HEAT AND TEMPERATURE - Antimatter | Life in a · Web viewIn this section of the course we will look at what heat and temperature are, the effects of adding and removing heat from materials

Heat and Temperature

1. INTRODUCTION..........................................................................................................................

2. TEMPERATURE SCALES...........................................................................................................

2.1. CELSIUS....................................................................................................................................2.2. FAHRENHEIT..............................................................................................................................2.3. KELVIN.....................................................................................................................................

2.3.1. Absolute Zero....................................................................................................................

3. SPECIFIC HEAT CAPACITY......................................................................................................

4. CHANGING STATE AND LATENT HEAT................................................................................

4.1. SPECIFIC LATENT HEAT.............................................................................................................4.1.1. Units.................................................................................................................................

5. THERMAL EXPANSION.............................................................................................................

5.1. LINEAR EXPANSION COEFFICIENT, A.........................................................................................5.1.1. Rivets................................................................................................................................5.1.2. Bimetallic Strip.................................................................................................................5.1.3. Thermostat........................................................................................................................

5.2. AREAL (OR SUPERFICIAL) EXPANSION........................................................................................5.3. VOLUME EXPANSION.................................................................................................................

6. HEAT TRANSFER........................................................................................................................

6.1. CONDUCTION.............................................................................................................................6.2. CONVECTION.............................................................................................................................6.3. RADIATION................................................................................................................................6.4. CONDUCTION, CONVECTION AND RADIATION.............................................................................

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1. IntroductionIn this section of the course we will look at what heat and temperature are, the effects of adding and removing heat from materials and methods of measuring temperature. Heat and temperature are two words that are commonly misinterpreted as many people assume them to mean the same thing. In science this is not so, they have quite distinct meanings

Heat is a form of energyquantities of energy are measured in joules, J

Temperature is the measure of how hot something is on a chosen scaletemperature can be measured in a variety of different units oC, oF, K etc..

Temperature and heat are obviously related. In many instances if heat is added to a substance its temperature will rise e.g. the water in a kettle which is supplied with heat from a heater element.. However, there are also many instances when supplying heat will not result in a temperature rise. For example, when a saucepan of boiling water is continuously heated, the temperature of the water remains at 100oC - adding extra heat doesn't raise the temperature. In fact, this heat is used to change the water to steam and so, with time, the amount of water in the saucepan will decrease.

Another distinction between heat and temperature can be made when comparing the heat content and temperature of the two objects shown in Figure 1.

mass, 1 kg

T = 50oCmass, 2 kg

T = 50oC

Figure 1:-Illustrating the difference between heat and temperature.

The temperature of both objects is the same, however, the heat content of the 2 kg mass is likely to be twice the heat content of the 1 kg mass provided that they are made from the same material.

Heat and temperature are NOT the same thing

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2. Temperature ScalesAs already mentioned, there are a number of different temperature scales including Celsius, Fahrenheit and Kelvin, each named after the scientist who developed them. Each temperature scale is calibrated with respect to two fixed temperatures. (Calibrate means to mark out a scale) We will take a closer look at the Celsius scale (formerly called Centigrade) to explain what this means.

2.1. CelsiusThe Celsius scale uses

melting point of pure ice = 0oCboiling point of pure water = 100oC

as the fixed point temperatures. These temperatures have to be measured at 760 mmHg (atmospheric pressure), as these temperatures vary with pressure. The boiling point of water is particularly prone to variation with air pressure. For example, air pressure falls with increasing altitude and it is found that the boiling point of water falls by roughly 1oC for every 300 m increase in altitude.

In 1742, in order to create the temperature scale which bears his name, Anders Celsius, a Swedish scientist, took the above two temperatures as his fixed points and divided the gap between them into 100 equal sections. As the division was by 100, the term centigrade was used (compare centimetre = one hundredth of a metre) to describe the size of each of these divisions. However, you should bear in mind that this decision was quite arbitrary and he could have accorded these fixed points any values he thought fit.

2.2. FahrenheitThe Fahrenheit scale was in common usage in Ireland up to the '70s and is still widely used in USA. This scale was developed by Daniel Fahrenheit, a German scientist, in the 18th century. In comparison to the centigrade scale, the choice of fixed point temperatures appears slightly odd. He used

melting point of an equal mixture of ice and salt = 0oFbody temperature = 90oF

These temperatures were later amended slightly, but the resulting scale now givesmelting point of pure ice = 32oFboiling point of pure water = 212oF

at 760 mmHg.Fahrenheit can be readily converted to Celsius using the relationship

C = 5/9 (F - 32)where

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C = temperature in CelsiusF = temperature in Fahrenheit

2.3. KelvinAs already mentioned, the choice of fixed points and the determination of the size of one degree were quite arbitrary. In the last century, it was obvious that the measurement of temperature needed to be put on a more scientific basis. A scale was developed based on the reference point of absolute zero.

2.3.1. Absolute Zero

Consider a gas in a cylinder fitted with a piston which is free to move up and down in the cylinder, shown in Figure 2. It was known that if heat was

moveable piston

gas

cylinder

Figure 2:- Schematic diagram of the basic apparatus required to ascertain the relationship between

volume and temperature for a gas.

supplied to the gas, it would tend to expand and push the piston out. Similarly, if the gas was cooled, the gas would contract and the piston would fall. Experiments were carried out on the relationship between the volume of the gas and its temperature. The results of such experiments are shown in Figure 3, over the range of temperatures available to scientists in the 18th and 19th centuries.

gas volume

temperature

Figure 3:- Relationship between temperature and volume for different gas pressures.

The different lines correspond to different gas pressures - in effect different weights of piston in Figure 2. What fascinated scientists about these lines was that if they were extended back to the temperature axis i.e. to a point where gas volume is zero,

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they all met up at the same point, read off the scale to be - 273oC. This is shown in Figure 4.

- 273 oC 0 oC 100 oCtemperature

gas volume

Figure 4:- Graphs from Figure 3 extrapolated to the temperature axis.

The graphs above show that the temperature and pressure in a gas are closely linked. In fact, both temperature and pressure depend on the motion of the molecules in the gas. The higher the average kinetic energy of the gas molecules, the higher its temperature and pressure. Cooling a gas, reduces the kinetic energy of the molecules (they move more slowly - remember the kinetic energy of a body depends on its speed). From this argument it is clear that there has to be a lower limit on temperature. As a gas is cooled, the molecules move more and more slowly until they eventually have no kinetic energy left - i.e. they are stationary. At this point the gas can no longer exert a pressure; as gases exert pressure due to the molecules moving around colliding with the container walls. Hence, the temperature at which the pressure of a gas is predicted to be zero must be the lowest temperature there is.

-273oC is the lowest possible temperature. It was decided to make this temperature the basis of a new scientific temperature scale, called the Kelvin scale after the British scientist, Baron Kelvin. - 273oC was assigned the temperature 0 K and was rechristened absolute zero. Another law in physics states that it is impossible to actually cool anything down to absolute zero. A lot of research effort is spent in trying to attain ever lower temperatures in order to investigate the behaviour of various materials at these temperatures. Each year records are set for the lowest temperature reached - in 1989 the record was 0.000000002 K.

Conversion of temperatures from Celsius to Kelvin is obtained simply by adding 273 to the Celsius temperature.

K = C + 273

whereK = Kelvin temperatureC = Celsius temperature

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Hence0oC = 273 K100oC = 373 KNote that

1oC change in temperature is equivalent to a 1 K change in temperature

The kelvin is the SI unit of temperature difference.

Example: (a) Convert the following Fahrenheit temperatures to Celsius(i) 78oF (ii) 10oF (iii) 300oF

(b) Convert the following Celsius temperatures to kelvin.(i) 20oC (ii) -65oC (iii) 227oC

Solution:(a) C = 5/9 (F - 32)(i) C = 5/9 (78 -32) = 5/9 (46) = 0.5555 x 46 = 25.56oC(ii) C = 5/9 (10 -32) = 5/9 (-22) = 0.5555 x -22 = -12.22oC(iii) C = 5/9 (300 -32) = 5/9 (268) = 0.5555 x 268 = 148.88oC(b) K = C + 273(i) K = 20 + 273 = 293 K(ii) K = -65 + 273 = 208 K(iii) K = 227 + 273 = 500 K

SAQQ1. Convert 44oF and -15oF to Celsius.Q2. Convert 200oC and -200oC to kelvin.Q3. Convert 82oF to kelvinQ4. A block of metal is heated from 30oC to 90oC. Calculate the change in temperature in (i) oC and (ii) K.

3. Specific Heat CapacityIf you had a number of different materials, in 1 kg portions, and supplied equal amounts of heat to each, you might be surprised to learn that the rise in temperature would vary from material to material. This is a result of the differences in specific heat capacity, c, between the materials.

The specific heat capacity of a material is defined as"the heat energy required to raise the temperature of 1 kg of a substance by 1oC without a change in state"

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SI units for specific heat capacity are J kg-1 K-1.

Typical values of specific heat capacity for different materials are shown in the table.

Substance Specific Heat Capacity / J kg-1 K-1

Water 4200Ice 2100

Glass 840Iron 460

Aluminium 900

Table 1:- Specific heat capacity values for a number of materials.

The specific heat capacity of water is 4 200 J kg-1 K-1, which means that it takes 4 200 J of heat energy to raise the temperature of 1 kg of water by 1oC. If the temperature of 2 kg of water had to be raised by 1oC, it should be clear that the energy required would be 2 x 4 200 = 8 400 J (i.e. 4 200 J is required by each kg of water). Additionally, if the temperature of 1 kg of water were to be raised by 2oC, then 2 x 4 200 = 8 400J would be required (4 200 to raise by 1oC then another 4 200 to raise it another oC). We can combine these to form an equation relating heat energy, mass of material and temperature change

Q = mcT

whereQ = heat energy supplied (J)m = mass of substance heated (kg)c = specific heat capacity of substance (J kg-1 K-1)T = change in temperature of substance (K)

Note: the Greek letter , capital delta, means "change in" something.

Example: Calculate the heat required to raise the temperature of 4 kg of water from 20oC to its boiling point. (Specific heat capacity of water = 4 200 J kg-1 K-1)

Solution: Q = ?, m = 4 kg, c = 4 200 J kg-1 K-1, T = 100 - 20 = 80oCNote - boiling point of water is 100oC.Heat requiredQ = mcT = 4 x 4 200 x 80 = 1 344 000 J

Note: The energy required doesn't depend on the actual temperatures involved - only on the temperature difference.

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Example: If 6.7 kJ of energy are supplied to 100 g of glass which is initially at 20oC, what is its final temperature? (Specific heat capacity of glass = 670 J kg-1 K-1)

Solution: Q = 6.7 kJ = 6 700 J, m = 100 g = 0.1 kg, c = 670 J kg-1 K-1, T = ?Q = mcT6 700 = 0.1 x 670 x T6 700 = 67 x T6 700 / 67 = T = 100Initial temp = 20oC, T = 100oC, henceFinal temperature = 20 + 100 = 120oC

Example: A 40 kg iron vat contains 10 kg of water at 15oC.(i) Calculate the heat required to raise the temperature to 75oC.(ii) How long will this take if a 5 kW heater is used.

(Specific heat capacity of water = 4 200 J kg-1 K-1)(Specific heat capacity of iron = 460 J kg-1 K-1)

Solution: The temperatures of both iron and water will rise on heating and it is usual to assume that the container and the liquid are always at the same temperature. (i) Heat required is the heat needed to raise both water and iron to 75oC.Can write this as Qtotal = Qwater + Qiron = (mcT)water + (mcT)iron

water ironm / kg 10 40 c / J kg-1 K-1 4 200 460T / K 75 - 15 = 60 75 - 15 = 60

Qtotal = (10 x 4 200 x 60) + (40 x 460 x 60) = 2 520 000 + 1 104 000Qtotal = 3 624 000 J = 3.624 x 106 J

(ii) The power of the heater is given, P = 5 kW = 5 000 W. (The term power is not used in the question but when a quantity is given with W as its unit, you then know that it is power)Power = energy / timeP = Q / t ----> Pt = Q ----> t = Q / Pt = 3.624 x 106 / 5 000 = 724.8 sDivide by 60 to find the number of minutes involved.

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724.8 / 60 = 12.08 mins.

SAQsQ1. Calculate the temperature rise when 70 000 J of heat are supplied to 2 kg of aluminium. (Specific heat capacity of aluminium = 900 J kg-1 K-1)Q2. How much heat is required to raise the temperature of a 5 kg block of ice at - 25oC to 0oC. (Specific heat capacity of ice = 2 100 J kg-1 K-1)Q3. Calculate the specific heat capacity of a material which undergoes a temperature rise of 10oC when 50 g of it are supplied with 0.75 kJ.Q4. A kettle, fitted with a 2 kW heater, is filled (2 kg) with water at 20oC. Calculate

(i) the heat required to raise the temperature of the water to boiling point

(ii) the time this will take.In this calculation you may ignore the heating of the kettle material itself.(Specific heat capacity of water = 4 200 J kg-1 K-1)Q5. A thermometer contains approximately 10 g of mercury and 100 g of glass. Calculate the heat required to raise the temperature of the thermometer from 15oC to 82oC.(Specific heat capacity of mercury = 140 J kg-1 K-1)(Specific heat capacity of glass = 670 J kg-1 K-1)

4. Changing State and Latent HeatSupplying heat to a material doesn't always lead to a rise in temperature. This is obvious to anyone who has ever boiled potatoes in a saucepan. Once the water has reached boiling point, it remains at 100oC, throughout the remainder of the cooking process. Meanwhile, heat is being supplied to the base of the saucepan - where does this extra energy go.

This problem is not peculiar to boiling water, any substance going through a change of state, from solid to liquid or liquid to gas, takes in heat without any change in temperature. This heat is referred to as latent heat, whereas heat which produces a change in temperature is termed sensible heat.

Latent heat - heat which produces a change in state without any change in temperature.Sensible heat - heat which produces a change in temperature without any change in state.

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If a substance is heated at a constant rate from the solid through to the gaseous state, a graph of temperature versus time would look as shown in Figure 5.

TIME

TEMPERATURE

melting point

boiling point

A

B

C

DE

Figure 5:- Temperature-time graph for the heating of a solid until it becomes gasesous.

The graph can be conveniently split into 5 sections.Section A - the solid is heated and its temperature rises until it reaches its melting point.Section B - at the materials melting point, the heat supplied is used to convert the solid to a liquid without a change in temperature (flat line). As heating proceeds more and more of the solid becomes liquid. At the left hand side of this line, substance is entirely solid and on the right the substance is completely liquid. Midway on this line, substance is half solid / half liquid.Section C - the liquid rises in temperature until it reaches its boiling point.Section D - at the materials boiling point, the heat supplied converts the liquid to a gas without a change in temperature.Section E - when all the liquid has been converted to gas, the heat supplied then results in temperature of the gas rising.

It should be clear from the description given above that:-Sections A, C, E are regions when heat supplied can be called sensibleSections B, D are regions where heat supplied can be called latent.Where a substance is experiencing a change in temperature, can always apply the equation Q = mcT in calculation.Where a substance is experiencing a change in state, as we will see in the next section, the equation Q = mL should be used.

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4.1. Specific Latent HeatThere are two separate definitions required for specific latent heat to cover the two possible changes of state.Solid to liquidSpecific latent heat of fusion, LF, is the energy required to convert 1 kg of solid at its melting point to liquid with no change in temperature.Liquid to gasSpecific latent heat of vaporisation, LV, is the energy required to convert 1 kg of liquid at its boiling point to gas with no change in temperature.

The SI unit for both these quantities is J kg-1 (read as joules per kilogramme). Typical values for water areLF = 3.30 x 105 J kg-1 LV = 2.25 x 106 J kg-1

The energy required for a change of state is given by Q = mL

whereQ = heat energy (J)m = mass (kg)L = specific latent heat of fusion or vaporisation (J kg-1)

Example: How much heat energy is required to convert 2 kg of ice at 0oC to water at the same temperature? (Specific latent heat of fusion for water = 3.3 x 106 J kg-1)

Solution: Q = ?, m = 2 kg, LF = 3.30 x 105 J kg-1

Q = m LF = 2 x 3.3 x 105 = 6.6 x 105 JNote: while the ice is melting, there is a mixture of water and ice. During the melting process, you should be clear that the temperature of both the ice and the water is 0oC.

Example: 5 kg of water at 80oC have to be converted to steam at 100oC. Determine(i) the quantity of heat required(ii) how long this will take with a 1.5 kW heater.

(Specific heat capacity of water = 4 200 J kg-1 K-1)(Specific latent heat of vaporisation of water = 2.25 x 106 J kg-1)

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Solution: This problem is slightly more complicated than the previous one. We now have water at 80oC which is to be converted to steam at 100oC - we have both a change in temperature and a change of state. This problem can only be solved by splitting the calculation into two parts. If you consider heating this water (1) initially all the heat added results in the temperature of the water rising. This continues until the water reaches its boiling point - use Q1 = mcT. (2) Once the water reaches its boiling point, the additional heat added results in the water being converted to steam without a change in temperature - use Q2 = m LV.

(i) Q1 = mcT, m = 5 kg, c = 4 200 J kg-1 K-1, T = 100 - 80 = 20 oC Q1 = 5 x 4 200 x 20 = 420 000 JQ2 = m LV, m = 5 kg, LV = 2.25 x 106 J kg-1

Q2 = 5 x 2.25 x 106 = 1.125 x 107 JTotal heat required, Q = Q1 + Q2 = 420 000 + 1.125 x 107 = 1.167 x 107 J

(ii) Power of heater, P = 1.5 kW = 1 500 W, Q = 1.167 x 107 J, t = ?Power, P = Q / tt = Q / P = 1.167 x 107 / 1 500 = 7780 s = 129.67 mins = 2.16 hrsThe SI unit of time is the second, so you may be wondering why I worked out my final answer in hours. The simple reason is that most people would have no idea how long 7780 s is. This is more clearly expressed at 2.16 hours.

In all questions involving changes in temperature and changes of state, it is useful to begin by clarifying in your own mind the various steps involved. The final example in this section deals with probably the most complex question on this topic as it covers two changes of state and a change in temperature for the three different states of water.

Example: How much heat energy is required to convert 3 kg of ice at - 20oC to steam at 140oC.(Specific heat capacity of ice = 2 100 J kg-1 K-1)(Specific heat capacity of water = 4 200 J kg-1 K-1)(Specific heat capacity of steam = 1 400 J kg-1 K-1)(Specific latent heat of fusion of water = 3.3 x 105 J kg-1)(Specific latent heat of vaporisation of water = 2.25 x 106 J kg-1)

Solution: Break the calculation up into the various stages of heating1 - heat raises temperature of the ice until it reaches its melting point - 0oC2 - ice melts to form water at 0oC3 - temperature of water rises from 0oC to until it reaches the boiling point 4 - water boils to form steam at 100oC5 - temperature of steam rises to a final value of 140oC

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Each stage corresponds to a different section of the graph shown in Figure 5.Q1, temperature of ice rises from -20oC to 0oC, use Q = mcT

m = 3 kg, c = 2 100 J kg-1 K-1, T = 20oCQ1 = 3 x 2 100 x 20 = 126 000 JQ2, ice converted to water at 0oC, use Q = m LF m = 3 kg, LF = 3.3 x 105 J kg-1

Q2 = 3 x 3.3 x 105 = 9.9 x 105 J Q3, temperature of water rises from 0oC to 100oC, use Q = mcTm = 3 kg, c = 4 200 J kg-1 K-1, T = 100oCQ3 = 3 x 4 200 x 100 = 1.26 x 106 JQ4, water converted to steam at 100oC, use Q = m LV m = 3 kg, LV = 2.25 x 106 J kg-1

Q4 = 3 x 2.25 x 106 = 6.75 x 106 J Q5, temperature of steam rises from 100oC to 140oC, use Q = mcTm = 3 kg, c = 1 400 J kg-1 K-1, T = 40oCQ5 = 3 x 1 400 x 40 = 168 000 J

Total heat required, Qtotal = Q1 + Q2 + Q3 + Q4 + Q5

Qtotal = 126 000 + 9.9 x 105 + 1.26 x 106 + 6.75 x 106 + 168 000= 9.294 x 106 J

You should note that by far the largest contribution to the total is Q4, which is the heat required to change the state of water from liquid to gas. This is generally true for most substances.

SAQsQ1. Calculate the heat energy required to melt 500 kg of copper at its melting point, given that its specific latent heat of fusion = 2.1 x 105 J kg-1.Q2. 4 kg of iron at 20oC is to be heated until it is fully molten. The melting point of iron is 1230oC. Determine the

(i) total heat energy required(ii) time this would take when heat is supplied at the rate of 20 kW.

(Specific heat capacity of iron = 460 J kg-1 K-1)(Specific latent heat of fusion of iron = 1.4 x 105 J kg-1)Q3. Redraw the temperature-time diagram in Figure 5. This time draw it to scale for heating of ice to water using the figures in the last worked example. Assume a heating rate of 2 kW. (Hint: from the values in the last example, determine how long each stage of heating would last)

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4.1.1. Units

This is a good point to review some of the units we have come across in the last few sections. The units for specific heat capacity and specific latent heat may have left you wondering "where have these come from?". These units arise from our basic idea about the equals sign. We are all used to ensuring that the numbers on either side of an equals sign are in fact equal e.g.7 + 2 = 9 correct5 porcupines = 5 kangaroos incorrectThe second example is incorrect because although the numbers on each side of the equals sign are equal, the units associated with the numbers are not the same. In any science equation, we must ensure that the quantities on each side of an equals sign are EXACTLY equal both numerically and in terms of units. Let us look at how the units for specific heat capacity may be determined from the basic equation.Q = mcTThe known units are:Q - J (joule), m - kg (kilogramme), T - K (kelvin)Normally we simply enter the numbers associated with the quantities into the equation. As we are trying to find the appropriate units for c in this equation we need to enter only the units of each quantity. We will denote the units of c by [c]J = kg x [c] x KRearrange leaving [c] on its ownJ / (kg x K) = [c] = J kg-1 K-1 Note that the units of c are joule divided by kilogramme and kelvin. These last two quantities can be written on the same line as the joule, so long as the power is made negative i.e.1 / kg = kg-1

1 / m2 = m-2

SAQ - Derive the unit for specific latent heat using the latent heat equation.

5. Thermal ExpansionWhen the temperature of a rod of material rises, we have already seen that the vibration of the constituent atoms increases. This tends to result in an increase in the length of the rod. This effect is called thermal expansion. In general, materials expand with rise in temperature and contract on cooling.

There are a small number of exceptions to this rule. The single most important exception is with water which actually expands as it cools between 4oC and 0oC.

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Hence, water tends to expand on cooling to its freezing point. This then leads to (i) the ice pushing up through the aluminium bottle top on a milk bottle on very cold mornings, (ii) a plastic container of milk which becomes more bloated looking when the milk freezes and (iii) pot holes. Pot holes start off as fine cracks in the road surface. In wet weather, these cracks fill with water which expands during frosty nights. This expansion causes further cracks to develop in the surrounding roadway. The vibration produced by passing cars, or repeated frosty nights, will eventually break up the road surface to form potholes.

It is important to note that water between 4oC and 0oC is an exception and that in general, a material will expand with increase in temperature and contract as temperature falls.

5.1. Linear Expansion Coefficient, The expansion a material undergoes as its temperature rises depends on such factors as the original length of the material and the temperature rise involved. However, it is important to note that the expansion is strongly dependent on the actual material involved. A measure of the degree to which a particular material will expand with temperature rise is given by the linear expansion coefficient, , of the material.

Linear expansion coefficient is defined as the fractional change in length of a substance per unit change in temperature. The SI unit of linear expansion coefficient is K-1 (said as "per degree kelvin"). Typical values of for a range of materials are shown in the table below. We will use this definition to determine a formula for the change in length (L) of a material of original length, Lo, undergoing a temperature change T (see Figure 6).

Lo Lo L

T To To + T

Figure 6:- Rod of material of length Lo expands as it experiences an increase in temperature.

= fractional change in length per unit change in temperature= fractional change in length / change in temperature

fractional change in length = change in length / original length = L / Lo = (L / Lo) / T

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Transfer T to right hand side of equationT = L / Lo

Transfer Lo to right hand side

Lo T = Lor

L = LoT

i.e. the change in length equals the product of the original length, the linear expansion coefficient and the change in temperature.

Example: The railway line from Waterford to Carlow is approximately 100 km long. Calculate the difference in the length of the line between a cold night (- 5oC) and a warm day (25oC).Linear expansion coefficient of iron = 12 x 10-6 K-1

Solution:Original length of track, Lo = 100 km = 100 000 mT = 25 - (-5) = 30 K = 12 x 10-6 K-1

L = ?L = LoT

= 100 000 x 12 x 10-6 x 30 = 36 m

The railway track will be 36 m longer at 25oC than at -5oC. This expansion has to be allowed for, otherwise the line will buckle during expansion. Expansion joints (Figure 7) are routinely used to allow the track to expand and contract with temperature without producing a distorted track which would be dangerous and unusable.

expansion gap

rail

Figure 7:- Expansion gap on a railway line.

Expansion joints are very common. For example, a concrete footpath will usually possess a layer of tar/rubber compound every couple of yards. As the concrete expands with rises in temperature, the rubber layer becomes compressed and rises above the footpath surface. At lower temperatures the concrete contracts and the layer becomes stretched. The advantage of such a joint is that it prevents areas of

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concrete expanding into each other which would produce stresses in the materials and may ultimate lead to cracking of the concrete.

Thermal expansion may cause problems for engineers and its effects must be allowed for but it can also be used to positive effect. In the next example, a ring which is too small to fit over a piston is heated causing it to expand and ultimately fit over the piston. The ring will then cool, contract and form an extremely tight seal around the piston. We will look at other beneficial uses of thermal expansion after this question.

Example: An iron ring of inner diameter 3.453 cm at 20oC is to fit on a piston cylinder of diameter 3.458 cm. To what temperature must the ring be heated so that it just fit over the piston.Linear expansion coefficient of iron = 12 x 10-6 K-1

Solution: It is very important to realise that all lengths expand according to our basic equation for thermal expansion. Hence, we can determine the expansion of the diameter of a circular piece of material using this formula. You may argue that if the ring of material expands then the hole in the centre should get smaller. This does not happen, as a ring of material is heated, the hole in the centre expands along with the ring itself.L = 3.458 - 3.453 = 0.005 cm = 5 x 10-3 cm = 5 x 10-5 mLo = 3.453 cm = 3.453 x 10-2 m = 12 x 10-6 K-1

T = ?L = LoT5 x 10-5 = 3.453 x 10-2 x 12 x 10-6 x T5 x 10-5 = 4.252 x 10-7 x T5 x 10-5 / 4.252 x 10-7 = TT = 117.6 K

Final temperature of ring = 20 + 117.6 = 137.6oC

5.1.1. Rivets

Rivets are metal fasteners used to hold steel plates tightly together (Figure 8). (i) The rivet is first heated and inserted into holes in both sheets to be fastened. (ii) Its end is then hammered over.(iii) As the rivet cools, it contracts and draws the two plates tightly together.

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hot rivet

steel plates

contracting rivet draws sheets together

(i) end hammered-over(ii)

(iii)

Figure 8:- The process by which rivets produce a tight seal between two metal sheets.

5.1.2. Bimetallic Strip

The bimetallic strip is a useful device formed by two thin strips of different metals strongly bonded together. The two metals most commonly used are brass and invar. You are probably familiar with brass and may know that it is formed from a combination of copper and zinc. Invar, however, is less well known. It is a steel/nickel alloy which has found numerous uses because it has a very small linear expansion coefficient. The name invar is an abbreviation of invariable - the length of a strip of invar barely changes even for large increases in temperature.Linear expansion coefficient of invar = 1 x 10-6 K-1 Linear expansion coefficient of brass = 18.7 x 10-6 K-1

As you can see from these figures, the expansion coefficient for brass is approximately 19 times that of invar. It is this difference in expansion coefficient that makes a brass/invar bimetallic strip so useful.

If equal lengths of invar and brass at 20oC are bonded together to form a bimetallic strip, the strip formed will be flat as shown below. If the temperature of the strip rises, the brass will look to expand 19 times as much as the invar, this cannot happen if the strip remains flat. The strip must bend with the brass on the (longer) outside and the invar on the inside.

brass

invar

brass

invar

Figure 9:- Bending of a flat bimetallic strip with increase in temperature.

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5.1.3. Thermostat

A thermostat is a device which is used to maintain the temperature of an enclosure (e.g. oven, room etc.) at a preset value. In a car a thermostat turns the fan on if the temperature of the radiator gets too hot. A simple thermostat is shown in Figure .

to heater circuit

electrical contacts

temperature control knob

bimetallic strip

invar brass

m s

block

Figure 10:- Schematic diagram of a thermostat.

When electrical contacts m and s (see Figure 10) are touching, this completes the electrical circuit and current flows in the heater circuit. This results in heat being supplied to the oven which produces a rise in temperature. As temperature rises the bimetallic strip will bend in such a way that at a particular preset temperature (indicated on the temperature control knob) the contact between m and s will be broken. This stops the current flow and effectively switches off the heaters. Over time the oven cools and the bimetallic strip straightens until the contacts meet once more, turning on the heater. Hence, the heaters are turned on and off as the temperature falls below and rises above the preset temperature. The preset temperature can be increased by rotating the control knob which pushes contact s to the left. This then increases the bend needed in the bimetallic strip to break contact, increasing the temperature at which the heaters will be switched off.

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5.2. Areal (or Superficial) ExpansionIn previous sections we looked at the expansion of a length of material with increasing temperature. We will now extend this approach to look at how the area of a sheet of material will change with change in temperature.Consider a square sheet of metal of side, Lo, at room temperature which undergoes an increase in temperature T. What is the new area of the sheet?

T

Lo

Lo

Lo + L

Lo + LAo A

Figure 11:- Expansion of a square sheet of material with increasing temperature.

As a result of the increase in temperature the sheet expands with each side of the sheet now of length

L = Lo + L = Lo + LoTHence the new area of the sheet

A = L x L = (Lo + LoT)(Lo + LoT)Multiplying out the brackets gives

A = Lo2 + Lo

2T + Lo2T + Lo

22T2

A = Lo2 + 2 Lo

2T + Lo22T2

The linear expansion coefficient of most materials is of the order of 10-5 K-1. The final term in the equation for A contains 2 which will have a value of approximately 10-10 (as 10-5 x 10-5 = 10-10). This value is so small that it will make the size of this final term insignificant allowing us to ignore it in the remainder of this analysis.

A = Lo2 + 2 Lo

2TWe can simplify this further

Lo2 = Ao = original area of the sheet

ThereforeA = Ao + 2 AoT

and A - Ao = A = 2 AoT

This formula looks quite similar to L = LoT

To retain the similarity a new factor is defined, called the coefficient of areal expansion, . This is defined as the fractional change in area per unit temperature

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change. As with the coefficient of linear expansion, the SI unit of measurement is the K-1.The formula for areal expansion becomes

A = AoT

whereA = change in area (m2)Ao = original area (m2) = coefficient of areal expansion (K-1)T = change in temperature (K)

Note that for the equationsA = AoT and A = 2 AoT to be equal, it should be obvious that

= 2 i.e.

coefficient of areal expansion = 2 x (coefficient of linear expansion)

Q. A sheet of steel measures 10 m x 12 m at a temperature of 20oC. Determine the likely increase in area of the sheet when its temperature rises to 100oC. Linear expansion coefficient of steel = 12 x 10-6 K-1.

Note - Only the linear expansion coefficient is given in the question. Because there is such a simple relationship between the linear and areal coefficients, only linear coefficients are usually quoted.

Solution:This question can be solved in two ways(i) Simply work with the individual lengths on each side of the sheet and determine the new area from the expansion in the lengths of each side.(ii) Use the areal expansion coefficient formula.We will use each technique in turn and compare the results. = 12 x 10-6 K-1, = 2 = 24 x 10-6 K-1, T = 100 - 20 = 80 K

(i) Steel measures 10 m x 12 m at 20oC, calculate each length at 100oC10 m L = LoT = 10 x 12 x 10-6 x 80 = 9.6 x 10-3 mNew length = 10 + 9.6 x 10-3 = 10.009 6 m12 m L = LoT = 12 x 12 x 10-6 x 80 = 0.011 52 mNew length = 12 + 0.011 52 = 12.011 52 m

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New area = 10.009 6 x 12.011 52 = 120.230 51 m2

Change in area = 120.23051 - 120 = 0.230 51 m2

(ii) Initial area, Ao = 10 m x 12 m = 120 m2

= 2a = 24 x 10-6 K-1, T = 80 K

A = AoT= 120 x 24 x 10-6 x 80 = 0.2304 m2

You should notice that the two results are not exactly equal. The answer to method (i) is in fact the more accurate of the two results. In deriving the formula A = AoT, we ignored the Lo

2T2 term. This means that the answer to method (ii) is slightly inaccurate. However, the difference between the results is only 0.000 11 m2 out of a total area of 120 m2 and this accuracy is acceptable in most situations.

5.3. Volume ExpansionIn the vast majority of situations, it is the expansion of 3-dimensional objects or quantities of liquids which are of interest to the engineer.. We can visualise a cube of material of side Lo which is heated through a temperature change T producing an expansion of length on each side of

L = LoTUsing a similar argument to that used to derive the formula for areal expansion, it can be shown that the change in volume is given by

V = VoT

WhereV = change in volume (m3)Vo = original volume (m3)

= coefficient of volume expansion (K-1)T = change in temperature (K)

Note that the SI unit of the volume expansion coefficient is K-1 as with linear and areal coefficients.

The volume expansion coefficient is again directly related to the linear expansion coefficient for solids

Volume expansion coefficient = 3 x (Linear expansion coefficient) = 3 x

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As we have already seen, linear expansion coefficients are normally quoted for solid materials. The volume expansion coefficient can then be obtained using the equation above. Liquids on the other hand do not have defined linear expansion coefficients as sensible linear expansion measurements cannot be carried out on a rod of liquid - liquid will always take up the shape of its container though obviously the volume of liquid will increase with temperature. As a result, the volume expansion coefficient is usually quoted for liquids.

Q. A steel tank can hold a volume of 4 m3 at 20oC. Calculate its likely capacity were its temperature to rise to 80oC. Linear expansion coefficient of steel = 12 x 10-6 K-1

Vo = 4 m3, = 3 x = 3 x 12 x 10-6 = 36 x 10-6 K-1, T = 80 - 20 = 60 KV = VoT

= 4 x 36 x 10-6 x 60= 8.64 x 10-3 m3

Capacity at 80oC = 4 + 8.64 x 10-3 = 4.008 64 m3

Q. A steel drum is filled to the brim with 60 litres of petrol early on a cold morning when the temperature is 0oC. During the heat of the midday sun the temperature of the drum and petrol rises to 35oC. Determine the volume of petrol, if any, which overflows.LInear expansion coefficient of steel = 12 x 10-6 K-1

Volume expansion coefficient of petrol = 950 x 10-6 K-1

SolutionPetrol will overflow is the expansion of the petrol is greater than the expansion of the drum capacity. Hence, we must determine the volume expansion of both steel drum and petrol.PetrolVo = 60 litres (note - there is no need to convert this to m3, simply remember that the unit for Vo must be the same as for V), = 950 x 10-6 K-1, T = 35 - 0 = 35 K

V = VoT= 60 x 950 x 10-6 x 35= 1.995 litres

Petrol expands by 1.995 litres

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Steel drumVo = 60 litres, = 3 x = 3 x 12 x 10-6 = 36 x 10-6 K-1, T = 35 - 0 = 35 K

V = Vo T= 60 x 36 x 10-6 x 35= 0.0756 litres

The steel drum expands its capacity by 0.0756

Obviously from the above, the volume of the petrol expands more than the capacity of the drum and so some petrol will overflow. The amount of overflow equals

Overflow = 1.995 - 0.0756 = 1.9194 litresThe reason why petrol will overflow is because it has a much larger volume expansion coefficient than that of steel. In general, liquids have a larger than metallic solids. However, if a plastic material had been used in the above example the overflow would have been greatly reduced as the linear expansion coefficient of polythene is 250 x 10-6 K-1 which means a volume expansion coefficient of 750 x 10-

6 K-1.

SAQsSAQ1 - A block of brass has a volume of 0.245 m3 at 20oC. Determine its new volume when placed in an oven at a temperature of 250oC.Linear expansion coefficient of brass = 18.7 x 10-6 K-1 SAQ2 - A tray of containing 560 cm3 of mercury at 15oC is heated until it reaches 80oC. Determine the new volume of the mercury.Volume expansion coefficient of mercury = 180 x 10-6 K-1 SAQ3 - A block of steel has dimensions 40 cm x 30 cm x 200 mm at 18oC. Determine the volume of the block when it is at 145oC.Linear expansion coefficient of steel = 12 x 10-6 K-1

SAQ4 - A homeowner orders 1000 litres of oil from a local supplier. The oil is delivered in the middle of the day when the temperature is 20oC. The owner checks the oil level later that evening, when the temperature has fallen to 0oC, to find that the oil volume in the storage tank is well below 1000 litres.

(i) Explain why the owner has not been 'done'.(ii) What was the likely volume of oil indicated when the owner checked the

tank (ignore the contraction of the steel tank).(iii) If the oil is checked a day later and the volume is now 1008 litres,

determine the oil temperature.Volume expansion coefficient of petrol = 950 x 10-6 K-1

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6. Heat TransferThere are three main ways in which heat can be transferred from one place to another.

These are

1) Conduction

2) Convection

3) Radiation

Conduction and convection are atomic (or molecular) processes, while radiation is

not.

6.1. ConductionConduction is the only way that heat can travel through opaque solids.A solid can be thought of as made up of atoms which act like hard spheres attached to each other by spring-like bonds (Figure 12).

Figure 12:- Solid materials can be represented as regularly spaced atoms connected by spring like

bonds.

At any given temperature all the atoms are vibrating to the same extent. If one part of the solid is now heated, the atoms at this point vibrate more (they are now at a higher temperature) and as they are linked to their neighbours by spring-like bonds, neighbouring atoms also vibrate more. These atoms then pass on this vibration to their neighbours and so, in this way, the heat is passed through the solid. Note that in conduction heat travels from hot to cold. The thermal conductivity, k, is a measure of how good the material is at conducting heat. Typical values are given in Table 2.

Substance k / W m-1 K-1

Copper 390Iron 46

Concrete 1.3

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Glass(typical) 0.84Water 0.57

Air 0.024

Table 2:- Thermal conductivities, k, of a range of materials.

From the table it is clear that metals are good conductors of heat. The diagram below, Figure 13, shows a rod of material along which heat will be conducted.

Figure 13:- Heat conduction though a rod of material.

The thermal conductivity of a material is defined as the rate of heat flow through 1 m2

area of material which is 1 m thick and across which there is a temperature difference of 1 kelvin.

The rate at which heat is conducted through a material depends on (1) thermal conductivity of material, k

directly proportional (2) area through which heat is conducted, A

directly proportional(3) difference in temperature across the material, T1 - T2 = T

directly proportional(4) thickness of material through which heat travels, L

inversely proportionalFrom this information, we can directly formulate a formula for the rate at which heat is conducted through a material.

Q/t = kAT / LWhere

Q/t = rate at which heat is conducted (i.e. heat conducted per sec) -Wk = thermal conductivity of material (W m-1 K-1)A = area through which heat is conducted (m2)T = temperature difference across the material (K)

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L = thickness of material (m)

Question 1. The temperature of a room is maintained at 20oC above the outside temperature. If all the heat losses occur through a plate glass window of area 2 m2 and thickness 3 mm, calculate the rate of heat flow out through the glass.Thermal conductivity of glass = 1.1 W m-1 K-1

Q/t = ? A = 2 m2 L = 3 mm = 0.003 m T = 20 Kk = 1.1 W m-1 K-1

Q/t = kAT / L= 1.1 x 2 x 20 / 0.003 = 14, 666 J/s

According to this calculation, heat is lost from the room at the rate of 14, 666 joules per second. This is an extremely high figure and is equivalent to the heat output of 7 electric heaters with two bars working continuously. It seems that the formula has not produced a realistic figure. The problem is not with the formula but with the figures we have used. The temperature difference of 20oC corresponds to the temperature difference between the air in the room and the outside air. However, the value for T in the formula should refer to the difference in temperature between the two surfaces of the glass.

3 mm thick windowpanestill air

20oC

0oC

T

inside outside

Figure 14:- Temperature profile across a window pane.

From Figure 14, it should be clear that the difference in temperature between the two surfaces of the glass is much less than the 20oC used in the calculation. In fact there are layers of still air on either side of the windowpane which act as insulators ensuring that the inner glass surface has a lower temperature than the room while the outer glass surface is at a higher temperature than the outside air.

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Question 2. An icebox is insulated with cork lining which is 50 mm thick. Its inner surface is kept at a temperature of 2oC while its outer surface is at 17oC. If the cork box has a surface area of 2.4 m2, determine how much heat flows into the box per hour.Thermal conductivity of cork = 0.046 W m-1 K-1

Question3: A copper bar has a length of 60 cm and is 20mm in diameter. If one end is heated such that its temperature is maintained at 100oC, and the other end is dipped in an ice/water mixture which is always at 0oC,

(i) Determine the rate at which hear is conducted through the metal(ii) How much ice will melt in 45 minutes.(Note: kCu = 400 W m-1 K-1 : L ice = 3.3 x 105 J kg-1)

6.1. Convection

Convection is the main way that heat can travel in liquids and gases.In Table 2 for thermal conductivities, it was clear that the values for air and water were very low. These figures deal only with heat transfer by conduction, it is obvious that heat can be easily dispersed through the air in a room or in the water in a kettle. Heat must be able to travel through gases and liquids by other means.A simple experiment (see Figure 15) -

(1) A piece of ice is floated on water in a test-tube and the tube is heated from the bottom.

Very soon the ice melts as expected. (2) A piece of ice is now weighed down so that it sinks in water and the top of

tube is heated.The water at the top of the tube boils before the ice melts!

ice Heat

(i) (ii) Heat

Figure 15:- Experiment demonstrating heat flow in liquids.

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This may surprise you. Heat transfer in liquids and gases occurs by a different process to that for solids and this process makes use of the fact that the molecules in a gas or liquid can themselves move around unlike in solids.

If heat is supplied to the bottom of a beaker of water, the water at the bottom, nearest the heat source, takes in the heat resulting in a temperature rise. This causes the water in this region to expand which reduces its density (remember : density = mass / volume, if the water expands its volume increases and so its density must fall). Lower density liquids always float on higher density liquids e.g. oil on water. The warmed water rises as its density falls to be replaced by the cooler water from the top. This water is now next to the heat source and so its temperature will increase, expand, density falls and it will rise up. This repetitive process sets up continuous movement of liquid in the beaker and eventually all the liquid is heated in turn by the heater. This circulation is called a convection current. This process explains why in any liquid or gas, the hottest material is to be found at the top - heat rises.

(1) (2)

(1) warm water, which is less dense, rises

(2) denser, cold water falls to replace (1) Heat

Figure 16:- Water circulation in a convection current.

The results of the experiment illustrated in Figure 15 can now be easily explained.(1) The heat delivered to the bottom of the test-tube heats the water here - this

water rises to the top and transfers the heat to the ice cube melting it.(2) The heat delivered to the top of the test-tube heats the water at the top

reducing its density. Because its density is lower than the cooler water below it, the hot water will remain at the top receiving more and more heat until it boils. The ice eventually melts because the test-tube glass itself heats up and passes the heat around the tube by conduction which is then conducted into the ice.

As a result of convection, heaters used to heat liquids or gases (including the water in a kettle or the air in a room) are generally located at the bottom of the container. An exception is the immersion heater used in a domestic hot water boiler which only

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heats the top portion of a quantity of water when small amounts of hot water are required.

6.2. Radiation Radiation is the only way that heat can be transferred through a vacuum such as outer space and as such is the only way we can receive heat from the Sun. Conduction and convection both require atoms/molecules to carry or pass the heat on.

All objects above 0 K radiate heat in the form of electromagnetic waves. This radiation travels in straight lines at the speed of light in whatever medium it is travelling. The nature of this radiation depends on the temperature of the object. For example the human body has a temperature of approximately 300 K and as a result radiates heat in the form of infra-red (IR) radiation which is invisible to the human eye but which can be detected using IR cameras. An object which is considerably hotter, e.g. an electric bar heater, glows ‘red hot’ when the temperature reaches approximately 1000 K and an object approaching 2000 K will glow ‘white hot’ (e.g. the filament in a bulb). As well as the type of radiation changing with temperature, the intensity of the radiation also increases very strongly with temperature. The exact dependence is given by Stefan’s Law which states that the intensity of the radiation is proportional to the fourth power of absolute temperature. In mathematical form, we have

Q/t T4 or Q/t = ε AT4

whereQ/t = rate at which heat is radiated from object, measured in Watts ε = emissivity = Stefan’s constant (= 5.67 x 10-8 W m-2 K-4)A = surface area of object (m2)T = absolute temperature of object in kelvin

The emissivity of a body is calibrated on a scale of 0 to 1. A body that is a perfect emitter has an emissivity of 1 and is also a perfect absorber. Such bodies are called blackbodies. A body that is a poor emitter (ε =0) is also a poor absorber, but is a good reflector of heat.

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Question:- A radiator has a surface area of 1.8 m2 and a surface temperature, when hot, of 60oC. Determine the rate at which heat is radiated by the radiator if it behaves as a blackbody.

Solution: Use Stefan’s equationQ/t = ε AT4 Q/t = ? = 5.67 x 10-8 W m-2 K-4 A = 1.8 m2 ε=1T = 60oC = 60 + 273 = 333 KNote - When using Stefan’s equation, all temperatures must be in kelvin.

Q/t = ε AT4 = (1)(5.67 x 10-8 ) x 1.8 x 3334

= 1.0206 x 10-7 x 1.23 x 1010 = 1 254 W

Note that this value is purely the rate at which radiant heat is emitted by the radiator. In fact, more heat is supplied to the room via convection from the radiator than by direct radiation. Convection occurs as air next to the radiator is heated, becomes less dense and rises being replaced by cooler, denser air. There is usually an upward flow of hot air above a hot radiator.

Question - An electric bar heater is 30 cm long and 2 cm in diameter. It radiates heat at the rate of 1 kW. Determine the temperature of the bar, assuming ε=1.

For any material, at a particular temperature, the nature of its surface will determine how good it is at radiating heat. Good radiators are good absorbers but poor reflectors of heat radiation.Dull, dark surfaces tend to be very good radiators of heat.

- wear dark clothes in winter - dark clothes are good absorbers of radiant heatBright shiny surfaces tend to be very poor radiators of heat.

- shiny surfaces are good reflectors of radiant heat - keep you cool

To summarise Radiant heat is emitted by all objects with temperature greater than 0 K The nature of the radiation depends on the temperature of the object Radiation travels at the speed of light Radiation can travel through a vacuum Radiation is absorbed by dull, dark surfaces and reflected by shiny, bright surfaces

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6.3. Conduction, Convection and RadiationIt is important to note that while there are three separate heat transfer processes, it is quite normal for all three to be occurring simultaneously. For example consider the case where a piece of hot glass is left on a table to cool. Glass will cool by conduction - heat will flow from the hot glass into the cool material of the table, the rate of cooling depends on what the table surface is made from. Glass will cool by convection - the air around the glass will be heated by the glass and rise taking heat away. Cooler air will then come in contact with the glass surface, taking heat away from the glass as its temperature increases and it rises. Heat loss to surrounding air will be enhanced if the glass is placed in a draught.Glass will cool by radiation - glass radiates heat in the form of light and, if it is hot enough, this light will be visible.

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HEAT AND TEMPERATURE - Antimatter | Life in a · Web viewIn this section of the course we will look at what heat and temperature are, the effects of adding and removing heat from materials

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