The ratio of energies of radiation emitted by | vedclass

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Question

(C) According to the Stefan-Boltzmann law,the net rate of energy emission $E$ by a black body at temperature $T$ in surroundings at temperature $T_S$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) According to the Stefan-Boltzmann law,the net rate of energy emission $E$ by a black body at temperature $T$ in surroundings at temperature $T_S$ is given by $E = \sigma A (T^4 - T_S^4)$.Given $T_1 = 600 \ K$,$T_2 = 900 \ K$,and $T_S = 300 \ K$.The ratio of energies is $\frac{E_1}{E_2} = \frac{T_1^4 - T_S^4}{T_2^4 - T_S^4}$.Substituting the values: $\frac{E_1}{E_2} = \frac{(600)^4 - (300)^4}{(900)^4 - (300)^4}$.Factoring out $(300)^4$: $\frac{E_1}{E_2} = \frac{(300)^4 [2^4 - 1^4]}{(300)^4 [3^4 - 1^4]}$.$\frac{E_1}{E_2} = \frac{16 - 1}{81 - 1} = \frac{15}{80} = \frac{3}{16}$.

The ratio of energies of radiation emitted by a black body at $600 \ K$ and $900 \ K$ when the surrounding temperature is $300 \ K$ is: (in $/16$)

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