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7.3Thermal energy transfer
There are three distinct methods to transfer thermal energy from one place to another: conduction, convection and radiation.
Conduction
Imagine a solid with one end kept at a high temperature, as shown in Figure 7.10. The electrons at the hot end of the solid have a high average kinetic energy. This means they move a lot. The moving electrons collide with neighbouring molecules, transferring energy to them and so increasing their average kinetic energy. This means that energy is being transferred from the hot to the cold side of the solid; this is conduction.

Figure 7.10: Conduction of thermal energy through a solid as a result of a temperature difference.

Collisions between electrons and molecules are the dominant way in which thermal energy is transferred by conduction, but if there are strong bonds between molecules there is another way. Molecules on the hot side of the solid vibrate about their equilibrium positions, stretching the bonds with neighbouring molecules. This stretching forces the neighbours to also begin to vibrate, and so the average kinetic energy of the neighbours increases. Energy is again transferred.
For a solid of cross-sectional area A, width Îx and temperature difference between its ends ÎT, experiments show that the rate at which energy is being transferred is:
ÎQÎt=kAÎTÎx
where k is called the conductivity and depends on the nature of the substance. Referring to Figure 7.10, ÎT=THâTC.
CHECK YOURSELF 5
Figure 7.11 shows the cross sections of three walls made of the same material and of thicknesses L, 2L and 4L. Thermal energy transfers from left to right through the same cross sectional area. In which case is the rate of transfer of thermal energy the greatest?

Figure 7.11: For check yourself question 5.

WORKED EXAMPLE 7.6
Two rods of different materials but of the same cross sectional area and length are joined as shown. The thermal conductivities are shown in Figure 7.12.

Figure 7.12: For worked example 7.6.

The left end is kept at constant temperature of 100 Â°C and the right end at a constant temperature 0 Â°C. What is the temperature at the point where the rods are joined?
Answer
Let us call the temperature at the joining point T. Then the rate at which thermal energy transfers in the left rod is Â kA100âTL and in the right rod it is (3k)ATâ0L. The two rates must be equal by energy conservation. Hence
kA100âTL100âTT=(3k)ATâ0L=3T=25Â°C
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Convection
Convection is a method of energy transfer that applies mainly to fluids, i.e. gases and liquids. If you put a pan of water on a stove, the water at the bottom of the pan is heated by conduction. As it gets hotter the water expands, it gets less dense and so rises to the top. In this way thermal energy from the bottom of the pan is transferred to the top. Similarly, air over a hot radiator in a room is heated, expands and rises, transferring warm air to the rest of the room. Colder air takes the place of the air that rose and the process repeats, creating convection currents. In other words convection is due to differences in density. Convection is responsible for winds in the atmosphere and for ocean currents.
Radiation
Both conduction and convection require a material medium through which thermal energy is to be transferred. The third method of thermal energy transfer, radiation, does not. Energy from the sun has been warming the earth for billions of years. This energy arrives at earth as radiation having travelled through the vacuum of space at the speed of light.
StefanâBoltzmann law
One of the great advances in physics in the nineteenth century was the discovery that all bodies that are kept at some absolute (kelvin) temperature T radiate energy in the form of electromagnetic waves. This is radiation created by oscillating electric charges in the atoms of the body. The power radiated by a body is governed by the StefanâBoltzmann law (obtained experimentally by J. Stefan and deduced theoretically by L. Boltzmann).
KEY POINT
The power radiated by a black body depends on its surface area A and the absolute temperature of the surface T and is given by P = ÏAT4. The constant Ï is the StefanâBoltzmann constant and equals Ï = 5.67 Ã 10â8 W mâ2 Kâ4.
The power radiated per unit area of the body, the radiated intensity, is then PA=ÏT4.
Strictly, this applies to a theoretical body called a black body. This is a perfect radiator as well as a perfect absorber. A black body will absorb all the radiation falling upon it, reflecting none. A black body at low temperature radiates very little, and since it absorbs all the radiation falling on it, it looks black. At high temperature it radiates a lot and looks very bright. A very good example of this is a piece of charcoal: dark when cold, glowing orange-red when hot. (In ChapterÂ 8 we will see how the StefanâBoltzmann law is modified to apply to real bodies as well.)
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Light incident on a shiny metallic surface will reflect with little light being absorbed. Light incident on glass will pass through the glass with very little being absorbed. So why does a black body absorb all the radiation incident on it, transmitting none and reflecting none? The incident light is an electromagnetic wave and so contains an electric field (see SectionsÂ 13.3 and 18.1). This field will set electrons in a surface oscillating. In the case of glass, the electrons are very tightly bound to their atoms and so cannot oscillate much. The light just goes through. In the case of the metal, there are lots of free electrons. The electric field makes them oscillate a lot which means they are accelerated. But accelerated charges radiate electromagnetic waves, and this is the reflected light. In a black body, the electric field also accelerates electrons which gain energy, but before they have a chance to radiate they collide with atoms giving them their kinetic energy. The incident radiation has been absorbed and the temperature of the body increases. As the temperature increases more and more the vibrations of the atoms get larger and larger. Now, the collisions between atoms and electrons are very violent and give electrons massive accelerations. The electrons now radiate; the body is emitting black body radiation.
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Figure 7.13 shows three black-body spectra emitted from the same surface at three different temperatures of the surface (T = 273 K, 320 K and 350 K). We see that with increasing temperature, the peak of the curve occurs at lower wavelengths and the height of the peak increases. The quantity B plotted on the vertical axis is the power emitted per unit area per unit wavelength and so has units of (W mâ2) mâ1 = W mâ3. It is called spectral intensity. The area under the graph is the radiated intensity, Ï T4.

Figure 7.13: Black-body spectra for a body at the three temperatures shown. Notice that as the temperature increases the peak gets higher and the wavelength at the peak shifts to the left.

The energy radiated is electromagnetic radiation and is distributed over an infinite range of wavelengths. The peak of the spectrum corresponds to a specific wavelength Î»max, Figure 7.14, called the peak wavelength.

Figure 7.14: For this graph, the peak wavelength is 0.25 Ã 10â6 m.

Wienâs law
The peak wavelength Î»0 depends on temperature according to Wienâs law.
Wienâs law states that Î»maxT = 2.90 Ã 10â3 K m.
So, for the graph of Figure 7.14, the temperature is T=2.90Ã10â30.25Ã10â6=11600â1.2Ã104Â K.
EXAM TIP
Increasing the temperature shifts the peak wavelength to the left and makes the curve taller.
CHECK YOURSELF 6
The area of the body in Figure 7.14 is doubled. How does the graph change?
WORKED EXAMPLE 7.7
A human body has temperature 37 Â°C, the average earth surface temperature is 288 K and the temperature of the sun is 5800 K. In each case, calculate the peak wavelength of the emitted radiation.
Answer
We just have to apply Wienâs law, Î»max T = 2.90 Ã 10â3 K m, and make sure we use kelvins in each case. So:
human body: Î»max=2.90Ã10â3273+37â9Ã10â6Â m, an infrared wavelength
earth surface: Î»max=2.90Ã10â3288â1Ã10â5Â m, an infrared wavelength
sun: Î»max=2.90Ã10â35800â5Ã10â7Â m, which is a visible wavelength.
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WORKED EXAMPLE 7.8
By what factor does the power emitted by a black body increase when the temperature is increased from 100 Â°C to 200 Â°C?
Answer
The temperature in kelvin increases from 373 K to 473 K. Since the emitted power is proportional to the fourth power of the temperature, power will increase by a factor:
(473373)4=2.59
EXAM TIP
Make sure the temperature is in kelvin.
TEST YOUR UNDERSTANDING
22Describe what is meant by
aconduction
bconvection.
23Half the floor of a room is covered by marble tiles and the other half by a carpet. Why does walking barefoot feel colder on the marble tiles?
24For most people, walking in 20 Â°C weather would be considered comfortable but swimming in 20 Â°C water would be uncomfortable. Suggest a reason for this.
25A pan of water placed on a stove comes to a boil. Discuss how thermal energy is transferred from the stove to the surface of the water.
26How does blowing over a hot cup of coffee help the coffee cool down?
27The road over a bridge freezes before the road on the ground. Suggest why this happens.
28Suggest whether there is any point in using a ceiling fan in winter.
29You stand on land by the sea on a hot summer day. Why is there a breeze from the sea towards the land?
30By what method does a spacecraft in the vacuum of space âloseâ thermal energy?
31A cylindrical solid tube is made out of two smaller tubes, X and Y, of different material. X and Y have the same length and cross-sectional area. The tube is used to conduct energy from a hot to a cold reservoir (see Figure 7.15).

Figure 7.15: For question 31.

State and explain whether the following are equal or not:
athe rates of transfer of energy through X and through Y
bthe temperature differences across X and across Y.
32In the previous problem the temperatures of the hot and cold reservoirs are constant at 80 Â°C and 20 Â°C respectively. The thermal conductivity of X is k and that of Y is 2k. Calculate the temperature where X and Y join.
33The temperature of a room in winter is kept constant at 24 Â°C by using a heater. The room is separated from the outside, where the temperature is â5.0 Â°C, by a wall of area 12 m2 and thickness 0.20 m. The material of the wall has a thermal conductivity of 0.15 W mâ1 Kâ1. Estimate the power of the heater.
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34A cylindrical metallic rod has radius 1.0 cm and length 25 cm. One end of the rod is placed into a large chunk of ice at 0 Â°C and the other in boiling water at 100 Â°C. The thermal conductivity of the rod is 350 W mâ1 Kâ1. Estimate how long it will take to melt 1 g of ice. It takes about 334 J to melt 1 g of ice at 0 Â°C.
35Calculate the ratio of the power radiated per unit area from two black bodies at temperatures 900 K and 300 K.
36aState what you understand by the term black body.
bGive an example of a body that is a good approximation to a black body.
cBy what factor does the rate of emitted radiation from a body increase when the temperature is raised from 50 Â°C to 100 Â°C?
37A star radiates like a black body. Most of the energy is emitted at a wavelength of 410 nm. What is the surface temperature of the star?
38Star X has peak wavelength 480 nm and star Y has peak wavelength 560 nm. Which star has the higher surface temperature?
39The surface area and kelvin temperature of a black body are both doubled. By what factor do:
athe radiated power
bthe radiated intensity
increase?
40A sphere of radius R and a cube of side R radiate like black bodies. The sphere and the cube are at the same temperature. Which body loses more energy per second?
41The graph (Figure 7.16) shows a black-body spectrum.

Figure 7.16: For question 41.

aDetermine the temperature of the black body.
bThe kelvin temperature of the black body in a is doubled. What is the peak wavelength?
cThe area of the body in a is reduced, but the temperature stays the same as that found in a. Draw a graph to show the new black-body spectrum. (Hint: what does the area represent?)
42The graph (Figure 7.17) shows the black-body spectrum of the sun plotted for the range of visible wavelengths only.
aFind the surface temperature of the sun.
bDo you agree with the statement that the color of the sun is white?

Figure 7.17: For question 42

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Luminosity and apparent brightness
In an example of lack of economy of symbols and terms, the quantity I=P4Ïd2 is called apparent brightness by astronomers and is given the symbol b. Thus, apparent brightness is the power received per unit area of the detector. Astronomers also call the total power radiated by a star luminosity, L and so in astronomy we have:
b=L4Ïd2
Luminosity is a major characteristic of a star. As we saw when we discussed black-body radiation, stars are excellent approximations to a black body and so the luminosity of a star is given by the Â­StefanâBoltzmann law:
L=ÏAT4
where Ï = 5.67 Ã 10â8 W mâ2 Kâ4 is the StefanâBoltzmann constant, A is the surface area of the star and T its surface temperature in kelvin.
Now, two stars of the same luminosity will not appear equally bright when observed from earth if they are at different distances from earth. Apparent brightness is a measure of how bright a star actually appears.
By combining the formula for luminosity with that of apparent brightness we see that
b=ÏAT44Ïd2
Apparent brightness is easily measured (with a charge-coupled device, or CCD). If we also know the luminosity then we can determine the distance.
WORKED EXAMPLE 7.9
The apparent brightness of a star is 3.4 Ã 10â8 W mâ2 and its luminosity is 5.1 Ã 1028 W. What is its distance?
Answer
b=L4Ïd2âd=L4Ïbâââââ
Therefore:
d=5.1Ã10284ÏÃ3.4Ã10â8ââââââââââââââ=3.5Ã1017Â m
WORKED EXAMPLE 7.10
aThe radius of star A is three times that of star B and its temperature is double that of B. Find the ratio of the luminosity of A to that of B.
bThe stars in a have the same apparent brightness when viewed from earth. Calculate the ratio of their distances.
Answer
aLALB=Ï4ÏR2ATA4Ï4ÏR2BTB4=R2AT4AR2BTB4=(3RB)2(2TB)4R2BTB4=32Ã24=144
bbAbB11âdAdB=1=(LA4Ïd2A)(LB4Ïd2B)=(LALB)(d2Bd2A)=LALBâââââ=144ââââ=12
WORKED EXAMPLE 7.11
The apparent brightness of a star is 6.4 Ã 10â8 W mâ2. Its distance is 15 ly (1 ly = 9.46 Ã 1015 m). Find its luminosity.
Answer
We use b=L4Ïd2 to find
L=b4Ïd2=(6.4Ã10â8Wm2)Ã4ÏÃ(15Ã9.46Ã1015)2Â m2=1.62Ã1028Â Wâ1.6Ã1028Â W
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WORKED EXAMPLE 7.12
A star has half the sunâs surface temperature and 400 times its luminosity. Estimate the ratio of the starâs radius to that of the sun. (The symbol â refers to the sun.)
Answer
We have that:
400R2R2â16R2R2â=LLâ=Ï4ÏR2T4Ï4ÏR2âT4â=R2(Tâ/2)4R2âT4â=R2R2â16=400=16Ã400âR=80Râ
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We have learned about the particle nature of matter, i.e. atoms and molecules. This poses the obvious question of what the nature of matter really is. What other particles do we discover as we go deeper and deeper into atoms? The production of steam through the boiling of water made the steam engine possible. In a similar way, pressurized steam created by using the energy produced in fission reactions makes the production of electricity possible by rotating a coil in a magnetic field. The concept of a black body made possible our understanding of how stars radiate through application of the StefanâBoltzmann law.
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TEST YOUR UNDERSTANDING
43The luminosity of a star is 4.5 Ã 1028 W and its distance from earth is 8.3 Ã 1017 m. Calculate the apparent brightness of the star.
44The luminosity of the sun is 3.9 Ã 1026 W and its distance from earth is 1.5 Ã 1011 m. Calculate the apparent brightness of the sun.
45The light from a star a distance of 6.6 Ã 1017 m away is received on earth with an apparent brightness of 3.0 Ã 10â8 W mâ2. Calculate the luminosity of the star.
46The apparent brightness of a star is 8.4 Ã 10â10 W mâ2 and its luminosity is 6.2 Ã 1032 W. Calculate the distance to the star.
47Two stars are the same distance from earth, and their apparent brightnesses are 9.0 Ã 10â12 W mâ2 (star A) and 3.0 Ã 10â13 W mâ2 (star B). Calculate the ratio of the luminosity of star A to that of star B.
48Two stars, A and B, have the same size, but A has a temperature that is four times larger than that of B.
aEstimate how much more energy per second A radiates.
bThe apparent brightness of the two stars is the same; determine the ratio of the distance of B to that of A.
49Two stars have the same luminosity. Star A has a surface temperature of 5000 K, and star B a temperature of 10 000 K.
aSuggest which is the larger star and by how much.
bThe apparent brightness of A is double that of B; calculate the ratio of the distance of A to that of B.
50Star A has apparent brightness 8.0 Ã 10â13 W mâ2 and its distance is 120 ly. Star B has apparent brightness 2.0 Ã 10â15 W mâ2 and its distance is 150 ly. The two stars have the same size. Calculate the ratio of the temperature of star A to that of star B.
51Take the surface temperature of our sun to be 6000 K and its luminosity to be 3.9 Ã 1026 W. Find, in terms of the solar radius, the radius of a star with:
atemperature 4000 K and luminosity 5.2 Ã 1028 W
btemperature 9250 K and luminosity 4.7 Ã 1027 W.
52Two stars, X and Y, have the same apparent brightness and temperature. The distance to X is double the distance to Y. What is the ratio of radii RxRy?
53aDescribe how the colour of the light from a star can be used to determine the surface temperature of the star.
bA star appears blue and another appears red. Which is hotter?
54Stars A and B emit most of their light at wavelengths of 650 nm and 480 nm, respectively. Star A has twice the radius of star B. Find the ratio of the luminosity of star A to that of B.
