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8.1Radiation from real bodies

Black bodies are idealized bodies. Real bodies radiate according to the modified Stefan–Boltzmann law: P = e σ A T 4. The constant e is known as the emissivity the ratio of the intensity radiated by a body to the intensity radiated by a black body of the same temperature 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 = eσ T 4.

The unit of intensity is W m−2.

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 the ratio of reflected to incident power or intensity; it has no unit of the body.

KEY POINT

Albedo is the ratio α= total reflected power total incident power

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 + α = 1.

Consider a body of emissivity e and surface temperature T whose surroundings have a temperature T s. The surroundings may be assumed to be a black body. The body radiates an intensity e σ T 4. The surroundings radiate an intensity σ T S 4. Of this intensity the body will reflect a fraction α σ T S 4 and so will absorb a fraction (1−α)σ T S 4, i.e. e σ T S 4. The net intensity for the body is then I net =e σ T 4−e σ T s 4. Thus the rate at which energy leaves the body (the power) of surface area A is

P net=e σ A(T 4−T S 4)

At equilibrium no net power leaves the body and so T = T s, 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 λ 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 °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 °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 T 1 and the temperature of the surroundings be T 2. Then:

P net =e σ A(T 1 4−T 2 4)

Substituting the values from the question:

P net =0.90×5.67×10−8×1.60×(310 4−273 4)

P net =301 W

So the energy lost in time t seconds is:

E=P net t

E=301×30×60

E=5.4×10 5 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 kg–1 K–1. This energy loss would result in a drop in body temperature of Δ T=5.4×10 5 60×4200=2.1 K.)

The solar constant

The sun may be considered to radiate as a black body. The sun emits a total power P = 3.9 × 1026 W. The average earth–sun distance is d = 1.50 × 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 (4π d 2) and so the power per unit area, i.e. the intensity the power of radiation received or emitted per unit area; its unit is W m−2, received by earth is:

I=P 4 π d 2

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 m−2

Substituting the numerical values for our sun gives:

I=3.9×10 26 4 π(1.50×10 11)2=1379≈1400 W m−2.

This is called the solar constant, S the intensity of solar radiation received at the upper atmosphere of the earth. It is the intensity of the solar radiation at the top of the earth’s atmosphere. S ≈ 1400 W m–2.

The radius of the sun is 7.0 × 108 m. The intensity emitted by the sun is

I=3.9×10 26 4 π(7.0×10 8)2=6.3×10 7 W m−2.

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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 × 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 °C. The temperature of the surroundings is 57 °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?

5 aDefine 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 °C and emissivity 0.90 assuming black-body surroundings at –15 °C.