Physics Friday 35

Suppose we have a spherical blackbody of radius R. Now let us shine radiation of intensity I on this sphere from one direction. Assuming the sphere is of high enough thermal conductivity to be treated as having a uniform temperature, and neglecting any other heat sources, what will the temperature T of the sphere be in equilibrium?

The sphere has some amount of power being absorbed from the incoming radiation; it is in equilibrium when the power it radiates as a blackbody equals this. (Note that if it is above this temperature, it will radiate more power than it absorbs, and will cool toward the equilibrium temperature. Similarly, if it is below this temperature, it absorbs more power than it radiates, and will heat up toward the equilibrium).

The absorbed power is the intensity of the incoming radiation times the area of the cross-section the absorbing surface presents to the radiation. As our object is a sphere, the cross section is a circle of radius R, and so the power absorbed is Pab = πR2·I.

Now, we found previously that the power radiated per unit of surface area for a blackbody is given by the Stefan-Boltzmann Law:
j = σT4, where σ is the Stephan-Boltzmann constant.
Thus the power emitted by the sphere is this times its surface area:
Pem = 4πR2·σT4

Setting these equal to find the equilibrium temperature:
Note that the radius of our spherical blackbody does not matter, only the intensity of the incoming radiation.

For a numerical example, the intensity of the sun’s radiation at the distance of the earth (solar constant) is approximately 1366 W/m². Using this in our formula, along with σ = 5.670400×10-8 W·m-2·K-4, we see:

(which is about 6° C). The actual average temperature of the earth (≈288 K, or 15° C) differs from this due to albedo (the planet is not a perfect absorber), atmospheric effects (such as the greenhouse effect), and internal heat generation.
(See also the Wikipedia entry on Effective temperature of a planet).


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