to cause cancers, and yet when used correctly, the same radiation can be directed on to cancer cells to kill them.
Decisions about the use of potentially harmful radiation have to balance risk with benefit. Consider a pregnant woman involved in a car crash who appears to have sustained a serious spinal injury. X-rays could damage the developing fetus, but without X-ray images to inform treatment, the woman could be paralysed.
Decisions like this have no easy answers. Whichever decision is made, the doctors will aim to minimise risk; for example, if they decide to use X-rays, they will shield the fetus with lead shields which block X-rays as much as they can.
As you will see in this chapter, electromagnetic radiation is all around us. Decisions about how to use these waves have to be taken by everybody, not just the scientists.

Discussion questions

A large sports arena concerned about security is considering installing Xray scanners to check supporters and their bags for dangerous items.
1 Discuss the risks and benefits of installing this system.
2 Consider who should make the final decision: the arena managers, supporters or scientists?

15.1 Electromagnetic waves

William Herschel was an astronomer. In 1799 he was investigating the spectrum of light from the Sun. He used a prism to create a spectrum then placed a thermometer at different parts of the spectrum. The thermometer reading increased when the light fell on it. The light energy was absorbed, creating a heating effect. Herschel noticed this effect was greatest for red light and smallest for violet.
Herschel then placed his thermometer past the red end of the spectrum. The temperature rise was even greater. He concluded that there must be some type of radiation, invisible to the human eye, beyond the red end of the spectrum. He called this infrared radiation (‘infra’ means below). Infrared radiation is the thermal radiation described in Chapter 11. It is the heat you can feel radiating out from any hot object.
Figure 15.2: A modern version of Herschel’s experiment. The spectrum of light from the Sun extends beyond the visible region, from infrared to ultraviolet. The thermometer placed beyond the red light detects more thermal radiation than any of the others.

Beyond the violet end of the spectrum

The discovery of radiation beyond the red end of the spectrum encouraged people to look beyond the violet end. In 1801, a German scientist called Johan Ritter used silver chloride to look for ‘invisible rays’. Silver salts are blackened
by exposure to sunlight (this is the basis of film photography). Ritter directed a spectrum of sunlight onto paper soaked in silver chloride solution. The paper became blackened and, to his surprise, the effect was strongest beyond the violet end of the visible spectrum. He had discovered ultraviolet radiation (‘ultra’ means beyond).
Both infrared and ultraviolet radiation were discovered by looking at the spectrum of light from the Sun. However, they do not have to be produced by an object like the Sun. Imagine a lump of iron that you heat in a Bunsen flame. At first, it looks dull and black. Take it from the flame and you will find that it is emitting infrared radiation. Put it back in the flame and heat it more. It begins to glow, first a dull red colour, then more yellow, and eventually white hot. It is emitting visible light. When its temperature reaches about 1000 C 1000 C 1000^(@)C1000^{\circ} \mathrm{C}, it will also be emitting ultraviolet radiation.
This experiment suggests that there is a connection between infrared, visible and ultraviolet radiation. A cool object emits only radiation at the cool end of the spectrum. The hotter the object, the more radiation it emits from the hotter end.
The Sun is a very hot object (Figure 15.3). Its surface temperature is about 5500 C C ^(@)C{ }^{\circ} \mathrm{C}, so it emits a lot of ultraviolet radiation. Most of this is absorbed in the atmosphere, particularly by the ozone layer. A small amount of ultraviolet radiation does get through to us. The ozone layer is depleted (decreased) by chemicals used in aerosols and refrigerants. Depletion lets more ultraviolet through, increasing the risk of skin cancer. In 1985, an international agreement called the Montreal Protocol regulated, then banned, the use of CFCs - the main chemicals involved in the depletion. In 2019, NASA reported a 20% decrease in ozone depletion. While there is still work to be done, this is an example of what global cooperation on climate issues can achieve.
Figure 15.3: The Sun is examined by several satellite observatories. This image was produced by the SOHO satellite using a camera that detects the ultraviolet radiation given off by the Sun. You can see some detail of the Sun’s surface, including giant prominences looping out into space. The different colours indicate variations in the temperature across the Sun’s surface.

Electromagnetic waves

We have seen that a spectrum is formed when light passes through a prism because some colours are refracted more than others. The violet end of the spectrum is refracted the most. Now we can deduce that ultraviolet radiation is refracted even more than violet light, and that infrared radiation is refracted less than red light.
To explain the spectrum, and other features of light, physicists developed the wave model of light.
Just as sound can be thought of as vibrations or waves travelling through the air (or any other material), so we can think of light as being another form of wave. Sounds can have different pitches - the higher the frequency, the higher the pitch. We can think of a piano keyboard as being a ‘spectrum’ of sounds of different frequencies. Light can have different colours, according to its frequency. Red light has a lower frequency than violet light. Visible light occurs
as a spectrum of colours, depending on its frequency.
A Scottish physicist, James Clerk Maxwell, described light as small oscillations in electric and magnetic fields which he called electromagnetic waves. His theory allowed him to predict that they could have any value of frequency. In other words, beyond the infrared and ultraviolet regions of the spectrum, there must be even more types of electromagnetic wave (or electromagnetic radiation). By the early years of the 20th century, physicists had discovered or artificially produced several other types of electromagnetic wave to complete the electromagnetic spectrum.
The electromagnetic spectrum is a family of transverse waves. Like all waves, they can be reflected, refracted or diffracted. They all travel at the same speed as light in a vacuum. The waves have different frequencies, and this means they have different effects on the materials with which they interact. These waves have many very important uses, and some can be hazardous.

The speed of electromagnetic waves

All types of electromagnetic wave have one thing in common: they travel at the same speed in a vacuum. They travel at the speed of light, which has a value close to 300000000 m / s ( 3 × 10 8 m / s ) 300000000 m / s 3 × 10 8 m / s 300000000m//s(3xx10^(8)(m)//s)300000000 \mathrm{~m} / \mathrm{s}\left(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right) in a vacuum and approximately the same speed in air. Like light, the speed of electromagnetic waves depends on the material through which they are travelling.

Wavelength and frequency

We can represent light as a transverse wave. Figure 15.4 compares red light with violet light. Red light has a greater wavelength than violet light. This means that there is a greater distance from one wave crest to the next. This is because both red light and violet light travel at the same speed, but violet light has a greater frequency, so it goes up and down more often in the same length.
a

b wavelength
Figure 15.4: Comparing red and violet light waves. Both travel at the same speed, but red light has a longer wavelength because its frequency is lower. The wavelength is the distance from one crest to the next (or from one trough to the next). Think of red light waves as longer, lazy waves. Violet light is made up of shorter, more rapidly vibrating waves.
The waves that make up visible light have very high frequencies: over one hundred million million hertz, or 10 14 Hz 10 14 Hz 10^(14)Hz10^{14} \mathrm{~Hz}. Their wavelengths are very small, from 400 nm (nanometres) for violet light to 700 nm for red light. ( 1 nm is onebillionth or one-thousand-millionth of a 400 mn = 400 1000000000 m 400 mn = 400 1000000000 m 400mn=(400)/(1000000000)m400 \mathrm{mn}=\frac{400}{1000000000} \mathrm{~m}. million waves of visible light fit into a metre.
Figure 15.5 shows the complete electromagnetic spectrum, with the wavelengths and frequencies of each region. In fact, we cannot be very precise about where each region starts and stops. This is similar to the light spectrum: it is hard to say where red finishes and orange begins. Even the ends of the visible light section are uncertain, because different people can see slightly different ranges of wavelengths, just as they can hear different ranges of sound frequencies.
Figure 15.5: The electromagnetic spectrum. Remember all waves travel at the same speed and because v = f λ v = f λ v=f lambdav=f \lambda. As frequency increases, wavelength decreases.

Questions

1 Draw and label two waves to show the difference between red light and violet light.
2 Write the names of the waves in the electromagnetic spectrum:
a in order of increasing wavelength
b in order of increasing frequency.
3 Describe and explain what happens when monochromatic green light of wavelength 540 nm passes through a prism.
4 Use the equation v = f λ v = f λ v=f lambdav=f \lambda to calculate the frequency of the green light in question 15.3.

Uses of electromagnetic waves

The different frequencies of the waves in the electromagnetic spectrum give them different properties and this means that they are useful in different ways. Here are some important examples.

Radio waves

Radio waves are used to broadcast radio and television signals. These waves are sent out from a transmitter a few kilometres away to be captured by an aerial on the roof of a house. Radio astronomy can be used to detect radio waves from objects such as stars, galaxies and black holes. Radio frequency identification (RFID) chips are microchips inserted underneath the skin. These can store vital medical information and could be used instead of a passport, or as a contactless bank card.
An RFID tag stores data. It transmits the data as radio waves, which is read by an electronic reader.
Figure 15.6: Tiny RFID tags can store and transmit data. A tag inserted in your body could be used to open doors or pay for shopping to track for your movements.

Microwaves

Used in satellite television broadcasting, because microwaves pass easily through the Earth’s atmosphere. They travel up to a broadcasting satellite, thousands of kilometres away in space. Then they are sent back down to subscribers on Earth. Microwaves are also used to transmit mobile phone (cellphone) signals between masts, which may be up to 20 km apart.
Microwaves ovens emit microwaves, which are absorbed by molecules in food, causing heating.

Infrared radiation

Infrared radiation is used in remote controls for devices such as televisions. A
beam of radiation from the remote control carries a coded signal to the appliance, which changes the settings on the television; for example, changing channel, increasing the volume. You may be able to use a smartphone camera to observe this type of radiation, which would otherwise be invisible to our eyes. Point a remote control at your phone and record what happens as you press buttons. Grills and toasters also use infrared radiation to cook food. Security alarms send out beams of infrared radiation and detect changes in the reflected radiation, which may indicate the presence of an intruder. Infrared red light is also used together with optical fibres. In Figure 15.9b an endoscope allows the doctor to see inside the patient. Visible light is passed down an optical fibre to illuminate the lungs of this COVID-19 patient. Reflected light travels back up to the doctor’s eye. Infrared radiation can also be used in medicine to detect heat which may indicate infection, or to speed up healing and reduce pain.

Visible light

Visible light provides us with information about the world, both directly through our eyes and via optical instruments such as cameras, telescopes and microscopes. Visible light is also vital for photosynthesis.
Figure 15.7: Microwaves allow signals to be transmitted around the curve of the earth’s surface.
Figure 15.8: Using an infrared scanner allows a doctor to see a patient’s veins. This means injections can be targetted much more precisely.
Figure 15.9a: This leaf has used visible light to grow. It is illuminated by the microscope lamp, and the reflected light is focused by the microscope lenses. b: An endoscope allows the doctors to see inside the patient. Visible light is passed down an optical fibre to illuminate the lungs of this COVID-19 patient. Reflected light travels back up to the doctor’s eye.

Ultraviolet (UV) light

UV light causes some chemicals to emit visible light. This includes body fluids such as sweat and saliva. Forensic scientists use this to find evidence at crime scenes which is invisible to the human eye.
Chemicals which glow in UV light can be used for security marking of valuable equipment and banknotes.
Ultraviolet light can be used to sterilise water. Exposure to UV radiation destroys DNA within any bacteria and viruses contained in the water. This makes the bacteria and viruses harmless and the water much safer to use.
Figure 15.10: Many banknotes have markings which are only visible in
ultraviolet light. This helps in the detection of forged banknotes.
X-rays
X-rays can penetrate solid materials and so they are used in security scanners at airports. They are also used in hospitals and clinics to see inside patients without the need for surgery. The X-rays are detected using electronic detectors. Bone absorbs X-rays more strongly than flesh, so bones appear as a shadow in the image. Similarly, a metal gun will appear as a shadow because it absorbs X-rays more strongly than the items around it in Figure 15.11.
Figure 15.11: X-ray scanners allow airport security to quickly detect hidden items.

Gamma rays

Gamma rays can damage or kill living cells. A targeted beam of gamma rays can be used to kill cancerous cells.
Surgical instruments can be sterilised by using gamma rays to kill any bacteria.
Gamma rays can also be used in the detection of cancer. You will learn more about gamma rays and how they are used in Chapter 23.
Figure 15.12: A nurse demonstrates a radiation helmet. The holes in the helmet allow the gamma rays to be targeted exactly on to a tumour.

Questions

5 Name two types of electromagnetic radiation that are used to cook food.
6 Name three types of electromagnetic radiation that have medical uses.
7 A girl phones her friend on her mobile phone while watching TV by a log fire. Explain how infrared radiation, microwaves, visible light and radio waves are involved in this.

15.2 Electromagnetic hazards

All types of radiation can be hazardous. Even bright light shining into your eyes can blind you. Infrared radiation can cause burns. UV radiation from the Sun can damage skin cells which may lead to sunburn and skin cancer. Sunbeds can also damage skin cells and should be used with care. UV radiation can also damage eye cells which is why is important to protect your eyes in bright sunlight by wearing sunglasses or a hat. In general, the higher the frequency of the wavelength, the greater the harm it can cause.
X-rays and gamma rays are the most dangerous part of the electromagnetic spectrum. They can cause cell mutations which may lead to cancer. People who work with X-rays and gamma rays are in danger of being exposed to too much radiation. They can protect themselves by standing well away when a patient is being examined, or by enclosing the equipment in a metal case, which will absorb rays.
Longer waves such as microwaves are much less harmful, but as we use them a lot more frequently in everyday life, we are exposed to much larger amounts of these waves. Microwaves are used to cook food in microwave ovens. This shows that they have a heating effect when absorbed. Telephone engineers, for example, must take care not to expose themselves to excessive amounts of microwaves when they are working on the masts of a mobile phone (cellphone) network. Domestic microwave ovens must be checked to ensure that no radiation is leaking out.
Mobile phones use radio waves and microwaves. Many people are concerned that these could be harmful. Scientists have researched this and the only effect they have found consistent evidence for is a slight heating effect. This is not believed to be harmful. If any risks are found, they would have more effect on children, as they are still developing. Any harm would be increased by using the phone for longer.
Figure 15.13a: The radiographer operates the machine from a separate room.
Figure 15.13b: All areas where X-rays or gamma rays are used have hazard warning labels.

Questions

8 The radiographer must leave the room before operating an X-ray machine. Why is it considered too dangerous for the radiographer but not for the patient?
9 A headline reads, ‘Scientists prove mobile phones are safe’. Explain why this headline is misleading.

ACTIVITY 15.1

Electromagnetic waves - friend or foe?
Each of the pictures in Figure 15.14 represents a use or a danger of a type of electromagnetic radiation. Identify the wave involved and discuss what the use or danger is.
Figure 15.14: Uses and dangers of electromagnetic radiation.
Find or draw as many pictures as you can showing the uses and dangers of electromagnetic waves.
Use your pictures to make a poster. Include a representation of the electromagnetic spectrum similar to Figure 15.5.

15.3 Communicating using electromagnetic

waves

Satellites

Satellites are objects which orbit the Earth. Earth has one natural satellite - the Moon. Earth also has many artificial satellites, many of which are used in the transmission of information carried by electromagnetic waves. Most artificial satellite communication uses microwaves.
Some of the satellites used for communication are in geostationary orbits. This means that they orbit at the same rate as the Earth turns and so they stay above one point on the Earth’s surface. They are about 35000 km above the Earth’s surface and are positioned above the Equator. These satellites are powerful and can transmit large amounts of data. This makes them suitable for satellite television and some satellite phones. The waves travel a long distance to the satellite, meaning there is a slight delay, making it more difficult to have a conversation.
Low Earth orbits are much closer. They can be as low as 2000 km above the Earth’s surface. This means there is no delay in conversation. These satellites orbit the Earth in as little as two hours. A lot more of these are needed than for geostationary orbits as they only cover a small area of the Earth’s surface (see Figure 15.15). They cannot transmit data as fast as geostationary satellites and so are not suitable for television transmission.
Figure 15.15: Low Earth orbits require many more satellites.

The right wave for the job

Mobile phones and wireless internet use microwaves, because they can pass through most walls and only a small aerial is needed.
Figure 15.16: This laptop is connected to the internet using microwaves.
Bluetooth - used for short range communication - uses radio waves. The signal is weakened when it passes through walls. This short range signal is useful for communication between electronic devices such as in hands-free mobile phone systems.
Figure 15.17: A woman wearing a bluetooth headset.
Optical fibres (for cable television and high speed internet) use infrared radiation and visible light. Optical fibres are made of glass which is transparent to light and infrared radiation. These waves have a higher frequency and can carry more data. This is vital for high speed broadband connections.
Figure 15.18: Optical fibres.

Analogue and digital signals

In telephone communication, the signal being transmitted starts off as a sound wave. This varies in amplitude and frequency as the people talk. A sound wave is an analogue signal - it can vary continuously. In traditional telephones, the sound wave is converted by a microphone to an electrical signal, which varies in the same way as the sound wave. The electrical signal is transmitted via copper wires to a receiver, where a speaker converts it back to a sound wave. The transmission process for analogue signals causes the wave to become distorted and unwanted vibrations are often heard as noise. This means the signal is not always clear.
A digital system is a sequence of pulses. Digital signals are either on or off. Digital systems using optical fibres give a much clearer signal. They are also faster and optical cables can carry much more data than a copper cables.
Figure 15.19: An analogue signal varies continuously while a digital signal is either on or off.
Figure 15.20: The use of regenerators means that digital signals can be transmitted over hundreds of kilometres without being distorted.

Making a digital phone call

The analogue sound wave from a caller is encoded by a converter which turns the sound wave to a series of pulses. The digital signal is carried along optical fibres by visible pulses or infrared waves. The signal passes through one or more regenerators which clean up the signal, removing any distortion. The regenerators also boost the signal if it has lost power. A second converter switches the signal back to an analogue signal which can be converted to a sound wave.
Digital signals can transmit data much more rapidly and accurately than analogue signals. Digital signals can also communicate directly with computers which only use digital data.

ACTIVITY 15.2

Guess who?
Work in a group of about four students. Write the names of different waves on identical cards (radio waves, microwaves, infrared, visible light, ultra violet, X-rays, gamma rays, red light, violet light, sound, ultrasound, water and seismic waves).
Place the cards face down in the middle. One player takes a card and looks at it without letting the others see. The other players then try to find out what the wave is by asking questions to which the answer can only be ‘yes’ or ‘no’. Each player can only ask one question.
When the wave has been identified, the next player takes their turn.

REFLECTION

There are two spectra which you need to remember in order: visible light and the electromagnetic spectrum. How easily can you remember them? What helps you with this? Mnemonics can be very useful here. The colours of the visible spectrum can be remembered using Richard Of York Gave Battle In Vain. The first letters of the words correspond to the first letters of the colours. Try to make a mnemonic to remember the order of the waves in the electromagnetic spectrum.

Questions

10 There are two types of communication satellite: geostationary and low
earth orbit. Explain which type:
a is best suited to transmitting TV signals
b gives the shortest delay for phone conversations
c takes least time to orbit the Earth.
11 State two benefits of converting a phone signal from analogue to digital.

PROJECT

Rainbow balloon debate

Balloon debates usually involve discussing which person should be thrown out of an overloaded hot air balloon basket, so that it does not crash. Each passenger explains why they should stay and then a vote is taken. In this activity, the passengers are the different waves of the electromagnetic spectrum.
Seven different wavelengths of radiation are travelling in a hot air balloon.
  • Radio wave
  • Microwaves
  • Infrared radiation
  • Visible light
  • Ultraviolet radiation
  • X-rays
  • Gamma rays
The balloon can only manage six waves, and so one of the waves must be thrown overboard, or the balloon will crash. As a class you will debate which wave should be sacrificed. Each wave will have a team who prepare to speak in favour of their wave and look at arguments why the other waves should go.
You will need to divide the class into seven teams (you could have fewer teams and leave some waves out if necessary). Your teacher will assign each team a wave.
You will need to prepare arguments why your given wave is so important that it must stay in the balloon. Think about what it is used for - medical or communication uses for example - and why losing this would be bad. Think about the harmful aspects of your wave - you will have to answer questions about these, so be prepared to defend your wave.
You will also need to prepare questions for the other waves. Your aim is to make them seem either harmful or not particularly useful so that they are more likely to be voted off.
Your teacher will tell you how long each team can speak for and how long the question portion will last.

PEER ASSESSMENT

After every team has spoken and has been questioned, you will all vote for which wave must go. To help with this, record your impressions of each group on a sheet similar to this.
Use your scoring and notes to decide how you will vote.

SUMMARY

The electromagnetic spectrum is a group of waves which have similar properties to light.
In order of increasing frequency, the waves in the electromagnetic spectrum are: radiowaves, microwaves, infrared, visible light, ultraviolet, X -rays and gamma rays.
All electromagnetic waves travel at the same speed - the speed of light.
The speed of light is 300000000 m / s 300000000 m / s 300000000m//s300000000 \mathrm{~m} / \mathrm{s}.
Electromagnetic waves have different wavelengths and this gives them different properties and makes them suitable for different uses.
High frequency electromagnetic radiation can be hazardous. It can damage cells and cause cells to mutate.
Radio waves, microwaves, visible light and infrared are used in communication systems.
Signals can be analogue or digital. Digital signals transmit data more accurately and faster.

EXAM-STYLE QUESTIONS

1 Which statement is correct when comparing gamma rays and microwaves?
A Microwaves have a higher speed in air and the same frequency.
B Microwaves have a higher frequency and the same wavelength.
C Microwaves have a longer wavelength and the same speed.
D Microwaves have a higher frequency and the same speed.
2 This is part of the electromagnetic spectrum:
radiowaves microwaves P P P\mathbf{P}
visible
light
visible light| visible | | :--- | | light |
ultraviolet Q Q Q\mathbf{Q}
gamma
rays
gamma rays| gamma | | :--- | | rays |
Increasing R
radiowaves microwaves P "visible light" ultraviolet Q "gamma rays" Increasing R | radiowaves | microwaves | $\mathbf{P}$ | visible <br> light | ultraviolet | $\mathbf{Q}$ | gamma <br> rays | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | Increasing R | | | | | | |
Which row correctly describes the labels P , Q P , Q P,Q\mathbf{P}, \mathbf{Q} and R R R\mathbf{R} ?
P Q R
A X-rays infrared radiation wavelength
B infrared radiation X-rays frequency
C X-rays infrared radiation frequency
D infrared radiation X-rays wavelength
P Q R A X-rays infrared radiation wavelength B infrared radiation X-rays frequency C X-rays infrared radiation frequency D infrared radiation X-rays wavelength| | P | Q | R | | :--- | :--- | :--- | :--- | | A | X-rays | infrared radiation | wavelength | | B | infrared radiation | X-rays | frequency | | C | X-rays | infrared radiation | frequency | | D | infrared radiation | X-rays | wavelength |
3 Which of the following statements is true for electromagnetic waves?
A They all have the same frequency.
B They cannot be refracted.
C They all travel at the same speed.
D They are harmless.
4 This diagram shows white light being dispersed by a prism. The dispersed light hits a screen.

a What will be seen at B ?
b What will be seen at C?
c What might be detected at A?
d What might be detected at D ?
e State one way in which the waves at A, B, C and D are alike.
f State one way in which the four waves differ from each other.
5 These are some waves on the electromagnetic spectrum:
X-rays
blue
light
blue light| blue | | :--- | | light |
yellow
light
yellow light| yellow | | :--- | | light |
X-rays "blue light" "yellow light" | | X-rays | blue <br> light | yellow <br> light | | | | | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
a Copy the diagram and add these labels in the correct order: radio waves, ultraviolet, infrared, gamma rays, microwaves, red light.
b Name two of the waves which can be used for cooking. … [2]
c Name the wave with the highest frequency. … [1]
d Name two waves our senses can detect from the Sun. … [2]
6 A student shines a beam of monochromatic red light at a prism.
a The red light has a wavelength of 700 nm . Use the equation v = f λ v = f λ v=f lambdav=f \lambda to calculate the frequency of the red light.
b Is the wavelength of infrared waves longer or shorter than 700 nm ?

SELF-EVALUATION CHECKLIST

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
I can
See
Topic...
See Topic...| See | | :--- | | Topic... |
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
Almost
there
Almost there| Almost | | :--- | | there |
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |
State the speed of electromagnetic
waves in a vacuum.
State the speed of electromagnetic waves in a vacuum.| State the speed of electromagnetic | | :--- | | waves in a vacuum. |
15.1
Describe the properties all
electromagnetic waves share,
including speed.
Describe the properties all electromagnetic waves share, including speed.| Describe the properties all | | :--- | | electromagnetic waves share, | | including speed. |
15.1
List the waves in the electromagnetic
spectrum in order of increasing
frequency or wavelength.
List the waves in the electromagnetic spectrum in order of increasing frequency or wavelength.| List the waves in the electromagnetic | | :--- | | spectrum in order of increasing | | frequency or wavelength. |
15.1
Describe how radio waves,
microwaves, visible light and infrared
are used in communications.
Describe how radio waves, microwaves, visible light and infrared are used in communications.| Describe how radio waves, | | :--- | | microwaves, visible light and infrared | | are used in communications. |
15.3
Describe how X-rays are used in
medicine and security.
Describe how X-rays are used in medicine and security.| Describe how X-rays are used in | | :--- | | medicine and security. |
15.2
Describe how gamma rays can be
harmful.
Describe how gamma rays can be harmful.| Describe how gamma rays can be | | :--- | | harmful. |
15.2
Consider the safety issues which may
arise when using microwaves.
Consider the safety issues which may arise when using microwaves.| Consider the safety issues which may | | :--- | | arise when using microwaves. |
15.2
Describe the advantages of using
digital signals.
Describe the advantages of using digital signals.| Describe the advantages of using | | :--- | | digital signals. |
15.3
I can "See Topic..." "Needs more work" "Almost there" "Confident to move on" "State the speed of electromagnetic waves in a vacuum." 15.1 "Describe the properties all electromagnetic waves share, including speed." 15.1 "List the waves in the electromagnetic spectrum in order of increasing frequency or wavelength." 15.1 "Describe how radio waves, microwaves, visible light and infrared are used in communications." 15.3 "Describe how X-rays are used in medicine and security." 15.2 "Describe how gamma rays can be harmful." 15.2 "Consider the safety issues which may arise when using microwaves." 15.2 "Describe the advantages of using digital signals." 15.3 | I can | See <br> Topic... | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :--- | :--- | :--- | :--- | | State the speed of electromagnetic <br> waves in a vacuum. | 15.1 | | | | | Describe the properties all <br> electromagnetic waves share, <br> including speed. | 15.1 | | | | | List the waves in the electromagnetic <br> spectrum in order of increasing <br> frequency or wavelength. | 15.1 | | | | | Describe how radio waves, <br> microwaves, visible light and infrared <br> are used in communications. | 15.3 | | | | | Describe how X-rays are used in <br> medicine and security. | 15.2 | | | | | Describe how gamma rays can be <br> harmful. | 15.2 | | | | | Consider the safety issues which may <br> arise when using microwaves. | 15.2 | | | | | Describe the advantages of using <br> digital signals. | 15.3 | | | |

Chapter 16

Magnetism

IN THIS CHAPTER YOU WILL:

  • describe magnetic forces between magnets, and between magnets and magnetic materials
  • distinguish between hard and soft magnetic materials, and nonmagnetic materials
  • describe an experiment to identify the pattern of magnetic field lines around a bar magnet
  • distinguish between the design and use of permanent magnets and electromagnets
    :)\rangle use magnetic field patterns between magnets to explain the forces between them and understand how to work out the relative strength of a magnetic field.

GETTING STARTED

Spend two minutes producing a mind map that answers the following questions before comparing notes with your neighbour for a further two minutes, adding or correcting your own work. Be prepared to share your thoughts with the class.
  • Which elements are magnetic?
  • Describe what life on Earth would be like if magnetism did not exist. This might include inventions that depend on magnetism.

FLIPPING FIELDS

A German polar researcher called Alfred Wegener (Figure 16.1) suggested that continents like South America and Africa used to fit together like a jigsaw and the pieces have drifted apart. Few people believed him but later scientists proved him right. Changes in the Earth’s magnetic field helped prove the Earth’s surface is made of plates that move in a process called plate tectonics.
Figure 16.1: Alfred Wegener, the man whose theory of continental drift was proved right by physics.
The Earth has a magnetic field, as if a giant bar magnet sits along its axis (Figure 16.2). But where does this field come from? The Earth’s core is made of iron. Like all metals, iron contains delocalised electrons, which are free to move. The spin of the Earth, and convection currents in the outer core move these electrons, which creates electric currents. These electric currents create magnetic fields just as in electromagnets, which you will meet later in the chapter. While many animals use the Earth’s magnetic field for navigation, we have mostly replaced the compass with the GPS.
Figure 16.2: A NASA image of what the Earth’s magnetic field looks like.
Molten (liquid) rock erupting at mid-ocean ridges pushes the tectonic plates apart at about the same speed as our fingernails grow (Figure 16.3). Grains of magnetite in the molten rock act like little compasses, which are free to change direction. As the rock cools, these little compasses line up with the Earth’s magnetic field and become fixed in place when the rocks crystallise (freeze). They show that the Earth’s magnetic field has reversed in the past. The magnetic pattern is symmetrical (a mirror image on each side of the mid-ocean ridge), which shows that the continents were once stuck together, proving Wegener right.
The Earth’s magnetic field acts as a shield, protecting our atmosphere and
life on Earth by deflecting the harmful solar wind (a stream of charged particles from the Sun). It also protects all the electric circuits developed over the last century. In the process of reversing, the field has to decrease before increasing again in the opposite direction. Some scientists believe that animal species can become extinct during this reversal process, which has happened roughly once every 300000 years in the last 20 million years. The last reversal was 780000 years ago.
Figure 16.3: The magnetic stripes on the ocean floor that show that the Earth’s magnetic field has reversed once every 300000 years on average.

Discussion questions

1 Describe what causes the Earth’s magnetism in less than 20 words.
2 In what ways is the Earth’s magnetic field beneficial?
3 Should we be concerned about magnetic reversals?

16.1 Permanent magnets

A compass needle is like a bar magnet. When it is free to rotate (Figure 16.4), it turns to point north-south. One end points north - this is the magnet’s north pole, pointing roughly in the direction of the Earth’s geographical North Pole. The other end is the magnet’s south pole. Sometimes, the north and south poles of a magnet are called the north-seeking pole and south-seeking pole, respectively.
When two magnets are brought close together, there is a force between them. The north pole ( N pole) of one will attract the south pole (S pole) of the other. Two north poles will repel each other, and two south poles will repel each other (Figure 16.5). This is summarised as:
  • like poles repel
  • unlike poles attract.
    ‘Like poles’ means poles that are the same - both north, or both south. ‘Unlike poles’ means opposite poles - one north and the other south. People often remember this rule more simply as ‘opposites attract’.
Figure 16.4: A freely suspended magnet turns so that it points north-south.
Figure 16.5a: Two like magnetic poles repel one another. b: Two unlike magnetic poles attract each other.
Since the north pole of the compass needle is attracted to the Earth’s North Pole, it follows that there must be a magnetic south pole up there, under the Arctic ice. It is easy to get confused about this. In fact, for a long time, mediaeval scientists thought that compass needles were attracted to the Pole Star. Eventually, an English instrument-maker called Robert Norman noticed that, if he balanced a compass needle very carefully at its midpoint, it tilted downwards slightly, pointing into the Earth. Now we know that the Earth itself is magnetised, rather as if there was a giant bar magnet inside it.

Magnetic materials

Magnetic materials are attracted by a magnet and can be magnetised. Though they can be magnetised, not all pieces of magnetic material are magnets. They first need to be magnetised. In contrast, non-magnetic materials are not attracted by a magnet and cannot be magnetised. Examples include plastic and rubber.
A compass needle is a permanent magnet. Like many bar magnets, it is made of hard steel. You have probably come across another type of magnetic material,
called ferrite. This is a ceramic material used for making fridge magnets and the magnets sometimes used to keep cupboard doors shut. There are also small rareearth magnets in the headphones used with mobile phones, which are based on elements such as neodymium.
Most magnetic materials (including steel and ferrite) contain iron, the most common magnetic element. For this reason, they are known as ferrous materials (from the Latin word ferrum meaning ‘iron’). Other magnetic elements include cobalt and nickel. (If a material contains iron, this is not a guarantee that it will be magnetic. Stainless steel contains a lot of iron, but magnets will not always stick to it.)
Magnetic materials may be classified as hard (permanent) or soft (temporary). Table 16.1 summarises the difference. A soft magnetic material such as soft iron can be magnetised and demagnetised easily.
Type of
magnetic
material
Type of magnetic material| Type of | | :--- | | magnetic | | material |
Description Examples Uses
hard
retains magnetism
well, but difficult to
magnetise in the first
place
retains magnetism well, but difficult to magnetise in the first place| retains magnetism | | :--- | | well, but difficult to | | magnetise in the first | | place |
hard steel
permanent magnets,
compass needles,
loudspeaker magnets
permanent magnets, compass needles, loudspeaker magnets| permanent magnets, | | :--- | | compass needles, | | loudspeaker magnets |
soft
easy to magnetise, but
readily loses its
magnetism
easy to magnetise, but readily loses its magnetism| easy to magnetise, but | | :--- | | readily loses its | | magnetism |
soft iron
cores for
electromagnets
(discussed later in the
chapter), transformers
and radio aerials
cores for electromagnets (discussed later in the chapter), transformers and radio aerials| cores for | | :--- | | electromagnets | | (discussed later in the | | chapter), transformers | | and radio aerials |
"Type of magnetic material" Description Examples Uses hard "retains magnetism well, but difficult to magnetise in the first place" hard steel "permanent magnets, compass needles, loudspeaker magnets" soft "easy to magnetise, but readily loses its magnetism" soft iron "cores for electromagnets (discussed later in the chapter), transformers and radio aerials"| Type of <br> magnetic <br> material | Description | Examples | Uses | | :--- | :--- | :--- | :--- | | hard | retains magnetism <br> well, but difficult to <br> magnetise in the first <br> place | hard steel | permanent magnets, <br> compass needles, <br> loudspeaker magnets | | soft | easy to magnetise, but <br> readily loses its <br> magnetism | soft iron | cores for <br> electromagnets <br> (discussed later in the <br> chapter), transformers <br> and radio aerials |
Table 16.1: Hard and soft magnetic materials. Hard steel is both hard to bend and difficult to magnetise and demagnetise. Soft iron is both easier to bend and easier to magnetise and demagnetise.

Induced magnetism

A bar magnet is an example of a permanent magnet. It can remain magnetised. Its magnetism does not disappear. Permanent magnets are made of hard magnetic materials.
A permanent magnet can attract or repel another permanent magnet. It can also
attract other unmagnetised magnetic materials. For example, a bar magnet can attract steel pins or paper clips, and a fridge magnet can stick to the steel door of the fridge.
What is going on here? Steel pins are made of a magnetic material. When the north pole of a permanent magnet is brought close to a pin, the pin is attracted (see Figure 16.6). The attraction tells us that the end of the pin nearest the magnetic pole must be a magnetic south pole, as shown in Figure 16.6. This is known as induced magnetism. The two objects will have opposite polarities. In this example the pin will have a south pole induced in the end nearest the north pole of a bar magnet. When the permanent magnet is removed, the pin will return to its unmagnetised state (or it may retain a small amount of magnetism).
Figure 16.6: A steel pin is temporarily magnetised when a permanent magnet is brought close to it.

Questions

1 Name three magnetic elements.
2 What is the rule about whether magnetic poles repel or attract?
3 Copy and complete Table 16.2.
Type of magnetic
material
Type of magnetic material| Type of magnetic | | :--- | | material |
Description Examples Uses
Hard
Soft
"Type of magnetic material" Description Examples Uses Hard Soft | Type of magnetic <br> material | Description | Examples | Uses | | :--- | :--- | :--- | :--- | | Hard | | | | | Soft | | | |
Table 16.2
4 Why is a permanent magnet made of steel rather than iron?

16.2 Magnetic fields

A magnet affects any piece of magnetic material that is nearby. We say that there is a magnetic field around the magnet. You have probably done experiments with iron filings or small compasses to illustrate the magnetic field of a magnet. Figure 16.7 shows the field of a bar magnet as revealed by iron filings.
Figure 16.7: The magnetic field pattern of a bar magnet is illustrated by iron filings. The iron filings cluster most strongly around the two poles of the magnet. This is where the field is strongest.
Figure 16.8: Field lines are used to represent the magnetic field around a bar magnet.
Figure 16.8 shows how we represent the magnetic field of a single bar magnet using magnetic field lines. Of course, the field fills all the space around the magnet, but we can only draw a selection of typical lines to represent it. The pattern tells us two things about the field:
  • Direction: the direction of a magnetic field line at any point is the direction of the force on the north pole of a magnet at that point. We use a convention that says that field lines come out of north poles and go in to south poles.
  • Strength: lines that are close together indicate a strong field.

Plotting field lines

Iron filings can illustrate the pattern of the magnetic field around a magnet. Place a magnet under a stiff sheet of plain paper or (preferably) clear plastic. Sprinkle filings over the paper or plastic. Tap the paper or plastic to allow the filings to move slightly so that they line up in the field. You should obtain a pattern similar to that shown in Figure 16.7.
An alternative method of doing this uses a small compass called a plotting compass. When a plotting compass is placed in a magnetic field, its needle turns to indicate the direction of the field. Activity 16.1 describes how to use a plotting compass to show the pattern of a magnetic field. When drawing magnetic field patterns, the field lines should never cross and they should include arrows, pointing from magnetic north to magnetic south.

ACTIVITY 16.1

Plotting field lines

In pairs, design an experiment to plot a field pattern around a bar magnet. Create a worksheet which gives instructions on how to do the experiment. There should be at least ten field lines (including at least two from each side of the magnet). In addition to the bar magnet, the only pieces of equipment needed for the experiment are a plotting compass, a pencil and a sheet of plain paper.
In your worksheet remember to include:
  • how a compass works and how it can reveal magnetic field lines
  • a clear step-by-step method for plotting the field lines (including helpful sketches).
You could provide an optional extension task such as: plot the field lines between two bar magnets that are attracting or repelling. (In this case, make sure you suggest a way to prevent the magnets from moving.)
You could also produce a very short podcast (maximum two minutes) to demonstrate how to do the experiment.
After you have created your worksheet, you will have the opportunity to look at the worksheets written by other pairs. You should note any physics you have learned, and any ideas for making your own worksheet or presentation clearer or more engaging in the future.

PEER ASSESSMENT

Provide feedback on the worksheets produced by other pairs. As you give feedback on the worksheets, think about these questions:
  • Is the physics of a how a compass works explained clearly?
  • Could you follow the steps in the method?
  • Are any steps missing or unclear?
Discuss any improvements that could be made. Each pair should note down the feedback and, if there is time, make any improvements.

Questions

5 a Sketch the field pattern around a bar magnet.
b Someone has drawn the field pattern but forgot to label the poles. Do the arrows point towards or away from the magnetic north pole?
c When looking at a field pattern, how would you know where the magnetic field is strongest?
d Where is the field strongest on a bar magnet?
6 Figure 16.9 shows the Earth’s magnetic field.
Figure 16.9: The Earth’s magnetic field.
A compass needle inside a compass is a lightweight bar magnet suspended at its centre of gravity on top of a tiny vertical column.
a What will the compass needle line up with when it is suspended (held at a point) so that it can move freely?
b What magnetic pole is at the geographical North Pole when the north pole of the compass needle points in that direction?
c Is the magnetic field stronger at the Equator or at the geographic North
Pole? Explain your answer.
d Why is a compass very difficult to use near to the north magnetic pole?

Interacting magnetic fields

Figure 16.10a: The attraction between two opposite magnetic poles shows up in their field pattern. b b b\mathbf{b} : The field pattern for two like poles repelling each other.
We can show the field patterns for two magnets attracting (Figure 16.10a) and repelling (Figure 16.10b) each other. Notice that there is a point between the two repelling magnets where there is no magnetic field.
When two bar magnets are placed close together their magnetic fields interact (affect each other) and produce a new pattern of magnetic lines of force. From these patterns, it is possible to say whether the magnets will attract or repel.
Notice that, in Figure 16.10b, the field lines from each magnet between the south poles are in the same direction. The field lines inside the box are all pointing downwards and are close together, so the field there is stronger than the field at the north poles. The magnets will move from where the field is stronger to where it is weaker - so like poles repel. A similar argument can be used to explain why
unlike poles attract. Activity 16.2 is an alternative model to explain magnetic attraction and repulsion.

ACTIVITY 16.2

An imaginary model to explain magnetic attraction and repulsion
Sometimes it helps to have an imaginary model to understand the physics which causes something to happen, for example, to understand the physics in magnetic attraction.
Imagine that magnetic field lines are like stretched elastic bands that want to pull the two ends closer together. Imagine also that there is pressure between the field lines pushing them apart and that, the closer the lines, the bigger this pressure is.
Sketch the field lines between and around:
  • a pair of like poles
  • a pair of unlike poles.
Use this imaginary model to explain the attraction and repulsion.
If you can visualise what is happening it should help you to remember how to sketch the patterns.

REFLECTION

Models are an important tool in physics.
Sketch the field lines between and around:
  • Did Activity 16.2 help you to visualise what is going on when magnets and magnetised objects attract and repel?
  • Did it help you to remember how to sketch the field patterns?
  • Can you suggest a better model?

Electromagnets

Using magnetic materials is only one way of making a magnet. An alternative method is to use an electromagnet. A typical electromagnet is made from a coil of copper wire. A coil like this is sometimes called a solenoid. When a current flows through the wire, there is a magnetic field around the coil (Figure 16.11). Copper wire is often used, because of its low resistance, though other metals will do. The coil does not have to be made from a magnetic material. The point is that it is the electric current that produces the magnetic field.
You can see that the magnetic field around a solenoid (Figure 16.11) is similar to that around a bar magnet (Figure 16.8). One end of the coil is a north pole, and the other end is a south pole. In Figure 16.11, the field lines emerge from the left-hand end, so this is the north pole.
There is no way to change the strength of a permanent magnet, but there are three ways to increase the strength of an electromagnet:
  • Increase the current flowing through it: the greater the current, the greater the strength of the field.
  • Increase the number of turns of wire on the coil: this does not mean making the coil longer, but packing more turns into the same space to concentrate the field.
  • Add a soft iron core: an iron core becomes strongly magnetised by the field, and this makes the whole magnetic field much stronger.
Figure 16.11: A solenoid. When a current flows through the wire, a magnetic field is produced. The field is similar in shape to that of a bar magnet. Note that the field lines go all the way through the centre of the coil.
Electromagnets have the great advantage that they can be switched on and off. Simply switch off the current and the field around the coil disappears. This is the basis of a number of applications. For example, the electromagnetic cranes that move large pieces of metal and piles of scrap around in a scrapyard (Figure 16.12). The current is switched on to start the magnet and pick up the scrap metal. When it has been moved to the correct position, the electromagnet is switched off and the metal is released.
Electromagnets are also used in electric doorbells, loudspeakers, electric motors, relays and transformers. These uses are described in detail later in Chapters 20 and 21.
Figure 16.12: Using an electromagnet in a scrapyard. With the current switched on, a steel object or pile of scrap can be lifted and moved. Then the current is switched off to release it.

Questions

7 a Name three ways to increase the strength of an electromagnet.
b What are the advantages of an electromagnet over a bar magnet.
c What does an electromagnet need that a permanent magnet does not?

ACTIVITY 16.3

How does the strength of an electromagnet depend on the number of turns?
Arun investigated how the strength of an electromagnet varies with the number of turns. Figure 16.13 shows the equipment he put together, and Table 16.3 shows the data he collected.
Arun made sure that the pole at the bottom of the electromagnet was the same as the top of the permanent magnet so that they would repel.
Figure 16.13: Apparatus for investigating the strength of an electromagnet.
Number of turns Mass /g
6 0.21
8 0.29
10 0.34
12 0.38
14 0.43
16 0.47
18 0.54
20 0.59
Number of turns Mass /g 6 0.21 8 0.29 10 0.34 12 0.38 14 0.43 16 0.47 18 0.54 20 0.59| Number of turns | Mass /g | | :---: | :---: | | 6 | 0.21 | | 8 | 0.29 | | 10 | 0.34 | | 12 | 0.38 | | 14 | 0.43 | | 16 | 0.47 | | 18 | 0.54 | | 20 | 0.59 |
Table 16.3: Experimental data
1 Plot a graph of mass against the number of turns.
2 The mass of the permanent magnet is not changing, so what is causing the apparent change in mass?
3 What does the graph tell Arun about the strength of the electromagnet as the number of turns increases?
4 Why is it important that the electromagnet and permanent magnet are repelling (and not attracting)? How could Arun have checked this before collecting data?
5 How else could Arun increase the strength of the electromagnet?

The field around a solenoid

When an electric current flows through a solenoid, a magnetic field is produced both inside and outside the coil (see Figure 16.11). The magnetic field inside an electromagnet is uniform. This means that the field lines are parallel and the same distance apart. The field around the outside of a solenoid is similar to that around a bar magnet:
  • One end of the solenoid is the north pole and the other end is the south pole. Field lines emerge from the north pole and go in to the south pole.
  • The field lines are closest together at the poles, showing that this is where the magnetic field is strongest.
  • The lines spread out from the poles, showing that the field is weaker in these regions.
    The strength of the field can be increased by increasing the current. The field can be reversed by reversing the direction of the current.

Question

8 a Sketch a diagram of the magnetic field pattern around a solenoid.
b How would the pattern change when the current through the solenoid is reversed?

EXPERIMENTAL SKILLS 16.1

Which magnetic pole sits below the geographical North Pole?
The red end of a compass needle points towards the North Pole of the Earth. If you look on the Internet, it is not clear whether the magnetic pole near the geographical North Pole is magnetic north or magnetic south. Therefore, you need to work out whether the red end of a compass needle is a north or south magnetic pole.
You need to ensure you know which way conventional current is flowing (from the positive to the negative of a battery or power pack).

You will need:

  • electromagnet
  • compass.

Method

1 Look along the electromagnet from either end.
If conventional current is flowing clockwise around the end loop, you are looking at the south pole like the arrows on the S S SS (as shown in Figure 16.14a).
If conventional current is running anticlockwise, like the arrows on the N (Figure 16.14b), you are looking at the north pole.
The magnetic poles in Figure 16.11 are drawn correctly. Imagine standing at the left-hand end of the electromagnet looking along it, as if you were looking down a long tunnel. Conventional current would be flowing anticlockwise, telling you that you were at the north pole.
a

b
Figure 16.14
2 Bring your compass towards either pole of the electromagnet. Use your observations to decide whether the red end of your compass needle is north or south. Use this information to decide which magnetic pole sits below the geographical North Pole.

PROJECT

Read this newspaper extract.

COMPASS CONFUSION: HIKER ALMOST KILLED BY HIS COMPASS

An experienced hiker from the USA who became lost in the Australian Outback was rescued by the police last week. He complained, ‘my compass stopped working’. A spokesman for the Western Australia Police Force was
quoted as saying, ‘this gentleman is. lucky to be alive. The hiker was using a compass designed for use in the USA and had not realised until it was too late that he needed to buy a compass here in Australia.’
You are a science journalist whose job it is to explain the science behind this story. You can decide whether you write an article, write a script for a television news item or produce a podcast.
Remember that your audience are not scientists. Sometimes explaining a problem needs ideas from two or more topics, as is the case here. You will need to explain why magnetic compasses stop working when used in the opposite hemisphere (for example, a compass designed for the northern hemisphere will not work in the southern hemisphere). You need to use diagrams or video clips to help your audience follow your explanation.
When using a compass, you should hold it flat (that is, parallel to the ground). The compass needle (sometimes called the card) should be parallel to the baseplate of the compass (when viewed from the side) and needs to be able to swing clockwise or anticlockwise when viewed from above. However, you should recall that a compass needle lines up with the magnetic field lines, which are not parallel to the ground in Australia.
You will need to explain why the magnetic field lines are not parallel to the ground in Australia. You need to sketch what the compass needle looks like when viewed from the side. It looks like a tiny see-saw balanced on a tiny pillar at its mid-point (look back at Chapter 4 if necessary). Use the principle of moments and the idea of equilibrium to explain how a compass needle is made to be parallel to the base plate (and the ground). Then explain why this is a particular problem when a compass designed for one hemisphere is taken and used in the other.

SUMMARY

Magnets have a north pole and a south pole.
Magnetic field lines are drawn with an arrow from north to south, which is the direction of the magnetic force.
Like poles repel and unlike poles attract.
Magnetic elements include iron, cobalt and nickel. Metal alloys (for example, steel) containing these elements are also magnetic.
Magnetic materials can be magnetised while non-magnetic materials (like rubber and glass) cannot be magnetised.
A permanent magnet can attract unmagnetised magnetic materials by inducing magnetism in them.
A hard magnetic material like steel is difficult to magnetise and demagnetise.
A soft magnetic material like soft iron is easy to magnetise and demagnetise.
A magnetic field is a region of space around a magnet or electric current in which a magnet will feel a force.
A magnetic field pattern around a (bar) magnet can be produced using a plotting compass.
A magnetic field line is the line (direction) of force on the north pole of the magnet, which is why a magnet (for example, a compass needle) lines up with it.
An electromagnet (or solenoid) is a magnet created when a current is passed through wire, which is usually shaped into a coil.
The strength of an electromagnet increases with the number of turns in the coil, the strength of the current and if it has a soft iron core.
The advantages of an electromagnet over a permanent magnet are that its strength can be changed, it can be switched on and off and it can be reversed.
The magnetic forces of attraction and repulsion between magnets are caused when their magnetic fields interact (affect each other) and produce a new pattern of magnetic lines of force.
The closer together magnetic field lines are to each other, the stronger the magnetic field.

EXAM-STYLE QUESTIONS

1 Some metals and alloys are magnetic. Which of these is magnetic?
A aluminium
B copper
C gold
D steel
2 Which arrangement of bar magnets leads to the field pattern revealed by the plotting compasses?
[1]
3 Which of these magnetic field patterns is correct?
4 a Copy this image of a bar magnet and draw the field lines you would expect to see around it.
N N NN S S SS
N S| $N$ | $S$ | | :--- | :--- |
b Explain how you could use a plotting compass to investigate the magnetic field around a bar magnet. You may draw a diagram to help your answer.
[Total: 5]
5 The diagram shows the magnetic field between magnetic poles X X XX and Y.

a Which pole is at X? [1]
b If the magnets are released, in which direction will they move?
[Total: 2]
6 Karamveer investigated how the strength of an electromagnet varied with current. He used an arrangement like the one in the diagram. He passed a current through an electromagnet so that it attracted a small steel plate. For each current he passed through the coil, he suspended more masses from the bottom of the steel plate until the steel was pulled away from the electromagnet.
Current / A Force / N
0
2 0.00
4 0.60
6 1.20
8 1.90
10 2.40
3.00
Current / A Force / N 0 2 0.00 4 0.60 6 1.20 8 1.90 10 2.40 3.00| Current / A | Force / N | | :---: | :--- | | 0 | | | 2 | 0.00 | | 4 | 0.60 | | 6 | 1.20 | | 8 | 1.90 | | 10 | 2.40 | | | 3.00 |
a Plot a graph of force against current.
b State how the strength of the electromagnet varies with the current passing through the coil.
c State two other ways that the strength of an electromagnet can be increased.
[2]
d An electromagnet can be switched on and off. Suggest one situation where this would be an advantage over the constant field of a permanent magnet.
e In terms of forces, state why the steel plate falls from the electromagnet once the suspended masses exceed a certain value.
[Total: 8]
7 a What is the difference between a magnetically soft and a magnetically hard material?
b Give an example of a magnetically hard material and suggest where it might be used.
b Give an example of a magnetically soft material and suggest where it might be used.

SELF-EVALUATION CHECKLIST

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
I can
See
Topic..
See Topic..| See | | :--- | | Topic.. |
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
Almost
there
Almost there| Almost | | :--- | | there |
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |
Recall the names of the two poles of a
magnet.
Recall the names of the two poles of a magnet.| Recall the names of the two poles of a | | :--- | | magnet. |
16.1
Recall the direction that the arrows on
a field line should point.
Recall the direction that the arrows on a field line should point.| Recall the direction that the arrows on | | :--- | | a field line should point. |
16.2
Recall what happens between like
poles and unlike poles.
Recall what happens between like poles and unlike poles.| Recall what happens between like | | :--- | | poles and unlike poles. |
16.2
Identify common magnetic materials. 16.1
Recall the difference between
magnetic and non-magnetic materials.
Recall the difference between magnetic and non-magnetic materials.| Recall the difference between | | :--- | | magnetic and non-magnetic materials. |
16.1
Account for induced magnetism. 16.2
Recall examples of hard and soft
magnetic materials and the difference
between them.
Recall examples of hard and soft magnetic materials and the difference between them.| Recall examples of hard and soft | | :--- | | magnetic materials and the difference | | between them. |
16.1
Draw the pattern of magnetic field
lines around a bar magnet and
describe an experiment to plot them,
including the direction arrows.
Draw the pattern of magnetic field lines around a bar magnet and describe an experiment to plot them, including the direction arrows.| Draw the pattern of magnetic field | | :--- | | lines around a bar magnet and | | describe an experiment to plot them, | | including the direction arrows. |
16.2
Recall why a compass needle or other
permanent magnet lines up with the
field lines when it is placed in a
magnetic field.
Recall why a compass needle or other permanent magnet lines up with the field lines when it is placed in a magnetic field.| Recall why a compass needle or other | | :--- | | permanent magnet lines up with the | | field lines when it is placed in a | | magnetic field. |
16.2
Describe electromagnets (and
solenoids).
Describe electromagnets (and solenoids).| Describe electromagnets (and | | :--- | | solenoids). |
16.2
Describe the three ways in which the
strength of an electromagnet can be
Describe the three ways in which the strength of an electromagnet can be| Describe the three ways in which the | | :--- | | strength of an electromagnet can be |
16.2
I can "See Topic.." "Needs more work" "Almost there" "Confident to move on" "Recall the names of the two poles of a magnet." 16.1 "Recall the direction that the arrows on a field line should point." 16.2 "Recall what happens between like poles and unlike poles." 16.2 Identify common magnetic materials. 16.1 "Recall the difference between magnetic and non-magnetic materials." 16.1 Account for induced magnetism. 16.2 "Recall examples of hard and soft magnetic materials and the difference between them." 16.1 "Draw the pattern of magnetic field lines around a bar magnet and describe an experiment to plot them, including the direction arrows." 16.2 "Recall why a compass needle or other permanent magnet lines up with the field lines when it is placed in a magnetic field." 16.2 "Describe electromagnets (and solenoids)." 16.2 "Describe the three ways in which the strength of an electromagnet can be" 16.2 | I can | See <br> Topic.. | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :--- | :--- | :--- | :--- | | Recall the names of the two poles of a <br> magnet. | 16.1 | | | | | Recall the direction that the arrows on <br> a field line should point. | 16.2 | | | | | Recall what happens between like <br> poles and unlike poles. | 16.2 | | | | | Identify common magnetic materials. | 16.1 | | | | | Recall the difference between <br> magnetic and non-magnetic materials. | 16.1 | | | | | Account for induced magnetism. | 16.2 | | | | | Recall examples of hard and soft <br> magnetic materials and the difference <br> between them. | 16.1 | | | | | Draw the pattern of magnetic field <br> lines around a bar magnet and <br> describe an experiment to plot them, <br> including the direction arrows. | 16.2 | | | | | Recall why a compass needle or other <br> permanent magnet lines up with the <br> field lines when it is placed in a <br> magnetic field. | 16.2 | | | | | Describe electromagnets (and <br> solenoids). | 16.2 | | | | | Describe the three ways in which the <br> strength of an electromagnet can be | 16.2 | | | |
changed.
Describe the advantages of an
electromagnet over a permanent
magnet.
Describe the advantages of an electromagnet over a permanent magnet.| Describe the advantages of an | | :--- | | electromagnet over a permanent | | magnet. |
16.2
Explain that magnetic forces are due
to interactions between magnetic
fields.
Explain that magnetic forces are due to interactions between magnetic fields.| Explain that magnetic forces are due | | :--- | | to interactions between magnetic | | fields. |
16.2
Use the spacing of the field lines to
work out the relative strength of the
magnetic field.
Use the spacing of the field lines to work out the relative strength of the magnetic field.| Use the spacing of the field lines to | | :--- | | work out the relative strength of the | | magnetic field. |
16.2
changed. "Describe the advantages of an electromagnet over a permanent magnet." 16.2 "Explain that magnetic forces are due to interactions between magnetic fields." 16.2 "Use the spacing of the field lines to work out the relative strength of the magnetic field." 16.2 | changed. | | | | | | :--- | :--- | :--- | :--- | :--- | | Describe the advantages of an <br> electromagnet over a permanent <br> magnet. | 16.2 | | | | | Explain that magnetic forces are due <br> to interactions between magnetic <br> fields. | 16.2 | | | | | Use the spacing of the field lines to <br> work out the relative strength of the <br> magnetic field. | 16.2 | | | |

> Chapter 17

Static electricity

IN THIS CHAPTER YOU WILL:

  • investigate the forces between positive and negative electric charges
  • explain static electricity in terms of gaining or losing electrons
  • distinguish between electrical conductors and insulators
    describe electric fields.

GETTING STARTED

In a small group, discuss what you know about electricity. Organise your ideas into a mind map. Include as many of the following words as possible, giving definitions and examples wherever you can: electron, charge, current, static, shock, voltage, conductor, insulator.
One member of the group should now stay with the map to explain it to other groups. The rest of the group should visit the maps of different groups. Having done this, return to your own group and add any additional information to your map.

A shocking phenomenon

Figure 17.1a: Lightning and electrical sparks are both caused by a large build-up of electric charge. b: electrical hazard symbol
The similarity between lightning bolts and the sign used for electrical hazards is not a coincidence. A large build-up of electrical charge can cause the air to break down and conduct electricity, potentially delivering a fatal electric shock. In December 2014, a potential difference of 1.3 billion volts
was recorded during a thunderstorm over Ooty in Southern India.
The US scientist Benjamin Franklin was lucky to survive his experiments into this kind of electricity. He suspected lightning was a form of electricity due to the similarity between a lightning flash and the sparks he produced in his laboratory. To investigate he flew a kite into a thundercloud. To avoid being electrocuted, he included a metal key at the bottom of the kite string, and attached a length of ribbon to the key. Holding the ribbon, he was relatively safe from electrocution (although other people were killed when they repeated his experiment). As a bolt of lightning struck the kite, Franklin saw the fibres of the kite string stand on end and a spark jumped from the key to the ground.
Franklin’s method was hazardous, and by flying a kite into the cloud, he may have brought on a lightning flash and thereby affected the situation he was investigating. Sunil Gupta and the team who measured the potential difference in the Ooty storm used a rather different method. They studied muons - a type of subatomic particle - as they passed through the storm cloud. The muons gained energy from the potential difference in the cloud, increasing the number which reached the scientists’ muon sensors. This allowed the team to calculate the potential difference within the cloud.

Discussion questions

1 Franklin made important discoveries about static electricity, but his experiment would be considered too dangerous to be carried out now. Should scientists be allowed to conduct hazardous investigations?
2 Gupta’s observations did not affect the situation he was studying. Why this is an important feature of scientific research?

17.1 Charging and discharging

We experience static electricity in a number of ways in everyday life, including lightning flashes. You may have noticed tiny sparks when you take off clothes made of synthetic fibres. You may have felt a small shock when getting out of a car. An electrostatic charge builds up on the car and then discharges through you when you touch the metal door. You may have rubbed a balloon on your clothes or hair and seen how it will stick to a wall or ceiling.
When you rub a plastic object (for example, a balloon) with a cloth, both are likely to become electrically charged. You can tell that this is so by holding the balloon or the cloth close to your hair - they attract the hair.
Figure 17.2: Objects which have electrostatic charge can attract light objects such as hair or paper.
You have observed that static electricity is generated by friction. You have also observed that a charged object may attract uncharged objects.
Now we have to think systematically about how to investigate this phenomenon. First, how do two charged objects affect one another? Figure 17.3 shows one way of investigating this. A plastic rod is rubbed with a cloth so that both become charged. The rod is hung in a cradle so that it is free to move. When the cloth is brought close to it, the rod moves towards the cloth (Figure 17.3a). When a second rod is rubbed in the same way and brought close to the first one, the hanging rod moves away (Figure 17.3b). Now we have seen both attraction
and repulsion. This suggests that there are two types of static electricity. Both rods have been treated in the same way, so we expect them to have the same type of electricity. The cloth and the rod must have different types.
The two types of static electricity are referred to as positive charge and negative charge. We can explain the experiments shown in Figure 17.3 by saying that the process of rubbing gives the rods one type of electric charge (say, negative), while the cloth is given the opposite type (say, positive). Figures 17.3c and 17.3 d show the two experiments with the charges marked.
From these experiments, we can also say something about the forces that electric charges exert on each other:
  • Two positive charges will repel each other.
  • Two negative charges will repel each other.
  • A positive charge and a negative charge attract each other.
Figure 17.3: Two experiments to show the existence of two, opposite, types of static electricity. a: The charged rod and cloth attract one another. b: The two charged rods repel one another. c: The rod and the cloth have opposite electric charges. d: The two rods have the same type of electric charge.
You can see that this rule is similar to the rule we saw for magnetic poles in
Chapter 16. But do not confuse magnetism with static electricity! Magnetism arises from magnetic poles - static electricity arises from electric charges. When you rub a plastic rod, you are not making it magnetic.

Questions

1 Copy and complete these sentences.
There are two types of charge called qquad\qquad and qquad\qquad .
Two objects with the same charge will qquad\qquad and two objects with opposite charges will qquad\qquad .
2 Figure 17.4 shows a child’s hair standing on end after the child has been on a trampoline.

Figure 17.4

a How does bouncing on the trampoline cause the child’s hair to become charged?
b The individual hairs are repelling each other. What does this tell you?
3 Explain why, after walking on a nylon carpet, you may get a small shock when you touch a metal door handle.

EXPERIMENTAL SKILLS 17.1

Investigating static electricity
In this experiment you will find out about static electricity by charging materials and observing how they behave.

You will need:

  • polythene and acetate or glass rods
  • cloths of different materials
  • paper stirrups with fine thread
  • clamp and stand
  • watch glass
  • balloons
  • tiny pieces of thin paper.

Getting started

Investigate how much you need to rub the rods in order to charge them.
Test this by using the rod to pick up little pieces of paper.
Figure 17.5: Set-up for investigation.

Method

1 You need to be sure that you can place the rods so that they can turn freely, either by hanging them in the paper stirrup, or by placing them on an upturned watch glass. Try this out with your rods.
2 Rub a polythene rod with a woollen cloth, making sure that you rub the full length of the rod. Hang the rod or place it on a watch glass.
3 Rub another polythene rod and bring one end close to an end of the first rod. Do they attract or repel?
4 Rub an acetate or glass rod and repeat the test. What do you observe?
5 Try different combinations of rods. Try a cloth of a different fabric. Given that a polythene rod, rubbed with a woollen cloth, gains a negative charge, what can you say about the charges gained by the cloth and by the other rods?
6 Blow up a balloon and rub it. Can you determine whether it gains positive or negative charge? (Hint: polythene becomes negative and acetate becomes positive.)

Questions

1 Some students investigate polythene and acetate rods. They find that the two rods attract each other. One student says this proves that the rods have opposite charge. Explain why the student is wrong.
2 Describe what the students would need to do in order to prove that the rods have opposite charges.

17.2 Explaining static electricity

The Ancient Greeks knew something about static electricity. Amber is a form of resin from trees, which hardens and becomes fossilised. It looks like clear, orange plastic. The Greeks knew that, when rubbed, amber could attract small pieces of cloth or hair.
Like the Ancient Greeks, Franklin had no idea about electrons - these particles were not discovered until 100 years later. However, that did not stop Franklin from developing a good understanding of static electricity. In the discussion that follows, we will talk about electrons. They make it much easier to understand what is going on in all aspects of electricity.
Figure 17.6: An insect trapped in resin as it hardened. Resin is a good insulator and easily becomes electrostatically charged.

Friction and charging

It is the force of friction that causes charging. When a plastic rod is rubbed on a cloth, friction transfers tiny particles called electrons from one material to the other. When the rod is made of polythene, usually electrons are transferred from the cloth to the rod.
Electrons are a part of every atom. They are negatively charged, and they are found on the outside of the atom. The nucleus in the centre of the atom has a positive charge. The positive and negative charges in an atom are equal so overall an atom has no electric charge - we say that it is neutral. Since the outer electrons are relatively weakly held in the atom, they can be pulled away by the force of friction. When an atom has lost an electron, it becomes positively charged.
Charging is always the result of gaining or losing electrons. Positive charge is not transferred.
  • An object which gains electrons becomes negative.
  • An object which loses electrons becomes positive.
Since a polythene rod becomes negatively charged when it is rubbed with a cloth, we can imagine electrons being transferred from the cloth to the rod (see Figure 17.8). It is difficult to explain why one material pulls electrons from another. The atoms that make up polythene contain positive charges, and these must attract electrons more strongly than those of the cloth.
Remember that it takes two different materials to generate static electricity. One material becomes positive, the other negative.

Conductors and insulators

You may have noticed that all the examples of objects becoming charged involve non-metals. Metals are electrical conductors, which means electrons can move through them and the metal doesn’t stay charged. Gold and copper are particularly good electrical conductors. Non-metals, such as glass, plastic and amber are electrical insulators.
So why can charge move through conductors but not through insulators? In
insulators, the electrons are tightly bound to their atoms and not easily removed. In conductors some of the electrons are free to move between atoms (these electrons are sometimes referred to as free electrons).
When you rub a polythene rod, it gains electrons from the cloth and so becomes negatively charged. The electrons cannot move through the polythene, so the end which was rubbed remains charged.
When a copper rod is rubbed, electrons are also transferred by friction, but these electrons are free to move, so they flow through the rod, through your hand and into the Earth. This means the copper rod does not become charged.
A metal object can be charged if it is held by an insulating handle. In this case the charge will spread evenly through the conductor.

EXPERIMENTAL SKILLS 17.2

Investigating conductors and insulators

In this experiment you will find test materials to find out which are conductors and which are insulators.

You will need:

  • cell
  • lamp
  • wires with crocodile clips
  • materials to test.

Getting started

Connect the cell and lamp in a simple circuit to make the lamp light.
Make a gap in the circuit by removing a wire. Explain why the lamp no longer lights and consider how placing materials in the gap will help you decide if they are conductors or insulators.

Method

1 Connect the circuit as shown in Figure 17.7.
Figure 17.7: Circuit for testing materials.
2 Using the crocodile clips, attach a material into the gap in the circuit.
3 Observe whether the lamp lights. If it does, it is a conductor, if not it is an insulator.
4 Record your results in a table.

Questions

1 The wires you used are made of copper covered in plastic. Explain why these materials were chosen.
Figure 17.8: When a polythene rod is rubbed with a cloth, electrons are
transferred from the cloth to the polythene. The rod now has an overall negative charge and the cloth has an overall negative charge.

Questions

4 Copy and complete these sentences.
When a polythene rod is rubbed, qquad\qquad move from the qquad\qquad to the
qquad\qquad .
This means the rod becomes qquad\qquad and the cloth becomes qquad\qquad .
5 a Draw a diagram similar to Figure 17.8 to show how an acetate rod becomes positively charged by losing electrons.
b Write sentences similar to those you completed in question 17.4 to explain how the acetate becomes charged.
6 When you hold a polythene rod and rub it with a cloth, it becomes charged. If the rod is made of metal, it will not become charged. Explain this difference by describing what happens to the electrons in each case.

ACTIVITY 17.1: WHAT HAVE I LEARNT?

Take two minutes to summarise what you have learnt so far in this chapter on one sheet of paper.
Now, take two more minutes to condense your notes. You may use ten words and four diagrams.
Compare summaries with a partner. Assume that you can use only four words and one diagram as a reminder in a test. Which diagram and four words would you choose?

REFLECTION

Which part of Activity 17.1 did you find most useful? Summarising, condensing or sharing? Consider how this might help you when revising for a test.

17.3 Electric fields

Figure 17.9: The comb has been charged by rubbing it. It attracts the water when it is held close to the water.
A charged object can affect other objects, both charged and uncharged, without actually touching them. For example, a charged plastic rod can exert a force on another charged rod placed close by.
We say that there is an electric field around a charged object. Any charged object placed in the field will experience a force on it.
There are similarities between electric fields and magnetic fields, but take care not to confuse electric fields with magnetic fields. A magnet does not attract electric charges. A charged object does not attract a magnet.

Representing an electric field

A charged object is surrounded by an electric field. If a charged object moves into the electric field of a charged object, it will experience a force - it will be attracted or repelled. Figure 17.10 shows how we represent an electric field by lines of force (or electric field lines). This is similar to the way we represent a magnetic field by magnetic field lines.
The lines of force are shown coming out of a positive charge and going in to a negative charge. This is because the lines indicate the direction of the force on a positive charge placed in the field. A positive charge is repelled by another positive charge and attracted by a negative charge.
Figure 17.10c shows the field between two oppositely charged parallel plates. The lines of force between the plates are straight and parallel to one another (except at the edges).
Figure 17.10: The electric field around a charged object is represented by lines of force. a: An isolated positive charge. b: A negatively charged sphere. c: Two parallel plates with opposite charges.

Question

7 Draw the magnetic field around:
a an isolated negative charge
b two vertical metal plates - one positive and the other negative.

What is electric charge?

In physics, we find it relatively easy to answer questions such as ‘What is a rainbow?’ or ‘How does an aircraft fly?’ It is much harder to answer an apparently simple question like ‘What is electric charge?’. As with energy, we have to answer it by explaining how objects behave when they have it (energy or charge).
Objects with the same sign of charge repel one another. Objects with opposite charges attract. This is not a very satisfying answer, because magnetic poles
behave in the same way: north poles repel north poles and attract south poles. Because electric charge is a fundamental property of matter, we have to get a feel for it, rather than having a clear definition.
The electric force between two charged objects is one of the fundamental forces of nature. (The force of gravity between two masses is another fundamental force.) The electric force holds together the particles that make up an atom. It holds atoms together to make molecules, and it holds molecules together to make solid objects. Just think: whenever you stand on the floor, it is the electric force between molecules that prevents you from falling through the floor. It is a very important force.

Charged particles

We have already seen that electrons are the charged particles that are transferred from one object to another when they are rubbed together. Electric charge is a property of the particles that make up atoms.
Charge is measured in coulombs ©, named after Charles-Augustin de Coulomb, a French physicist. He discovered that the force between two charged objects depends on how big their charges are and on how far apart they are.
An electron is a negatively charged particle. It is much smaller than an atom, and only weakly attached to the outside of the atom. It is held there by the attraction of the positively charged nucleus of the atom. The nucleus is positively charged because it contains positively charged particles called protons.
An electron has a very tiny amount of electric charge. The electron charge is so small that it takes more than 6 million million million electrons to make 1 C of charge:
electron charge = 0.00000000000000000016 C or 1.6 × 10 19 C  electron charge  = 0.00000000000000000016 C  or  1.6 × 10 19 C {:[" electron charge "=-0.00000000000000000016C],[" or "-1.6 xx10^(-19)C]:}\begin{aligned} & \text { electron charge }=-0.00000000000000000016 \mathrm{C} \\ & \text { or }-1.6 \times 10^{-19} \mathrm{C} \end{aligned}
A proton has exactly the same size of charge as an electron, but positive, so the proton charge is:
proton charge = + 0.00000000000000000016 C or + 1.6 × 10 19 C  proton charge  = + 0.00000000000000000016 C  or  + 1.6 × 10 19 C {:[" proton charge "=+0.00000000000000000016C],[" or "+1.6 xx10^(-19)C]:}\begin{aligned} & \text { proton charge }=+0.00000000000000000016 \mathrm{C} \\ & \text { or }+1.6 \times 10^{-19} \mathrm{C} \end{aligned}
No-one knows why these values are exactly the same size (or even if they are exactly the same size), but it is fortunate that they are because it means that an atom that contains, say, six protons and six electrons is electrically neutral. If all
the objects around us were made of charged atoms, we would live in a shocking world!

Question

8 Calculate the number of electrons needed to give a charge of one coulomb.

PROJECT

Shocking shopping
Figure 17.11: A carpet in a shop can cause a shock.
Walking across a carpet can cause friction, which leads to people becoming electrostatically charged. Touching another person, or a conducting object such as a metal stair bannister, can cause a shock as the charge flows through to earth, discharging the person. These shocks are not usually dangerous but can be uncomfortable.
A manager of a large store has received a lot of complaints from customers who are experiencing painful shocks as they shop. The manager thinks it may be due to the carpet on the shop floor which is made of a hard-wearing material called polypropylene. The store manager wants to investigate whether she should invest in a new carpet that is less likely to create static. Until she does this, she wants to be able to advise customers on how to avoid shocks.
Your tasks as her scientific consultant are:
1 Explain what is happening. As an interim measure, she would like
leaflets to give to shoppers to help them understand what is happening and what they can do to minimise shocks. Could customers discharge regularly before too much charge builds up? Or would it help if they picked up their feet rather than shuffling?
2 Design the leaflet. It must have no more than 50 words and it should be illustrated.
3 Consider how new carpets could be tested to find out which create the most static. Testing by shocking people is not easy or ethical. Instead you should look at charging the carpets by friction and investigating the force of attraction or repulsion they produce. You will need to give step-by-step instructions for a fair test investigation.

PEER ASSESSMENT

Exchange leaflets and investigation plans with another student. Give them written feedback by answering the following points:
  • Does the leaflet make it clear how charging happens?
  • Does the leaflet give clear advice on avoiding shocks?
  • Will the investigation provide valid data?
  • Does the plan control all variables in order to give a fair test?

SUMMARY

There are two types of charge: positive and negative.
Like charges repel, opposite charges attract.
Conductors allow charge to flow. Insulators do not allow charge to flow.
Insulators can be charged by friction, which causes the loss or gain of electrons.
An electric field is the area around a charged object in which a charge will experience a force.
Electric fields can be represented by field lines which show the direction of the force on a positive charge.
Insulators are materials in which electrons are fixed in place. Conductors have free electrons.

EXAM-STYLE QUESTIONS

1 Which of these describes how an object becomes positively charged?
A It gains positive charge.
B It loses positive charge.
C It gains negative charge.
D It loses negative charge.
2 When an object is charged by friction, which particle is transferred?
A proton
B electron
C neutron
D atom
3 A positively charged rod is suspended so it can move freely. A negatively charged rod is brought close to it. What will happen to the positive rod?
A It will repel.
B Nothing will happen.
C It isn’t possible to predict what will happen from the information given.
D It will attract.
4 What does the arrow on an electric field line show?
A The force on a positive charge.
B The force on a neutral particle.

C The force on an electron.
D The force on a negative particle.

5 A student investigates charging polythene by friction. She takes two identical strips and rubs each with a cloth. She holds the two strips together and they move apart.

a Explain why the strips move apart
The student repeats the experiment using two strips of aluminium.
b State and explain what the student would see with the aluminium strips
c A plastic rod is charged by rubbing it with a cloth.
Which statement described what has happened to the cloth?
A The cloth becomes negatively charged because electrons transfer from the cloth to the rod
B The cloth becomes positively charged because electrons transfer from the cloth to the rod
C The cloth becomes negatively charged because protons transfer from the rod to the cloth
D The cloth becomes positively charged because protons transfer from the rod to the cloth.
6 A student combs his hair with a plastic comb. The comb becomes negatively charged.
a Explain how this happens.
[2]
b Name the particles which are transferred.
c The student notices his hair is now standing on end. Explain why this happens to his hair.
[Total: 5]
7 a the term ‘electric field’.
b Copy these diagrams and draw the electric field around the charged objects:

[Total: 7]
8 A student rubs a balloon so that it becomes positively charged. He places the balloon on top of an insulating sheet on an electronic balance.

a Explain, in terms of movement of electrons, how the balloon becomes positively charged.
b The student now brings a charged plastic rod close to the balloon. The reading on the balance increases to 11.63 g . State and explain what this tells you about the plastic rod.
c The student repeats the experiment, but this time he places the balloon directly on to the metal pan of the balance. This time when he brings the charged rod close to the balloon, the reading on the scale decreases to 6.94 g. Explain why.
9 A painter is spray painting a metal fence. The nozzle of the sprayer is negatively charged.
a State and explain what happens to the paint droplets as they pass through the nozzle.
b Describe two ways in which this makes the spray painting more efficient.[2]
c What charge should be given to the fence to make the process even more efficient?

SELF-EVALUATION CHECKLIST

After studying this chapter, think about how confident you are with the different topics. This will help you to see any gaps in your knowledge and help you to learn more effectively.
I can
See
Topic...
See Topic...| See | | :--- | | Topic... |
Needs
more
work
Needs more work| Needs | | :--- | | more | | work |
Almost
there
Almost there| Almost | | :--- | | there |
Confident
to move
on
Confident to move on| Confident | | :--- | | to move | | on |
Name the two types of charge. 17.1
Say how objects with identical or
opposite charges affect each other.
Say how objects with identical or opposite charges affect each other.| Say how objects with identical or | | :--- | | opposite charges affect each other. |
17.1
Describe experiments to show how
charges can be produced and detected.
Describe experiments to show how charges can be produced and detected.| Describe experiments to show how | | :--- | | charges can be produced and detected. |
17.2
Explain how charging by friction
happens by describing what happens
to the electrons.
Explain how charging by friction happens by describing what happens to the electrons.| Explain how charging by friction | | :--- | | happens by describing what happens | | to the electrons. |
17.2
Define and give examples of
conductors and insulators.
Define and give examples of conductors and insulators.| Define and give examples of | | :--- | | conductors and insulators. |
17.2
Name the unit of electrical charge.
Name the unit of electrical charge.| Name the unit of electrical charge. | | :--- |
17.3
Define what is meant by an electric
field.
Define what is meant by an electric field.| Define what is meant by an electric | | :--- | | field. |
17.3
Draw the field between two charged
plates or around a point charge of a
charged sphere, using arrows to show
the direction of the field.
Draw the field between two charged plates or around a point charge of a charged sphere, using arrows to show the direction of the field.| Draw the field between two charged | | :--- | | plates or around a point charge of a | | charged sphere, using arrows to show | | the direction of the field. |
17.3
Describe the difference between
conductors and insulators in terms of a
simple electron model.
Describe the difference between conductors and insulators in terms of a simple electron model.| Describe the difference between | | :--- | | conductors and insulators in terms of a | | simple electron model. |
17.3
I can "See Topic..." "Needs more work" "Almost there" "Confident to move on" Name the two types of charge. 17.1 "Say how objects with identical or opposite charges affect each other." 17.1 "Describe experiments to show how charges can be produced and detected." 17.2 "Explain how charging by friction happens by describing what happens to the electrons." 17.2 "Define and give examples of conductors and insulators." 17.2 "Name the unit of electrical charge." 17.3 "Define what is meant by an electric field." 17.3 "Draw the field between two charged plates or around a point charge of a charged sphere, using arrows to show the direction of the field." 17.3 "Describe the difference between conductors and insulators in terms of a simple electron model." 17.3 | I can | See <br> Topic... | Needs <br> more <br> work | Almost <br> there | Confident <br> to move <br> on | | :--- | :---: | :--- | :--- | :--- | | Name the two types of charge. | 17.1 | | | | | Say how objects with identical or <br> opposite charges affect each other. | 17.1 | | | | | Describe experiments to show how <br> charges can be produced and detected. | 17.2 | | | | | Explain how charging by friction <br> happens by describing what happens <br> to the electrons. | 17.2 | | | | | Define and give examples of <br> conductors and insulators. | 17.2 | | | | | Name the unit of electrical charge. | 17.3 | | | | | Define what is meant by an electric <br> field. | 17.3 | | | | | Draw the field between two charged <br> plates or around a point charge of a <br> charged sphere, using arrows to show <br> the direction of the field. | 17.3 | | | | | Describe the difference between <br> conductors and insulators in terms of a <br> simple electron model. | 17.3 | | | |

> Chapter 18

Electrical quantities

IN THIS CHAPTER YOU WILL:

  • learn about electric current, resistance and voltage
  • describe an experiment to determine resistance using ammeters and voltmeters
  • learn how the resistance of a wire relates to its length and diameter
  • understand that energy is transferred from the power source (for example, a battery) to the circuit components
  • calculate resistance, electrical power, energy and the cost of electrical energy
    learn that electrons flow in the opposite direction to conventional current
sketch and explain current-voltage characteristics > > >> calculate electric current and potential difference.

GETTING STARTED

Does anything get used up going around a circuit?
What do batteries and cells have that gets used up?
Why can birds stand on power lines without being electrocuted?
Do you know what happens when we charge a mobile (cell) phone?

INSPIRED BY AN EEL? USING PHYSICS FOR LAW ENFORCEMENT

Jean Richer, a French astronomer, met an electric eel (Figure 18.1) during an expedition to the Amazon basin in 1671. He described it as being ‘as fat as a leg’ and it made his arm go numb for 15 minutes after he touched it with his finger. About 80 % 80 % 80%80 \% of an electric eel is basically a battery, with a positive terminal at its head and a negative terminal at its tail. The charge is carried by positive ions instead of electrons. It has about 5000 to 6000 stacked plates (called electroplaques). Each pair of plates is like an electric cell that can produce a voltage of about 0.15 V so that an eel can deliver a shock of 900 V and a current of up to 1 A . To save energy it can send out low voltage pulses ( 75 V ). This makes the muscles of potential prey twitch, which can be detected by the eel. If it detects prey, the eel can send out bigger shocks when it wants to stun its victim. Some eels were taken back to be studied in Europe where they inspired the work of Italian scientists.
Luigi Galvini used frogs to show that electricity made their legs move. Alessandro Volta invented the first battery based on the electric eel.
Figure 18.1: An electric eel.
NASA researcher Jack Cover finished developing the TASER in 1974. He named it after the book Tom Swift and His Electric Rifle. TASERs are used by law enforcement officers to subdue suspected criminals without using lethal force (Figure 18.2). The TASER fires two thin insulated copper wires at the target. Once the barbs at the end of the wires hook into the skin, the person becomes part of the circuit. When a current is passed through the circuit, the person loses control of their muscles, like the victims of the electric eel.
Figure 18.2: Even without the cartridge that sends the copper wires to its target, a TASER still passes an electrical charge.

Discussion questions

1 Describe how an electric eel is like a battery.
2 A TASER is like an electric eel but why does the eel not need the wires used in a TASER?

18.1 Current in electric circuits

We use electric circuits to transfer energy from a battery or power supply to components in the circuit, which then transfer the energy to their surroundings. For an electric current to flow, two things are needed: a complete circuit for it to flow around, and something to ‘push’ it around the circuit. The push might be provided by a cell, battery or power supply. A battery is simply two or more cells connected end-to-end. In most familiar circuits, metals such as copper or steel provide the circuit for the current to flow around.
Figure 18.3a shows how a simple circuit can be set up in the laboratory. Once the switch is closed, there is a continuous metal path for the current to flow along. Current flows from the positive terminal of the battery (or cell). In the circuit symbol for a cell, the longer line represents the positive terminal (see Figure 18.3b). Current flows through the switch and the filament lamp, back to the negative terminal of the battery. A current that flows in the same direction all the time is called direct current (d.c.). You will meet alternating current (a.c.) in Chapter 21 when you learn about transformers. Alternating current is when current reverses direction. In many countries, mains electricity has a frequency of 50 Hz so it changes direction 100 times per second.
Figure 18.3b shows the same circuit as represented by a circuit diagram. Each component has its own standard symbol. If you imagine the switch being pushed so that it closes, it is clear from the diagram that there is a continuous path for the current to flow around the circuit.
Figure 18.3a: A simple electric circuit, set up in a laboratory. b: The same circuit represented as a circuit diagram.
It is obvious how the switch in Figure 18.3a works. You push the springy metal downwards until it touches the other metal contact. Then the current can flow through it. Most switches work by bringing two pieces of metal into contact with one another, though you cannot usually see this happening. It is worth having a look inside some switches to see how they work. (Of course, they must not be connected in a circuit when you examine them!)
Similarly, take a look at some filament lamps, like the one in Figure 18.3a. Every lamp has two metal contacts, for the current to flow in and out. Inside, one fine wire carries the current up to the filament (which is another wire), and a second wire carries the current back down again. Notice also how the circuit
symbols for these and many other components have two connections for joining them into a circuit.

Good conductors, bad conductors

The wires we use to connect up circuits are made of metal because metals are good conductors of electric current. The metal is usually surrounded by plastic, so that, if two wires touch, the electric current cannot pass directly from one to another (causing a short circuit). Plastics (polymers) are good electrical insulators.
  • Most metals, including copper, silver, gold and steel are good conductors.
  • Polymers (such as Perspex® or polythene), minerals and glass are good insulators.
In between, there are many materials that do conduct electricity, but not very well. For example, liquids may conduct, but they are generally poor conductors.
People can conduct electricity - that is what happens when you get an electric shock. A current passes through your body and, if it is big enough, it makes your muscles contract violently. Your heart may stop, and you may get burns. Our bodies conduct because the water in our tissue is quite a good electrical conductor.

What is electric current?

When a circuit is complete, an electric current flows. Current flows from the positive terminal of the supply, around the circuit, and back to the negative terminal.
What is actually travelling around the circuit? The answer is electric charge. The battery or power supply in a circuit provides the push needed to make the current flow. This push is the same force that causes electric charges to attract or repel one another.
A current is a flow of electric charge. In a metal, the current is a flow of electrons. These are the negatively charged particles you learned about in Chapter 17.

Measuring electric current

To measure electric current, we use an ammeter. There are two types, as shown in Figure 18.4.
Figure 18.4: Ammeters measure electric current, in amps (A). There are two types: analogue (on the left) and digital (on the right).
  • An analogue meter has a needle, which moves across a scale. You have to make a judgement of the position of the needle against the scale.
  • A digital meter gives a direct read-out in figures. There is no judgement involved in taking a reading.
A galvanometer is sometimes used instead of an ammeter when tiny currents need to be measured. It has a different circuit symbol - an upward pointing arrow to represent a needle. An ammeter is connected into a circuit in series, that is, between other components in the circuit. This circuit is called a series circuit, where components are connected in a line between other components. If the meter is connected the wrong way round, it will give negative readings. To add an ammeter to a circuit, the circuit must be broken (see Figure 18.5).
Figure 18.5: Adding an ammeter to a circuit. The ammeter is connected in series so that the current can flow through it.
In a simple series circuit like the one shown in Figure 18.5, it does not matter where the ammeter is added, since the current is the same all the way round the series circuit. It does not get used up as it flows through the lamp or other components in the circuit.
The reading on an ammeter is in amperes (shortened to amps (A), which is the SI unit of current. Smaller currents may be measured in milliamps (mA) or microamps ( μ A μ A muA\mu \mathrm{A} ):
1 mA = 0.001 A = 10 3 A 1 μ A = 0.000001 A = 10 6 A 1 mA = 0.001 A = 10 3 A 1 μ A = 0.000001 A = 10 6 A {:[1mA=0.001A=10^(-3)A],[1muA=0.000001A=10^(-6)A]:}\begin{aligned} & 1 \mathrm{~mA}=0.001 \mathrm{~A}=10^{-3} \mathrm{~A} \\ & 1 \mu \mathrm{~A}=0.000001 \mathrm{~A}=10^{-6} \mathrm{~A} \end{aligned}

Questions

1 a What instrument is used to measure electric current?
b How should it be connected in a circuit?
c Draw its circuit symbol.
2 A circuit is set up in which a cell makes an electric current flow through a lamp. Two ammeters are included, one to measure the current flowing into the lamp, the other to measure the current flowing out of the lamp.
a Draw a circuit diagram to represent this circuit.
b Add an arrow to show the direction of the current around the circuit.
c What can you say about the readings on the two ammeters?
3 a Name two materials that are good electrical conductors.
b Name two materials that are good electrical insulators.

Two pictures: current and electrons

Metals are good electrical conductors because they contain electrons that can move about freely. (This has already been mentioned in Chapters 11 and 17.) The idea is that, in a bad conductor, such as most polymers, all of the electrons in the material are tightly bound within the atoms or molecules, so that they cannot move. Metals are different. While most of the electrons in a metal are tightly bound within their atoms, some are free to move about within the material. These are called conduction electrons (see Figure 18.6). A voltage, such as that provided by a battery or power supply, can start these conduction electrons moving in one direction through the metal, and an electric current flows. Since electrons have a negative electric charge, they are attracted to the positive terminal of the battery.
Figure 18.6: In a metal, some electrons are free to move about. In copper, there is one conduction electron for each atom of the metal. The atoms, having lost an electron, are positively charged. A battery pushes the conduction electrons through the metal. The force is the attraction between unlike charges that was discussed in Chapter 17.
Conventional electric current flows from positive to negative. Figure 18.7 shows the direction of the flow of charge around a simple circuit. We picture positive charge flowing out of the positive terminal, around the circuit and back into the cell at the negative terminal. Now, we know that in a metal it is the negatively charged electrons that move. They leave the negative terminal of the cell, and flow around to the positive terminal, in the opposite direction to the current. Hence we have two different pictures of what is going on in a circuit.
Figure 18.7: Two ways of picturing what happens in an electric circuit: conventional current flows from positive to negative; electrons flow from negative to positive.
We can think of conventional current, a flow of positive charge, moving from positive to negative. Conventional current is rather like a fluid moving through the wires, just like water moving through pipes. This picture does not tell us anything about what is going on inside the wires or components of a circuit. However, it is perfectly good for working out many things to do with a circuit: what the voltage will be across a particular component, for example, or how much energy will be transferred to a particular lamp.
Alternatively, we can think of electron flow as, a movement of conduction
electrons, from negative to positive. This picture can allow us to think about what is going on inside the components of a circuit: why a resistor gets warm when a current flows through it, for example, or why a diode allows current to flow in one direction only.
These two pictures are both models. The electron flow picture is a microscopic model, since it tells us what is going on at the level of very tiny particles (electrons). The conventional current picture is a macroscopic or large scale model.
The electrons in a circuit flow in the opposite direction to the electric current. It is a nuisance to have to remember this. It stems from the early days of experiments on static electricity. Benjamin Franklin realised that there were two types of electric charge, which he called positive and negative. He had to choose which type he would call positive. He chose to say that when amber was rubbed with a silk cloth, the amber acquired a negative charge. Franklin was setting up a convention, which other scientists then followed - hence the term ‘conventional current’. He had no way of knowing that electrons were being rubbed from the silk to the amber, but his choice means that we now say that electrons have a negative charge. Remember that conventional current and electron flow move in opposite directions around a circuit.

Current and charge

An ammeter measures the rate at which electric charge flows past a point in a circuit. The electric current is defined as the charge passing a point in the circuit per unit of time (usually per second). We can write this relationship between current and charge as an equation using the quantities and symbols given in Table 18.1:
current ( A ) = charge ( C ) time ( s ) I = Q t current ( A ) = charge ( C ) time ( s ) I = Q t {:[current(A)=(charge(C))/(time(s))],[I=(Q)/(t)]:}\begin{aligned} \operatorname{current}(\mathrm{A}) & =\frac{\operatorname{charge}(\mathrm{C})}{\operatorname{time}(\mathrm{s})} \\ I & =\frac{Q}{t} \end{aligned}

KEY EQUATION

current ( A ) = charge ( C ) time ( s ) I = Q t current ( A ) =  charge  ( C ) time ( s ) I = Q t {:[current(A)=(" charge "(C))/(time(s))],[I=(Q)/(t)]:}\begin{aligned} \operatorname{current}(\mathrm{A}) & =\frac{\text { charge }(\mathrm{C})}{\operatorname{time}(\mathrm{s})} \\ I & =\frac{Q}{t} \end{aligned}
Quantity
Symbol for
quantity
Symbol for quantity| Symbol for | | :--- | | quantity |
Unit Symbol for unit
current I I II amps A
charge Q Q QQ coulombs C
time t t tt seconds s
Quantity "Symbol for quantity" Unit Symbol for unit current I amps A charge Q coulombs C time t seconds s| Quantity | Symbol for <br> quantity | Unit | Symbol for unit | | :--- | :---: | :--- | :---: | | current | $I$ | amps | A | | charge | $Q$ | coulombs | C | | time | $t$ | seconds | s |
Table 18.1: Symbols and units for some electrical quantities.
So a current of 10 A passing a point means that 10 C of charge flows past that point every second. You may find it easier to recall this relationship in the following form:
charge ( C ) = current ( A ) × ( C ) = current ( A ) × (C)=current(A)xx(\mathrm{C})=\operatorname{current}(\mathrm{A}) \times time ( s ) ( s ) (s)(\mathrm{s})
Q = I t Q = I t Q=It\mathrm{Q}=I t
So, if a current of 10 A flows around a circuit for 5 s , then 50 C of charge flows around the circuit.
Worked Example 18.1 shows how to calculate the charge that flows in a circuit.

WORKED EXAMPLE 18.1

1 A current of 150 mA flows around a circuit for one minute. How much electric charge flows around the circuit in this time?
Step 1: Write down what you know, and what you want to know. Put all quantities in the units shown in Table 18.1.
I = 150 mA = 0.15 A ( or 150 × 10 3 A ) t = 1 minute = 60 s Q = ? I = 150 mA = 0.15 A  or  150 × 10 3 A t = 1  minute  = 60 s Q = ? {:[I=150mA=0.15A(" or "150 xx10^(-3)(A))],[t=1" minute "=60s],[Q=?]:}\begin{aligned} & I=150 \mathrm{~mA}=0.15 \mathrm{~A}\left(\text { or } 150 \times 10^{-3} \mathrm{~A}\right) \\ & t=1 \text { minute }=60 \mathrm{~s} \\ & Q=? \end{aligned}
Step 2: Write down an appropriate form of the equation relating Q , I Q , I Q,IQ, I
and t t tt. Substitute values and calculate the answer.
Q = I t Q = 0.15 A × 60 s = 90 C Q = I t Q = 0.15 A × 60 s = 90 C {:[Q=It],[Q=0.15Axx60s=90C]:}\begin{aligned} & Q=I t \\ & Q=0.15 \mathrm{~A} \times 60 \mathrm{~s}=90 \mathrm{C} \end{aligned}

Answer

90 coulombs of charge flow around the circuit.

Questions

4 a In which direction does conventional current flow around a circuit?
b In which direction do electrons flow around a circuit?
5 a What is the unit of electric current?
b What is the unit of electric charge?
6 a How many milliamps are there in 1 amp ?
b How many microamps are there in 1 amp ?
7 Which of the following equations shows the correct relationship between electrical units?
1 A = 1 C s 1 A = 1 C s 1A=1(C)/(s)1 \mathrm{~A}=1 \frac{\mathrm{C}}{\mathrm{s}}
1 C = 1 A s 1 C = 1 A s 1C=1((A))/((s))1 \mathrm{C}=1 \frac{\mathrm{~A}}{\mathrm{~s}}
8 Calculate the missing values a-d in Table 18.2. Show all your working.
Charge Current Time
charge current time
220 C 2 A a a a\mathbf{a}
57.6 C b b b\mathbf{b} 3 hours
c c c\mathbf{c} 0.5 A 9 minutes
5.4 C 70 mA d d d\mathbf{d}
Charge Current Time charge current time 220 C 2 A a 57.6 C b 3 hours c 0.5 A 9 minutes 5.4 C 70 mA d| Charge | Current | Time | | :---: | :---: | :---: | | charge | current | time | | 220 C | 2 A | $\mathbf{a}$ | | 57.6 C | $\mathbf{b}$ | 3 hours | | $\mathbf{c}$ | 0.5 A | 9 minutes | | 5.4 C | 70 mA | $\mathbf{d}$ |
Table 18.2

18.2 Voltage in electric circuits

Figure 18.8 shows a circuit in which a cell pushes a current through a resistor. The cell provides the voltage needed to push the current through the resistor. Here, ‘voltage’ is a rather loose term, and we should say that there is a potential difference (p.d.) across the resistor. Potential difference is defined as the work done by a unit charge passing through a component (a resistor, in this case). It is measured in volts ( V V V\mathbf{V} ) using a voltmeter. The p.d. is also the difference in electrical potential between two points: the point where the current enters a component and where it leaves a component. This is rather like the difference in height that makes a ball roll downhill.
Figure 18.8: The cell provides the p.d. needed to push the current around the circuit. The amount of current depends on the p.d. and the resistance of the resistor. The ammeter measures the current flowing through the resistor. The voltmeter measures the p.d. across it. This circuit can therefore be used to find the resistance of the resistor. The ammeter is connected in series in the circuit. The voltmeter is connected in parallel in the circuit.
Voltmeters are always connected in parallel with a component, that is, across a component. This circuit is called a parallel circuit, where components are connected in branches across the circuit. Voltmeters can either have an analogue display (with a needle) or a digital display (with a numerical value). The reading
on a voltmeter is in volts (V) but they have different ranges. Smaller voltages may be measured in millivolts ( mV ) or microvolts ( μ V μ V muV\mu \mathrm{V} ). Take care not to confuse italic, V V VV, which is used as the symbol for an unknown potential difference or voltage, with upright, V , which is used as the symbol for the unit, volts. You can tell the difference in books, but you cannot easily tell the difference when they are written.
There is a special name for the p.d. across a cell. It is called the electromotive force (e.m.f.) of the cell, and it is also measured in volts. The term can be misleading since e.m.f. is a voltage, not a force. Any component that pushes a current around a circuit is said to be a source of e.m.f., for example, cells, batteries, power supplies and dynamos. The e.m.f. is defined as the electrical work done by a source in moving a unit charge around a complete circuit.

Questions

9 a What do the letters p.d. stand for?
b What meter is used to measure p.d.?
c Draw the symbol for this meter.
10 a What name is given to the p.d. across a cell or battery?
b What unit is this measured in?

Combining e.m.f.s

Many battery-operated electrical appliances need more than one cell to make them work. For example, a radio may need four 1.5 V cells. Each has an e.m.f. of 1.5 V and, when connected together in series, they give a combined e.m.f. of 6 V . You can see that, when cells are connected in series, their e.m.f.s add up.
Figure 18.9 shows some examples of this. In general, if cells with e.m.f.s E 1 E 1 E_(1)E_{1} and E 2 E 2 E_(2)E_{2} are connected in series, their combined e.m.f. E E EE is given by:
E = E 1 + E 2 E = E 1 + E 2 E=E_(1)+E_(2)E=E_{1}+E_{2}
You can understand why e.m.f.s add up like this if you think about what happens when electric charge passes through. For four 1.5 V cells in series, each cell does electrical work on each unit charge as it passes through, so their combined e.m.f. must be 6V.

a

b
C
Figure 18.9: The e.m.f.s of cells or other supplies add up when they are connected in series. Here, each individual cell has an e.m.f. of 1.5 V . a, b: More cells give a higher combined e.m.f. c: When one cell is connected the wrong way round, the combined e.m.f. is reduced.

Question

11 Three 12 V batteries are connected in series.
a Draw a diagram to show how these batteries could be connected to a lamp.
b Calculate the combined e.m.f. of the batteries.

What is a volt?

Why do we use high voltages for our mains supply? The reason is that a supply with a high e.m.f. does a lot of work on the charge that it pushes around the circuit. A 230 V mains supply does 230 J of work on each coulomb of charge that travels round the circuit.
This gives us a clue as to what we mean by a volt. A supply with an e.m.f. of 1 V does 1 J of work on each coulomb of charge it pushes round a circuit. In other words, a volt is a joule per coulomb.
The chemical energy supplied by the cell is what pushes electrons around a circuit. The e.m.f. tells you how much work is done on each coulomb of charge as it passes through the cell. This is described by the equation:
e.m.f. ( E ) = work done on the charge (J) charge (C) ( E ) =  work done on the charge (J)   charge (C)  (E)=(" work done on the charge (J) ")/(" charge (C) ")(E)=\frac{\text { work done on the charge (J) }}{\text { charge (C) }}
The bigger the e.m.f. of a cell, the more strongly electrons are pushed around the circuit, just like pressure determines how strongly water is pushed through a pipe.
The potential difference across a device such as a lamp is a measure of how
much electrical work is done by each coulomb as it passes through a component. This is described by the equation:
p.d. ( V ) = work done by the charge ( J ) charge (C) ( V ) =  work done by the charge (  J  )   charge (C)  (V)=(" work done by the charge ( "J" ) ")/(" charge (C) ")(\mathrm{V})=\frac{\text { work done by the charge ( } \mathrm{J} \text { ) }}{\text { charge (C) }}
Both word equations above can be summarised as the word equation:
voltage ( V ) = work done J charge C ( V ) =  work done  J  charge  C (V)=(" work done "J)/(" charge "C)(\mathrm{V})=\frac{\text { work done } \mathrm{J}}{\text { charge } \mathrm{C}}
Or, the symbol equation:
V = W Q V = W Q V=(W)/(Q)V=\frac{W}{Q}

KEY EQUATIONS

p.d. = work done by the charge charge V = W Q  p.d.  =  work done by the charge   charge  V = W Q " p.d. "=(" work done by the charge ")/(" charge ")quad V=(W)/(Q)\text { p.d. }=\frac{\text { work done by the charge }}{\text { charge }} \quad V=\frac{W}{Q}
e.m.f. = work done by the charge charge E = W Q  e.m.f.  =  work done by the charge   charge  E = W Q " e.m.f. "=(" work done by the charge ")/(" charge ")quad E=(W)/(Q)\text { e.m.f. }=\frac{\text { work done by the charge }}{\text { charge }} \quad E=\frac{W}{Q}
The voltage or potential difference is the work done (or energy transferred) per unit charge.
As electric current flows through a circuit, the chemical energy of the cell is transferred to the components as internal energy (in a resistor) or kinetic energy (in a motor) or transferred by light (in a lamp) and so on. Though we defined e.m.f. and p.d. in terms of work, we could have talked about the energy transferred instead. Recall from Chapter 8 that work done = energy transferred.

Questions

12 Rearrange the equation V = W Q V = W Q V=(W)/(Q)V=\frac{W}{Q} to make:
a W W quad W\quad W the subject of the equation
b Q Q quad Q\quad Q the subject of the equation.
13 Calculate the e.m.f. of a battery that gives 60 J of energy to a charge of 5 C .
14 The p.d. across a lamp is 12 V . The lamp is connected for 10 seconds. Calculate how many joules of energy are transferred when:
a a charge of 1 C passes through it
b a charge of 5 C passes through it
c a current of 2 A flows.
(Recall the equation that links charge, current and time.)
15 A circuit consists of two 1.5 V cells in series. How much energy does 2 C of charge gain on going through the cells?
16 A set of party lights consists of 20 identical lamps connected in series to a 240 V mains supply. What is the p.d. across each lamp?
17 a Draw a circuit consisting of a 1.5 V cell, a resistor and a lamp all in series. Arrange the cell so that conventional current flows clockwise round the circuit. Add a voltmeter across the cell. A charge of 6 C flows through the cell.
b Calculate how much energy this amount of charge gains from the cell.
c This amount of charge transfers 6 J of its energy to the lamp. What is the voltage across the lamp?
d Work out the voltage across the resistor.

ACTIVITY 18.1

The cake monster

How does energy travel from a cell to a lamp in a circuit? Figure 18.10 shows a model of how an electrical circuit works. During this activity, think carefully about how well it represents what goes on in an electric circuit.
Figure 18.10:
A road runs through a factory that bakes cakes. Each truck on the road can carry only one cake. A cake monster can take only one bite out of a cake. The cake inspector (in the white laboratory coat) measures the ‘cake difference’ (p.d. for short), which is the amount a monster bites out of each cake. A traffic inspector (not shown) counts the number of trucks that pass him in a given time. The trucks return any uneaten cake to the factory.
1 Copy and complete Table 18.3 to identify what each component (part) in the cake circuit represents in an electrical circuit.
Component
What it represents in an electric
circuit
What it represents in an electric circuit| What it represents in an electric | | :--- | | circuit |
cake factory
road
truck
cake monster
cake inspector
traffic inspector
Component "What it represents in an electric circuit" cake factory road truck cake monster cake inspector traffic inspector | Component | What it represents in an electric <br> circuit | | :--- | :--- | | cake factory | | | road | | | truck | | | cake monster | | | cake inspector | | | traffic inspector | |
Table 18.3
2 How might you change the cake monster model to represent (show) the following changes in an electrical circuit:
a increasing the number of lamps
b a lamp with a bigger resistance.
3 In the cake monster model of a series circuit, compare the ‘cake difference’ across the cake monster (or monsters) and the ‘cake difference’ across the cake factory.

Optional

4 Show how you would change the cake monster model to represent a parallel circuit. Using the model, explain why the current passing through the cell/battery is the sum of the currents passing through the individual branches of a parallel circuit.

REFLECTION

The cake monster activity is a model for the way electric circuits work. Did it help you understand how a circuit works and give meaning to the different components? For example, did you get the idea that the cake represents energy that is being transferred from the factory (cell) to the monsters (lamps)? Can you suggest a better model for the way electric circuits work?

18.3 Electrical resistance

If you use a short length of wire to connect the positive and negative terminals of a cell (a battery) together, you can do a lot of damage. The wire and the cell may both get hot, as a large current will flow through them. There is very little electrical resistance in the circuit, so the current is large. The current flowing in a circuit can be controlled by adding components with electrical resistance to the circuit. The greater the resistance, the smaller the current that will flow.

Defining resistance

How much current can a cell push through a resistor? The electrical resistance of a component is measured in ohms ( Ω Ω Omega\Omega ). It is defined as the potential difference across the component divided by the current passing through it:
resistance ( Ω ) = potential difference ( V ) current ( A ) R = V I  resistance  ( Ω ) =  potential difference  ( V ) current ( A ) R = V I {:[" resistance "(Omega)=(" potential difference "(V))/(current(A))],[R=(V)/(I)]:}\begin{aligned} & \text { resistance }(\Omega)=\frac{\text { potential difference }(\mathrm{V})}{\operatorname{current}(\mathrm{A})} \\ & R=\frac{V}{I} \end{aligned}
The circuit shown in Figure 18.11 illustrates how we can measure the resistance of a resistor (or of any other component). We need to know the current flowing through the resistor, measured by the ammeter. We also need to know the p.d. across it, and this is measured by the voltmeter connected in parallel across it.
KEY EQUATION resistance = p . d current R = V I  KEY EQUATION   resistance  = p . d  current  R = V I {:[" KEY EQUATION "],[{:[" resistance "=(p.d)/(" current ")],[R=(V)/(I)]:}]:}\begin{aligned} & \text { KEY EQUATION } \\ & \begin{aligned} \text { resistance } & =\frac{\mathrm{p} . \mathrm{d}}{\text { current }} \\ R & =\frac{V}{I} \end{aligned} \end{aligned}
A voltmeter is always connected across the relevant component because it is measuring the potential difference between the two ends of the component.
  • Ammeters are connected in series so that the current can flow through them.
  • Voltmeters are connected in parallel to measure the p.d. across the component.
    Worked Example 18.2 and Figure 18.11 show how to calculate the resistance of
    a resistor from measurements of current and p.d. Notice that we can show the current as an arrow entering (or leaving) the resistor. The p.d. is shown by a double-headed arrow to indicate that it is measured across the resistor. The resistance is simply shown as a label on or next to the resistor - it does not have a direction.

WORKED EXAMPLE 18.2

A resistor allows a current of 0.02 A to flow through it when there is a p.d. of 10.0 V between its ends. What is its resistance?
Figure 18.11: Sketch for diagram.
Step 1: Write down what you know, and what you want to know. You may prefer to write these quantities on a sketch of the situation (see Figure 18.11).
current I = 0.02 A I = 0.02 A I=0.02AI=0.02 \mathrm{~A}
p.d. V = 10.0 V V = 10.0 V V=10.0VV=10.0 \mathrm{~V}
resistance R = R = R=R= ?
Step 2: Write down the equation for R . Substitute values and calculate the answer.
R = V I = 10.0 V 0.02 A = 500 Ω R = V I = 10.0 V 0.02 A = 500 Ω {:[quad R=(V)/(I)],[=(10.0(V))/(0.02(A))],[=500 Omega]:}\begin{aligned} & \quad R=\frac{V}{I} \\ & =\frac{10.0 \mathrm{~V}}{0.02 \mathrm{~A}} \\ & =500 \Omega \end{aligned}

Answer

So, the resistance of the resistor is 500 Ω 500 Ω 500 Omega500 \Omega.

What is an ohm?

Let us think about the equation that defines what we mean by resistance:
We can see that it takes a p.d. of 10 V to make a current of 1 A flow through a 10 Ω 10 Ω 10 Omega10 \Omega resistor. It takes 20 V to make 1 A flow through a 20 Ω 20 Ω 20 Omega20 \Omega resistor, and so on. Hence resistance (in Ω Ω Omega\Omega ) tells us how many volts are needed to make 1 A flow through that resistor. To put it another way: one ohm is one volt per amp.
1 Ω = 1 V A 1 Ω = 1 V A 1Omega=1((V))/((A))1 \Omega=1 \frac{\mathrm{~V}}{\mathrm{~A}}
In the case of Worked Example 18.2, it would take 500 V to make 1 A flow through the 500 Ω 500 Ω 500 Omega500 \Omega resistor.

Changing current

You can think of an electric circuit as an obstacle race. The current (or flow of charge) comes out of the positive terminal of the cell and must travel around the circuit to the negative terminal. Along the way, it must pass through the different components. The greater their resistance, the harder it will be for the charge to flow, and so the current will be smaller.
The greater the resistance in the circuit, the smaller the current that flows. However, we can make a bigger current flow by increasing the p.d. that pushes it. A bigger p.d. produces a bigger current. The greater the p.d. in a circuit (or across a component), the greater the current that flows.

Resistance and thickness

This idea of an obstacle race can help us to think about the resistance of wires of different shapes. A long, thin wire has more resistance than a short, fat one. Imagine an obstacle course that includes pipes of different sizes through which the runners have to pass. It is easy to get through a short pipe with a large diameter. It is much harder when the pipe is long and narrow.
  • The longer a wire, the greater its resistance.
  • The greater the diameter of a wire, the less its resistance.

Questions

18 a What is the resistance of a lamp if a current of 5.0A flows through it when it is connected to a 240 V supply?
b When the p.d. across the lamp is increased, will the current flowing increase or decrease?
19 A student cuts two pieces of wire, one long and one short, from a reel.
a Which piece of wire will have the greater resistance?
b Draw a circuit diagram to show how you would check your answer by measuring the resistances of the two pieces of wire.

Measuring resistance

The circuit shown in Figure 18.12 can be used to find the resistance of a resistor. The circuit has a variable power supply, which can be adjusted to give several different values of p.d. For each value, the current is measured, and results like those shown in Table 18.4 are found. The last column in Table 18.4 shows values for R R RR, calculated using R = V / I R = V / I R=V//IR=V / I. These can be averaged to find the value of R R RR.
Figure 18.12: A circuit for investigating how the current through a resistor changes as the voltage across it varies. The power supply can be adjusted to give a range of values of p.d. (typically from 0 V to 12 V ).
p.d. V / V V / V V//VV / \mathrm{V} Current I / A I / A I//AI / \mathrm{A} Resistance R / Ω R / Ω R//OmegaR / \Omega
2.0 0.08 25.0
p.d. V//V Current I//A Resistance R//Omega 2.0 0.08 25.0 | p.d. $V / \mathrm{V}$ | Current $I / \mathrm{A}$ | Resistance $R / \Omega$ | | :---: | :---: | :---: | | 2.0 | 0.08 | 25.0 | | | | |
4.0 0.17 23.5
6.0 0.24 25.0
8.0 0.31 25.8
10.0 0.40 25.0
12.0 0.49 24.5
4.0 0.17 23.5 6.0 0.24 25.0 8.0 0.31 25.8 10.0 0.40 25.0 12.0 0.49 24.5| 4.0 | 0.17 | 23.5 | | :---: | :---: | :---: | | 6.0 | 0.24 | 25.0 | | 8.0 | 0.31 | 25.8 | | 10.0 | 0.40 | 25.0 | | 12.0 | 0.49 | 24.5 |
Table 18.4:Typical results for an experimental measurement of resistance.

EXPERIMENTAL SKILLS 18.1

How length and thickness affects the resistance of a wire
Understanding how to measure resistance is really important because it introduces the measurements of voltage and current, which are essential for understanding circuits. In more sophisticated (complicated) circuits, resistance can change so that a device can respond to the environment (for example, a light can come on when it gets dark).

You will need:

  • power supply
  • 5 insulated (coloured) wires
  • 2 crocodile clips
  • metre ruler
  • 2 lengths of resistance wire of different diameter
  • masking tape
  • heatproof mat to go underneath the resistance wire
  • ammeter
  • voltmeter.
If a power supply is not available then use a suitable cell or series of cells but include a switch and only close the switch when you take measurements (to avoid draining the cell or battery). If two different gauges
of wire are not available, lightly twist two lengths of the same gauge wire together to double the effective cross-section.
Safety: If the insulation on the wires melts or gives off poisonous fumes, reduce the voltage you use. Hot wires have the potential to burn skin. Avoid connecting the positive terminal of the power supply directly to the negative terminal (ensure that the current has to pass through the resistance wire). It may be necessary to place a resistor (fixed or variable) in series with the resistance wire in order to reduce the current through it. The voltages involved are too low to cause an electric shock.

Getting started

  • State what devices you will use to measure voltage and current and describe how they should be wired into the circuit.
  • What variable should be kept the same (control variable)?
  • Why is it important to avoid taking measurements when the wire has zero length?
  • Predict how the resistance of the wire will vary with the length of the wire.
  • Predict how the resistance of the wire will vary with the diameter of the wire.
  • Identify the independent and dependent variables.
  • State two reasons why it is important to plot a graph of your results.
  • You will be plotting a graph of your results. What should be plotted along on each axis?

Method 1

1 Set up the circuit as shown in Figure 18.13a. Set the power supply to 12 V .

b
Figure 18.13a:Circuit diagram for the experiment. b: How to attach the wire to the ruler.
2 Draw a table like Table 18.5, but remember to add more rows as you need them.
Length of
resistance wire /
cm
Length of resistance wire / cm| Length of | | :--- | | resistance wire / | | cm |
Current / A Voltage / V Resistance / Ω Ω Omega\Omega
Thin
10
20
30
Thick
10
20
30
"Length of resistance wire / cm" Current / A Voltage / V Resistance / Omega Thin 10 20 30 Thick 10 20 30 | Length of <br> resistance wire / <br> cm | Current / A | Voltage / V | Resistance / $\Omega$ | | | :--- | :--- | :--- | :--- | :---: | | Thin | | | | | | 10 | | | | | | 20 | | | | | | 30 | | | | | | Thick | | | | | | | | | | | | 10 | | | | | | 20 | | | | | | 30 | | | | |

Table 18.5

3 If your teacher has not already done so, attach the resistance wire along a metre ruler with insulating (or masking) tape at both ends and at the centre (see Figure 18.13b).
4 Put crocodile clips at the ends of the insulated (coloured) wires that will attach to the resistance wire.
5 Attach the first crocodile clip to the resistance wire where it crosses the 0.0 cm mark on the ruler and leave this in place throughout the experiment.
6 Attach the second crocodile clip where the resistance wire crosses the ruler at 10.0 cm and record the current and voltage values in the table.
7 Move the second crocodile clip at 10.0 cm intervals, each time recording the current and voltage values in the table.
8 Calculate the resistance values using R = V I R = V I R=(V)/(I)R=\frac{V}{I} and record the results in the table.
9 Plot a graph of resistance against length of the resistance wire (with the resistance along the horizontal axis) and label it ‘thin’.
10 Repeat the experiment with thicker resistance wire (or loosely twist two wires of the same diameter together but, if you do, ensure the teeth of the crocodile clips are in contact with both wires). Plot the graph on the same axes for easy comparison and label it ‘thick’.

Method 2

If you do not have access to resistance wires of five different diameters, twist wires of the same diameter together. Keep the length of wire the same (say, 50 cm ).
1 Use a micrometer to determine the diameter of the wire or look up the wire diameter corresponding to the SWG (standard wire gauge) value on the reel that the wire comes from.
2 Use this information to work out the cross-sectional area of the wire or wires.
3 Using wires of the same length but increasing cross-section, record the voltage and current values, and calculate the resistance.
4 Plot a graph of resistance against the cross-sectional area.
5 Describe the relationship between resistance and cross-section.
6 How would you show that there is an inverse relationship?
Questions
1 Copy each statement choosing the correct answer from the brackets.
a The resistance of a resistance wire {increases / decreases} when its length increases.
b The resistance of a resistance wire {doubles / halves} when its length doubles.
c This means that the resistance is {inversely / directly } proportional to its length. A straight-line graph passing through the origin will show this.
d The resistance of a resistance wire {increases / decreases } } }\} when its diameter increases.
e The resistance of a resistance wire {doubles / halves} when its cross-sectional area doubles.
f This means that the resistance is {inversely / directly } proportional to its cross-sectional area.
2 Extrapolate your resistance against length graph (Method 1) towards the origin. What is the resistance when the length of the resistance wire is zero?
3 Use your answer to the previous question to suggest why a direct connection between the terminals of the power supply is not a good idea.

18.4 More about electrical resistance

The equation R = V I R = V I R=(V)/(I)R=\frac{V}{I} is used to calculate the resistance of a component in a circuit. We can rearrange the equation in two ways so that we can calculate current or p.d.:
I = V R V = I R I = V R V = I R I=(V)/(R)quad V=IRI=\frac{V}{R} \quad V=I R
So, for example, we can calculate the current that flows through a 20 Ω 20 Ω 20 Omega20 \Omega resistor when there is a p.d. of 6.0 V across it. The current I I II is:
I = 6.0 V 20 Ω = 0.30 A I = 6.0 V 20 Ω = 0.30 A I=(6.0(V))/(20 Omega)=0.30AI=\frac{6.0 \mathrm{~V}}{20 \Omega}=0.30 \mathrm{~A}

Questions

20 Calculate the missing values a-d in Table 18.6. Show all your working.
Potential difference / V Current / A Resistance / Ω Ω Omega\Omega
240 2 a
12 b b b\mathbf{b} 3000
c 0.5 15
120 80 d d d\mathbf{d}
Potential difference / V Current / A Resistance / Omega 240 2 a 12 b 3000 c 0.5 15 120 80 d| Potential difference / V | Current / A | Resistance / $\Omega$ | | :---: | :---: | :---: | | 240 | 2 | a | | 12 | $\mathbf{b}$ | 3000 | | c | 0.5 | 15 | | 120 | 80 | $\mathbf{d}$ |

Table 18.6

21 a What p.d. is needed to make a current of 2.0 A flow through a 30 Ω Ω Omega\Omega resistor?
b Without calculation, what p.d. is required when the resistance is doubled?
22 a A p.d. of 240 V across a resistor causes a current of 80 mA to flow through it. What is the resistance of theresistor?
b What p.d. would cause a current of 40 mA to flow through the resistor?
23 What current flows when a p.d. of 7.5 V is connected across a 2 k תresistor?

Current-voltage characteristics

We can use the data shown in Table 18.4 to plot a graph of current against potential difference for a resistor. This graph is shown in Figure 18.14 and is known as a current-voltage characteristic.
Figure 18.14: The current-voltage characteristic for the data in Table 18.4.
  • The p.d. V V VV is on the x x xx-axis, because this is the quantity we vary. It is the independent variable.
  • The current I I II is on the y y yy-axis, because this is the quantity that varies as we change V V VV. It is the dependent variable.
In this case, the graph is a straight line that passes through the origin. This is what we expect because the quation I = V R I = V R I=(V)/(R)I=\frac{V}{R} shows that the current I I II is proportional to the p.d., V V VV.
A resistor with a current-voltage characteristic like this is called an ohmic resistor. It is easy to predict the current that will flow through an ohmic resistor because the current is directly proportional to the p.d. across it. Double the voltage gives double the current, and so on.
Figure 18.15 shows what happens if we use a filament lamp instead of an ohmic resistor. You can see that the current-voltage characteristic for the filament lamp is curved.
Figure 18.15: The current-voltage characteristic for a filament lamp. The graph is not a straight line through the origin, showing that the lamp is not an ohmic resistor.
  • At first, for low voltages, the graph is straight, showing that the current increases at a steady rate as the voltage is increased.
  • At higher voltages, the graph starts to curve over. The current increases more and more slowly as the voltage is increased. Current is not proportional to voltage.
    The graph shows that the lamp is not an ohmic resistor. Why is this? At first, when the voltage and current are small, the lamp behaves like an ohmic resistor. However, as the voltage increases, the current causes the filament to get hot and glow brightly. At high temperatures, the filament has a higher resistance and so the current does not increase as rapidly as it would do if the filament had remained cool.
This means that the resistance of the tungsten filament lamp increases as the current increases. But why does increasing the current make the tungsten hotter? This requires a more detailed explanation.
Increasing the current leads to an increase in the number of electrons flowing through the tungsten wire and this increases the number of collisions between the electrons and the lattice (the regular arrangement of atoms in the metal).
Some of the kinetic energy from the electrons is transferred into internal energy, which makes the lattice vibrate more. This increases the resistance because it increases the number of collisions between the electrons and the lattice. In a similar way, it is more difficult to move through a crowd where people are moving in random directions compared to a crowd of people who are standing still.
Figure 18.16 shows the typical shapes of the current-voltage characteristics for ohmic resistors, for a filament lamp and a diode. In Figure 18.16a, resistor Q has a higher resistance than resistor P. We can tell this because the current flowing through Q is always less than the current through P , for any voltage.

Figure 18.16: Typical current-voltage characteristics. a: For two ohmic resistors. b: For a filament lamp. c: For a diode.

Notice that these graphs show both positive and negative voltages. A negative current means one flowing in the opposite direction. This is achieved by connecting the cell or power supply the other way round. Figures 18.16a and b are symmetrical, showing that, whichever way round the components are connected, the current will be the same for a given voltage.
Figure 18.16c is the I V I V I-VI-V characteristic of a diode, which is not symmetrical. A diode acts as a switch and only allows current to flow in one direction. The arrow on the diode symbol indicates the direction that conventional current can flow. A diode is a semiconductor so it behaves like both an insulator and a conductor. A diode behaves like an insulator until it is given enough voltage (energy per unit charge) to make it behave as a conductor. For a silicon diode this threshold voltage is 0.7 V .

Questions

24 Look at the graph shown in Figure 18.16a. How can you tell from the graph that the resistors are both ohmic?
25 Look at the graph shown in Figure 18.16b. How can you tell from the graph that the lamp’s resistance increases as the p.d. across it increases?

Length and area

We have seen that the resistance of a wire depends on its length and its diameter. In fact, it is the cross-sectional area of the wire that matters.
  • The resistance of a wire is proportional to its length.
  • The resistance of a wire is inversely proportional to its cross-sectional area.
Suppose that we have a 4.0 metre length of wire. Its resistance is 100 Ω 100 Ω 100 Omega100 \Omega. What will be the resistance of a 2.0 metre length of wire with twice the cross-sectional area? Notice that making the wire shorter will reduce its resistance, and increasing its area will also reduce its resistance.
  • Halving the length gives half the resistance = 50 Ω = 50 Ω =50 Omega=50 \Omega.
  • Doubling the area halves the resistance again = 25 Ω = 25 Ω =25 Omega=25 \Omega.

Question

26 A 2.0 metre length of wire has a resistance of 4.0 Ω 4.0 Ω 4.0 Omega4.0 \Omega.
a What is the resistance of a piece of the same wire of length 20.0 metres?
b What is the resistance of a 4.0 metre wire with half the cross-sectional area, made of the same material?

18.5 Electrical energy, work and power

We use electricity because it is a good way of transferring energy from place to place. In most places, if you switch on an electric heater, you are getting the benefit of the energy released as fuel that is burned in a power station, which may be over 100 km away.
When you plug in an appliance to the mains supply, you are connecting up to quite a high voltage, something like 110 V or 230 V , depending on where you live. This high voltage is the e.m.f. of the supply. Recall that e.m.f. is the name given to the p.d. across an electrical source component such as a cell or power supply that pushes current around a circuit.
By the principle of conservation of energy, the energy given to the charges by the cell must equal the sum of the energies given by the charges to the various components. This means that the e.m.f. across the cell is equal to the sum of the potential differences across all the components round the circuit. The total voltage across the cell equals the sum of the voltages across the lamps. This is something you will meet again in Chapter 19.
Batteries and power supplies give energy to the charges in a circuit. Similarly, we can think about other components in a circuit. For example, a small lamp may have a p.d. of 1.5 V across it. This means that each coulomb of charge does 1.5 J of electrical work to pass through the lamp and will transfer 1.5 J of energy to the lamp. Remember, work done is the same as energy transferred.

Electrical power

Most electrical appliances have a label that shows their power rating. An example is shown in Figure 18.17.
Figure 18.17: A label from the back of a microwave oven.
Power ratings are indicated in watts ( W ) or kilowatts ( kW ). The power rating of an appliance shows the rate at which it transfers energy, and indicates the maximum electrical power the appliance draws from the mains supply when it is operating at full power.
Electrical power is the rate at which energy is transferred:
power ( W ) = energy transferred ( J ) time taken ( s ) ( W ) =  energy transferred  ( J )  time taken  ( s ) (W)=(" energy transferred "(J))/(" time taken "(s))(\mathrm{W})=\frac{\text { energy transferred }(\mathrm{J})}{\text { time taken }(\mathrm{s})}
P = Δ E t P = Δ E t P=(Delta E)/(t)P=\frac{\Delta E}{t}
The symbol, E E EE, represents energy transferred. You should recognise this definition of power from Chapter 8.
This equation also reminds us of the definition of the unit of power, the watt: one watt is one joule per second.
1 W = 1 J / s 1 W = 1 J / s 1W=1J//s1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}

Voltage and energy

We have seen that the e.m.f. (voltage) of a supply tells us how much energy it transfers to charges flowing around the circuit. The greater the current flowing around the circuit, the faster that energy is transferred. Hence the rate at which
energy is transferred in the circuit (the power, P P PP ) depends on both the e.m.f., E E EE, of the supply and the current, I I II, that it pushes round the circuit. The following equation shows how to calculate the electrical power:
power ( W ) = current ( A ) × ( W ) = current ( A ) × (W)=current(A)xx(\mathrm{W})=\operatorname{current}(\mathrm{A}) \times p.d. ( V ) ( V ) (V)(\mathrm{V})
P = I V P = I V P=IVP=I V

KEY EQUATION

power = current × p.d. P = I V  power  =  current  ×  p.d.  P = I V {:[" power "=" current "xx" p.d. "],[P=IV]:}\begin{aligned} \text { power } & =\text { current } \times \text { p.d. } \\ P & =I V \end{aligned}
You may prefer to remember this as an equation relating units:
watts = amps × volts  watts  =  amps  ×  volts  " watts "=" amps "xx" volts "\text { watts }=\text { amps } \times \text { volts }

Calculating energy

Since energy transferred = = == power × × xx\times time, we can use the equation P = I V P = I V P=IVP=I V to give an equation for electrical energy transferred E E EE in terms of current and voltage:
energy transferred ( J ) = current ( A ) × p.d. ( V ) × time ( s )  energy transferred  ( J ) = current ( A ) ×  p.d.  ( V ) ×  time  ( s ) " energy transferred "(J)=current(A)xx" p.d. "(V)xx" time "(s)\text { energy transferred }(\mathrm{J})=\operatorname{current}(\mathrm{A}) \times \text { p.d. }(\mathrm{V}) \times \text { time }(\mathrm{s})
E = I V t E = I V t E=IVtE=I V t

KEY EQUATION

energy transferred = current × p.d. × time E = I V t  energy transferred  =  current  ×  p.d.  ×  time  E = I V t {:[" energy transferred "=" current "xx" p.d. "xx" time "],[E=IVt]:}\begin{aligned} \text { energy transferred } & =\text { current } \times \text { p.d. } \times \text { time } \\ E & =I V t \end{aligned}
Worked Example 18.3 shows how to calculate the power of a device and how much energy is transferred in a given time.

WORKED EXAMPLE 18.3

An electric fan runs from the 230 V mains supply. The current flowing through it is 0.40 A . At what rate is electrical energy transferred by the fan? How much energy is transferred in one minute?
Step 1: First, we have to calculate the rate at which electrical energy is
transferred. This is the power, P P PP. Write down what you know and what you want to know.
V = 230 V V = 230 V V=230VV=230 \mathrm{~V}
I = 0.40 A I = 0.40 A I=0.40AI=0.40 \mathrm{~A}
P = P = P=P= ?
Step 2: quad\quad Write down the equation for power, which involves V V VV and I , substitute values and solve.
P = I V P = I V P=IVP=I V
P = 0.40 A × 230 V = 92 W P = 0.40 A × 230 V = 92 W P=0.40Axx230V=92WP=0.40 \mathrm{~A} \times 230 \mathrm{~V}=92 \mathrm{~W}
Step 3: To calculate the energy transferred in 1 minute, use E = P t E = P t E=PtE=P t (or E = E = E=E= I V t ) I V t ) IVt)I V t). Recall that time, t t tt, must be in seconds.
E = 92 W × 60 s = 5520 J E = 92 W × 60 s = 5520 J E=92Wxx60s=5520JE=92 \mathrm{~W} \times 60 \mathrm{~s}=5520 \mathrm{~J}

Answer

So, the fan’s power is 92 W , and it transfers 5520 J of energy each minute.

Questions

27 Write down an equation linking watts, volts and amps.
28 Calculate the missing values a-d in Table 18.7. Show all your working.
Voltage / V Current / A Power / P
240 2 a a a\mathbf{a}
12 b b b\mathbf{b} 60
c c c\mathbf{c} 0.5 15
120 80 d d d\mathbf{d}
Voltage / V Current / A Power / P 240 2 a 12 b 60 c 0.5 15 120 80 d| Voltage / V | Current / A | Power / P | | :---: | :---: | :---: | | 240 | 2 | $\mathbf{a}$ | | 12 | $\mathbf{b}$ | 60 | | $\mathbf{c}$ | 0.5 | 15 | | 120 | 80 | $\mathbf{d}$ |

Table 18.7

29 A 12 V power supply pushes a current of 6.0 A through a resistor. At what rate is energy transferred to the resistor?
30 A tropical fish tank is fitted with an electric heater, which has a power
rating of 50 W .
The heater is connected to a 240 V supply. What current flows through the heater when it is switched on?
31 How much energy is transformed by an electric lamp in an hour if a current of 20 mA flows through it when it is connected to a 120 V supply?
The equation E = I V t E = I V t E=IVtE=I V t could be used to calculate the chemical energy stored in a battery. A battery delivers a current at its rated voltage for a given time (until it discharges). The chemical energy stored in a battery can also be found by multiplying its charge by its rated voltage.

Units of electrical energy

Like other stores of energy, we could calculate the amount of energy transferred electrically in joules (J), but it is much more convenient to use kWh . This is because 1 kW = 1000 W 1 kW = 1000 W 1kW=1000W1 \mathrm{~kW}=1000 \mathrm{~W} and 1 h = 3600 s 1 h = 3600 s 1h=3600s1 \mathrm{~h}=3600 \mathrm{~s}, so 1 kWh = 1000 W × 3600 s = 3.6 × 1 kWh = 1000 W × 3600 s = 3.6 × 1kWh=1000Wxx3600s=3.6 xx1 \mathrm{kWh}=1000 \mathrm{~W} \times 3600 \mathrm{~s}=3.6 \times 10 6 J 10 6 J 10^(6)J10^{6} \mathrm{~J} (a big number). 1 kWh is sometimes called a unit of electricity. It is not; it is a unit of energy. It is kWh that are measured using an electricity meter, like the one shown in Figure 18.18.
Figure 18.18a: An electricity meter. b: A typical smart electricity meter, which can be accessed remotely so that there is no need for someone to come to your house to read your meter.
All devices in homes have a power rating. For example, a fan heater might have a power output of 2 kW . If left running for two hours it would use 4 kWh (four units of electrical energy). This is easier to imagine than 1.44 × 10 7 J 1.44 × 10 7 J 1.44 xx10^(7)J1.44 \times 10^{7} \mathrm{~J}.
The equation for working out the number of units of electrical energy that are being used is:

KEY EQUATION

energy transferred = = == power × × xx\times time
(kWh) (kW) (hours)
Worked Example 18.4 shows how the number of units a device uses depends on how much power it transfers and the length of time it operates.

WORKED EXAMPLE 18.4

Marcus switches on a water heater for two hours. The power of the heater is 3.5 kW . How much energy is transferred in kWh (units)?
Step 1: Start by writing down what you know, and what you want to know.
power = 3.5 kW = 3.5 kW =3.5kW=3.5 \mathrm{~kW}
time = 2 = 2 =2=2 hours
energy transferred (kWh) = ?
Step 2: Now write down the equation for energy transferred in kWh . energy transferred ( kWh ) = ( kWh ) = (kWh)=(\mathrm{kWh})= power ( kW ) × ( kW ) × (kW)xx(\mathrm{kW}) \times time (hours)
Step 3: Substitute the values of the quantities on the right-hand side and calculate the answer.
energy transferred ( kWh ) = 3.5 kW × 2 ( kWh ) = 3.5 kW × 2 (kWh)=3.5kWxx2(\mathrm{kWh})=3.5 \mathrm{~kW} \times 2 hours = 7 kWh = 7 kWh =7kWh=7 \mathrm{kWh}
Answer
The water heater uses 7 kWh (or 7 units).
Worked Example 18.5 shows how to calculate the cost of electricity.
Zara checks her electricity bill for a three month period. The meter reading at the start was 2531 kWh and at the end it was 2647 kWh . Electricity costs 16p per unit. What is her bill for electricity?
Step 1: Start by writing down what you know, and what you want to know.
metre reading at start = 2531 kWh = 2531 kWh =2531kWh=2531 \mathrm{kWh}
meter reading at end = 2647 kWh = 2647 kWh =2647kWh=2647 \mathrm{kWh}
cost per unit = 16 = 16 =16=16 p
units used = ?
total cost of electricity = = == ?
Step 2: Work out how many units were used.
Units used = 2647 kWh 2531 kWh = 116 kWh  Units used  = 2647 kWh 2531 kWh = 116 kWh {:[" Units used "=2647kWh-2531kWh],[=116kWh]:}\begin{aligned} \text { Units used } & =2647 \mathrm{kWh}-2531 \mathrm{kWh} \\ & =116 \mathrm{kWh} \end{aligned}
Step 3: Now write down the equation for the total cost of electricity. total cost of electricity = = == number of units × × xx\times cost per unit
Step 4: Substitute the values of the quantities on the right-hand side and work out the answer. total cost of electricity = = ==
116 kWh × 16 p = £ 18.56 116 kWh × 16 p = £ 18.56 116kWhxx16p=£18.56116 \mathrm{kWh} \times 16 \mathrm{p}=£ 18.56

Answer

Zara’s bill is £ 18.56 £ 18.56 £18.56£ 18.56.

Questions

32 A 3 kW air conditioning unit used 216 kWh of electricity. Calculate how many days it was switched on for.
33 Eshan recorded the readings on his family’s electricity meter at the beginning and end of one month. The readings were 990987 and 991013.
Electricity costs 0.5 dirhams per unit. Calculate the cost of the electrical energy used by Eshan’s family in this month.
34 Use the information in Table 18.8 to work out the missing values a j a j a-j\mathbf{a}-\mathbf{j}.
Assume that electricity costs 16 p per unit.
Appliance Power Time kWh / unit Cost
desktop computer a a a\mathbf{a} 3 hours 1.3 b b b\mathbf{b}
microwave oven 870 W c c c\mathbf{c} 0.87 d d d\mathbf{d}
television e e e\mathbf{e} 12 hours 1.0 f f f\mathbf{f}
energy efficient lamp 9 W g g g\mathbf{g} 4.5 × 10 2 4.5 × 10 2 4.5 xx10^(-2)4.5 \times 10^{-2} h h h\mathbf{h}
kettle 2 kW 3 minutes i i i\mathbf{i} j j j\mathbf{j}
Appliance Power Time kWh / unit Cost desktop computer a 3 hours 1.3 b microwave oven 870 W c 0.87 d television e 12 hours 1.0 f energy efficient lamp 9 W g 4.5 xx10^(-2) h kettle 2 kW 3 minutes i j| Appliance | Power | Time | kWh / unit | Cost | | :--- | :---: | :--- | :---: | :---: | | desktop computer | $\mathbf{a}$ | 3 hours | 1.3 | $\mathbf{b}$ | | microwave oven | 870 W | $\mathbf{c}$ | 0.87 | $\mathbf{d}$ | | television | $\mathbf{e}$ | 12 hours | 1.0 | $\mathbf{f}$ | | energy efficient lamp | 9 W | $\mathbf{g}$ | $4.5 \times 10^{-2}$ | $\mathbf{h}$ | | kettle | 2 kW | 3 minutes | $\mathbf{i}$ | $\mathbf{j}$ |

Table 18.8

PROJECT

Shining the light on what lamp to use
Design a public awareness and education campaign to persuade people to switch to using LED lamps.
Your campaign must include an eye-catching and scientifically-accurate poster or leaflet that can be understood by the public. One way to encourage people to read your leaflet is to provide a guide for choosing the correct bulb, including whether they need a screw or bayonet fitting.
You must include a table that compares the cost of the different light bulbs. This cost should include the purchase price plus the running cost over 20 years.
If you are working in pairs or small teams, you could also produce a podcast.

Background

Many people are already aware that energy-saving light bulbs can reduce carbon emissions and save them money. However, energy-saving light bulbs are often more expensive to buy and some people think they prefer the colour of light emitted by tungsten lamps or prefer the look of the filament that can be seen through the glass when the lamp is off (Figure

  1. KEY EQUATION
    work done = = == force × × xx\times distance moved in the direction of the force