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A solar collector is placed in direct sunlight where it absorbs energy at the rate of 830 J/s for each square meter of its surface. The emissivity of the solar collector is e = 0.65. What equilibrium temperature does the collector reach? Assume that the only energy loss is due to the emission of radiation.
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Added Sun, 26 Jul '15
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The rate of loss of heat by emission, for each square meter of surface is

esAT4 = (0.65)(5.67*10^-8)(1)(T4)

In equilibrium, this is equal to the rate at which the collector absorbs heat, so

(0.65)(5.67*10^-8)(1)(T4) = 830

T = 387.38 K = 114.38oC
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Added Sun, 26 Jul '15
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