Two bodies A and B have thermal emissivities | vedclass

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(D) According to Stefan-Boltzmann law,the total radiant power $P = e A \sigma T^4$. Since $P_A = P_B$ and $A_A = A_B$,we have $e_A T_A^4 = e_B T_B^4$.

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

Vedclass · Answer

(D) According to Stefan-Boltzmann law,the total radiant power $P = e A \sigma T^4$. Since $P_A = P_B$ and $A_A = A_B$,we have $e_A T_A^4 = e_B T_B^4$.Given $e_A = 0.01$,$e_B = 0.81$,and $T_A = 5802\;K$.$T_B = T_A \left( \frac{e_A}{e_B} ight)^{1/4} = 5802 	imes \left( \frac{0.01}{0.81} ight)^{1/4} = 5802 	imes \left( \frac{1}{81} ight)^{1/4} = 5802 	imes \frac{1}{3} = 1934\;K$.From Wien's displacement law,$\lambda_A T_A = \lambda_B T_B = b$ (constant).$\lambda_A = \frac{b}{T_A}$ and $\lambda_B = \frac{b}{T_B}$.Given $\lambda_B - \lambda_A = 1.00\;\mu m$.$\frac{b}{T_B} - \frac{b}{T_A} = 1.00\;\mu m$.$b \left( \frac{1}{1934} - \frac{1}{5802} ight) = 1.00\;\mu m$.$b \left( \frac{3 - 1}{5802} ight) = 1.00\;\mu m \Rightarrow b = \frac{5802}{2} = 2901\;\mu m \cdot K$.Now,$\lambda_B = \frac{b}{T_B} = \frac{2901}{1934} = 1.5\;\mu m$.Thus,both $(a)$ and $(b)$ are correct.

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