Parallel rays of light of intensity I = 912 | vedclass

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(A) In the steady state,the power absorbed by the black body equals the power radiated by it.The power absorbed by the spherical black body of radius

Vedclass · Accepted Answer

$330$

Vedclass · Answer

(A) In the steady state,the power absorbed by the black body equals the power radiated by it.The power absorbed by the spherical black body of radius $R$ is $P_{abs} = I 	imes A_{proj} = I \pi R^2$.The power radiated by the black body to the surroundings at temperature $T_0$ is $P_{rad} = \sigma A_{surf} (T^4 - T_0^4) = \sigma (4 \pi R^2) (T^4 - T_0^4)$.Equating the two: $I \pi R^2 = 4 \pi R^2 \sigma (T^4 - T_0^4)$.Simplifying,we get: $I = 4 \sigma (T^4 - T_0^4)$.Substituting the given values: $912 = 4 	imes (5.7 	imes 10^{-8}) 	imes (T^4 - 300^4)$.$T^4 - 300^4 = \frac{912}{4 	imes 5.7 	imes 10^{-8}} = \frac{912}{22.8 	imes 10^{-8}} = 40 	imes 10^8$.$T^4 = 40 	imes 10^8 + 81 	imes 10^8 = 121 	imes 10^8$.$T = (121 	imes 10^8)^{1/4} = (11^2 	imes 10^8)^{1/4} = 11^{1/2} 	imes 10^2 \approx 3.316 	imes 100 \approx 331.6 \ K$.Thus,the temperature is close to $330 \ K$.

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