Download HEAT AND TEMPERATURE

Heat and Temperature
1. INTRODUCTION .............................................................................................................................. 2
2. TEMPERATURE SCALES ............................................................................................................... 3
2.1. CELSIUS ......................................................................................................................................... 3
2.2. FAHRENHEIT .................................................................................................................................. 3
2.3. KELVIN .......................................................................................................................................... 4
2.3.1. Absolute Zero ........................................................................................................................ 4
3. SPECIFIC HEAT CAPACITY ......................................................................................................... 6
4. CHANGING STATE AND LATENT HEAT ................................................................................... 9
4.1. SPECIFIC LATENT HEAT ............................................................................................................... 11
4.1.1. Units .................................................................................................................................... 14
5. THERMAL EXPANSION ............................................................................................................... 14
5.1. LINEAR EXPANSION COEFFICIENT, ............................................................................................ 15
5.1.1. Rivets ................................................................................................................................... 17
5.1.2. Bimetallic Strip ................................................................................................................... 18
5.1.3. Thermostat .......................................................................................................................... 19
5.2. AREAL (OR SUPERFICIAL) EXPANSION ......................................................................................... 20
5.3. VOLUME EXPANSION ................................................................................................................... 22
6. HEAT TRANSFER .......................................................................................................................... 25
6.1. CONDUCTION ............................................................................................................................... 25
6.2. CONVECTION ............................................................................................................................... 28
6.3. RADIATION .................................................................................................................................. 30
6.4. CONDUCTION, CONVECTION AND RADIATION .............................................................................. 31
1
Heat and Temperature
1.
Introduction
In 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 energy
quantities of energy are measured in joules, J
Temperature is the measure of how hot something is on a chosen scale
temperature 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.
T = 50 oC
T = 50 oC
mass, 2 kg
mass, 1 kg
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
2
Heat and Temperature
2.
Temperature Scales
As 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.
Celsius
The Celsius scale uses
melting point of pure ice = 0oC
boiling 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.
Fahrenheit
The 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 = 0oF
body temperature = 90oF
These temperatures were later amended slightly, but the resulting scale now gives
melting point of pure ice = 32oF
boiling point of pure water = 212oF
at 760 mmHg.
Fahrenheit can be readily converted to Celsius using the relationship
C = 5/9 (F - 32)
where
3
Heat and Temperature
C = temperature in Celsius
F = temperature in Fahrenheit
2.3.
Kelvin
As 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
4
Heat and Temperature
extended back to the temperature axis i.e. to a point where gas volume is zero, they all
met up at the same point, read off the scale to be - 273oC. This is shown in Figure 4.
gas volume
- 273 oC
0 oC
100 oC
temperature
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
where
K = Kelvin temperature
C = Celsius temperature
5
Heat and Temperature
Hence
0oC = 273 K
100oC = 373 K
Note 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
SAQ
Q1. Convert 44oF and -15oF to Celsius.
Q2. Convert 200oC and -200oC to kelvin.
Q3. Convert 82oF to kelvin
Q4. A block of metal is heated from 30oC to 90oC. Calculate the change in
temperature in (i) oC and (ii) K.
3.
Specific Heat Capacity
If 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"
6
Heat and Temperature
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
Water
Ice
Glass
Iron
Aluminium
Specific Heat Capacity / J kg-1 K-1
4200
2100
840
460
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 = mcT
where
Q = 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 = 80oC
Note - boiling point of water is 100oC.
Heat required
Q = mcT = 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.
7
Heat and Temperature
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 = mcT
6 700 = 0.1 x 670 x T
6 700 = 67 x T
6 700 / 67 = T = 100
Initial temp = 20oC, T = 100oC, hence
Final 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 = (mcT)water + (mcT)iron
m / kg
c / J kg-1 K-1
T / K
water
10
4 200
75 - 15 = 60
iron
40
460
75 - 15 = 60
Qtotal = (10 x 4 200 x 60) + (40 x 460 x 60) = 2 520 000 + 1 104 000
Qtotal = 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 / time
P=Q/t
---->
Pt = Q
----> t = Q / P
6
t = 3.624 x 10 / 5 000 = 724.8 s
Divide by 60 to find the number of minutes involved.
8
Heat and Temperature
724.8 / 60 = 12.08 mins.
SAQs
Q1. 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 Heat
Supplying 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.
9
Heat and Temperature
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.
boiling point
E
D
TEMPERATURE
C
melting point
B
A
TIME
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 sensible
Sections 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 = mcT 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.
10
Heat and Temperature
4.1.
Specific Latent Heat
There are two separate definitions required for specific latent heat to cover the two
possible changes of state.
Solid to liquid
Specific 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 gas
Specific 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 are
LF = 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
where
Q = 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 J
Note: 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)
11
Heat and Temperature
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 = mcT. (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 = mcT, 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 J
Q2 = m LV, m = 5 kg, LV = 2.25 x 106 J kg-1
Q2 = 5 x 2.25 x 106 = 1.125 x 107 J
Total 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 / t
t = Q / P = 1.167 x 107 / 1 500 = 7780 s = 129.67 mins = 2.16 hrs
The 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 heating
1 - heat raises temperature of the ice until it reaches its melting point - 0oC
2 - ice melts to form water at 0oC
3 - temperature of water rises from 0oC to until it reaches the boiling point
4 - water boils to form steam at 100oC
5 - temperature of steam rises to a final value of 140oC
12
Heat and Temperature
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 = mcT
m = 3 kg, c = 2 100 J kg-1 K-1, T = 20oC
Q1 = 3 x 2 100 x 20 = 126 000 J
Q2, 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 = mcT
m = 3 kg, c = 4 200 J kg-1 K-1, T = 100oC
Q3 = 3 x 4 200 x 100 = 1.26 x 106 J
Q4, 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 = mcT
m = 3 kg, c = 1 400 J kg-1 K-1, T = 40oC
Q5 = 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.
SAQs
Q1. 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)
13
Heat and Temperature
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
correct
5 porcupines = 5 kangaroos
incorrect
The 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 = mcT
The 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 K
Rearrange leaving [c] on its own
J / (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 Expansion
When 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.
14
Heat and Temperature
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 + T
To
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
15
Heat and Temperature
Transfer T to right hand side of equation
T
= L / Lo
Transfer Lo to right hand side
LoT = L
or
L = LoT
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 m
T = 25 - (-5) = 30 K
 = 12 x 10-6 K-1
L = ?

L
= LoT
-6
= 100 000 x 12 x 10 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.
rail
expansion gap
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 concrete
16
Heat and Temperature
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 m
Lo = 3.453 cm = 3.453 x 10-2 m
 = 12 x 10-6 K-1
T = ?

L
= LoT
5 x 10-5 = 3.453 x 10-2 x 12 x 10-6 x T
5 x 10-5 = 4.252 x 10-7 x T
5 x 10-5 / 4.252 x 10-7 = T
T = 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.
17
Heat and Temperature
hot
rivet
steel plates
(i)
(ii)
end hammered-over
contracting rivet draws
sheets together
(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
brass
invar
invar
Figure 9:- Bending of a flat bimetallic strip with increase in temperature.
18
Heat and Temperature
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
block
bimetallic
strip
invar
brass
temperature
control knob
m
s
electrical
contacts
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.
19
Heat and Temperature
5.2.
Areal (or Superficial) Expansion
In 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?
Lo + L
Lo
T
Lo
Ao
A
Lo + L
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 + LoT
Hence the new area of the sheet
A = L x L = (Lo + LoT)(Lo + LoT)
Multiplying out the brackets gives
A = Lo2 + Lo2T + Lo2T + Lo22T2
A = Lo2 + 2 Lo2T + Lo22T2
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 Lo2T
We can simplify this further
Lo2 = Ao = original area of the sheet
Therefore
A = Ao + 2 AoT
and
A - Ao = A = 2 AoT
This formula looks quite similar to
L = LoT
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
20
Heat and Temperature
change. As with the coefficient of linear expansion, the SI unit of measurement is the
K-1.
The formula for areal expansion becomes
A = AoT
where
A = change in area (m2)
Ao = original area (m2)
 = coefficient of areal expansion (K-1)
T = change in temperature (K)
Note that for the equations
A = AoT and A = 2 AoT 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 100oC
10 m L = LoT = 10 x 12 x 10-6 x 80 = 9.6 x 10-3 m
New length = 10 + 9.6 x 10-3 = 10.009 6 m
12 m
L = LoT = 12 x 12 x 10-6 x 80 = 0.011 52 m
New length = 12 + 0.011 52 = 12.011 52 m
21
Heat and Temperature
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
= AoT
= 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 = AoT,
we ignored the Lo2T2 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 Expansion
In 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 = LoT
Using 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 = VoT
Where
V = 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)
=3x
22
Heat and Temperature
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 K


V
= VoT
= 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
Solution
Petrol 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.
Petrol
Vo = 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
= VoT
= 60 x 950 x 10-6 x 35
= 1.995 litres
Petrol expands by 1.995 litres
23
Heat and Temperature
Steel drum
Vo = 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 litres
The 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.
SAQs
SAQ1 - 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
24
Heat and Temperature
6.
Heat Transfer
There 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.
Conduction
Conduction 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
Iron
Concrete
390
46
1.3
25
Heat and Temperature
Glass(typical)
0.84
Water
Air
0.57
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 proportional
From this information, we can directly formulate a formula for the rate at which heat
is conducted through a material.
Q/t = kAT / L
Where
Q/t = rate at which heat is conducted (i.e. heat conducted per sec) -W
k = thermal conductivity of material (W m-1 K-1)
A = area through which heat is conducted (m2)
T = temperature difference across the material (K)
26
Heat and Temperature
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
k = 1.1 W m-1 K-1
Q/t
L = 3 mm = 0.003 m
T = 20 K
= kAT / 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
windowpane
still air
20oC
inside
outside
T
0oC
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.
27
Heat and Temperature
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.
28
Heat and Temperature
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
heats the top portion of a quantity of water when small amounts of hot water are
required.
29
Heat and Temperature
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
where
Q/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.
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 equation
30
Heat and Temperature
Q/t = ε AT4
Q/t = ?
 = 5.67 x 10-8 W m-2 K-4
A = 1.8 m2
ε=1
o
T = 60 C = 60 + 273 = 333 K
Note - 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 heat
Bright 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
6.3.
Conduction, Convection and Radiation
It 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.
31
Heat and Temperature
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.
32