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8.1Radiation from real bodies
Black bodies are idealized bodies. Real bodies radiate according to the modified StefanâBoltzmann law: P = eÏAT4. The constant e is known as the emissivity of the surface. Its value is between 0 and 1; the value e = 1 corresponds to the idealized black body.
Emissivity is the ratio of the power per unit area radiated by a body to the power per unit area radiated by a black body of the same temperature.
KEY POINT
The power radiated per unit area is called (radiated) intensity and so equals I = eÏ T4.
The unit of intensity is W mâ2.
Unlike a black body that absorbs all the radiation incident on it, reflecting none, a real body will reflect some of the incident radiation. The ratio of the reflected intensity (or power) to the incident intensity (or power) is called the albedo of the body.
KEY POINT
Albedo is the ratio Î±=Â total reflected powerÂ Â total incident powerÂ 
Albedo has no unit.
For a body that can only absorb or reflect (i.e. it cannot transmit) the emissivity and albedo are related by e + Î± = 1.
Consider a body of emissivity e and surface temperature T whose surroundings have a temperature Ts. The surroundings may be assumed to be a black body. The body radiates an intensity eÏT4. The surroundings radiate an intensity ÏT4S. Of this intensity the body will reflect a fraction Î±ÏT4S and so will absorb a fraction (1âÎ±)ÏT4S, i.e. eÏT4S. The net intensity for the body is then Â InetÂ =eÏT4âeÏT4s. Thus the rate at which energy leaves the body (the power) of surface area A is
Pnet=eÏA(T4âT4S)
At equilibrium no net power leaves the body and so T = Ts, as we might expect. Table 8.1 gives values for the emissivity of various surfaces.
Surface
Emissivity
black body
1
ocean water
0.8
ice
0.1
dry land
0.7
land with vegetation
0.6
Table 8.1: Emissivity of various surfaces.
Figure 8.1 shows the variation of the spectral intensity B with wavelength Î» from various different surfaces kept at the same temperature (300 K). The difference in the curves is due to the different emissivities (e = 1.0, 0.8 and 0.2). The curves are identical apart from an overall factor that shrinks the height of the curve as the emissivity decreases. The peak wavelength is the same for all three curves because the temperature is the same.
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Figure 8.1: The spectra of three bodies with different emissivities at the same temperature (300 K). The vertical axis shows spectral intensity.

CHECK YOURSELF 1
The kelvin temperature and radius of a spherical black body are both doubled. By what factor does a the radiated power and b the radiated intensity increase?
WORKED EXAMPLE 8.1
The emissivity of the naked human body may be taken to be e = 0.90. Assuming a body temperature of 37 Â°C and a body surface area of 1.60 m2, calculate the total amount of energy lost by the body when exposed to a temperature of 0.0 Â°C for 30 minutes.
Answer
The net power lost is the difference between the power emitted by the body and the power received. Let the body temperature be T1 and the temperature of the surroundings be T2. Then:
PnetÂ =eÏA(T41âT42)
Substituting the values from the question:
PnetÂ =0.90Ã5.67Ã10â8Ã1.60Ã(3104â2734)
PnetÂ =301Â W
So the energy lost in time t seconds is:
E=PnetÂ t
E=301Ã30Ã60
E=5.4Ã105Â J
(What does this mean for the human body? For the purposes of an estimate, assume that the body has mass 60 kg and is made out of water, with specific heat capacity c = 4200 J kgâ1 Kâ1. This energy loss would result in a drop in body temperature of ÎT=5.4Ã10560Ã4200=2.1Â K.)
The solar constant
The sun may be considered to radiate as a black body. The sun emits a total power P = 3.9 Ã 1026 W. The average earthâsun distance is d = 1.50 Ã 1011 m. Imagine a sphere of radius d centred at the sun. The power of the sun is distributed over the area of this sphere (4Ïd2) and so the power per unit area, i.e. the intensity, received by earth is:
I=P4Ïd2
We generalise the definition of intensity given earlier to:
KEY POINT
Intensity is the power of radiation received or emitted per unit area. Its unit is W mâ2
Substituting the numerical values for our sun gives:
I=3.9Ã10264Ï(1.50Ã1011)2=1379â1400Â WÂ mâ2.
This is called the solar constant, S. It is the intensity of the solar radiation at the top of the earthâs atmosphere. S â 1400 W mâ2.
The radius of the sun is 7.0 Ã 108 m. The intensity emitted by the sun is
I=3.9Ã10264Ï(7.0Ã108)2=6.3Ã107Â WÂ mâ2.
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TEST YOUR UNDERSTANDING
1The graph (Figure 8.2) shows the variation with wavelength of the spectral intensity of radiation emitted by two bodies of identical shape and size.

Figure 8.2: For question 1.

aExplain why the temperature of the two bodies is the same.
bThe upper line corresponds to a black body. Calculate the emissivity of the other body.
2The total power radiated by a surface of area 5.00 km2 and emissivity 0.800 is 1.35 Ã 109 W. Assume that the surface radiates into a vacuum at temperature 0 K. Calculate the temperature of the surface.
3A sphere of emissivity 0.80 has radius 0.50 m and temperature 27 Â°C. The temperature of the surroundings is 57 Â°C. At what rate does the body emit radiation and at what rate does it absorb radiation?
4A sphere of emissivity 0.70 has radius 0.60 m and temperature 300 K. The temperature of the surroundings is assumed to be 0 K.
aWhat is the intensity of the emitted radiation?
bWhat is the intensity of the received radiation at a distance of 2.2 m from the centre of the sphere?
5aDefine the term intensity in the context of radiation.
bEstimate the net intensity of radiation of a naked human body of surface area 1.60 m2, temperature 37 Â°C and emissivity 0.90 assuming black-body surroundings at â15 Â°C.
