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State Stefan's law of radiation. Derive an expression for the rate of loss of radiant energy per unit area by a perfect blackbody in a cooler surroundings. |
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Answer» Solution :Stefan's law of radiation : The quantity of radiant energy emitted by a perfect blackbody per unit time per unit surface area of the body is directly proportional to the fourth POWER of its ABSOLUTE temperature. Consider a perfect blackbody at absolute temperature T. We shall assume its surroundings also to act as a perfect blackbody at absolute temperature `T_(0)`, where `T_(@) lt T`. The energy radiated per unit time per unit surface area by a blackbody at temperature T is its emissive power `E_(b)` at that temperature. According to Stefan's law, `E_(b)rhoT^(4)` is the Stefan CONSTANT. At the same time, the body absorbs radiant energy from the surroundings. The radiant energy absorbed per unit time per area by the blackbody is `rhoT_(0)^(4)`. Therefore, the rate of loss of radinat energy per unit area by the blackbody is `rho(T^(4)-T_(0)^(4))`. |
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