Two identical plates P and Q,radiating as per | vedclass

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(B) For Fig. $1$,the power transferred per unit area is given by Stefan-Boltzmann law as:$W_0 = \sigma(T_P^4 - T_Q^4)$For Fig. $2$,let the temperature

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$3$

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(B) For Fig. $1$,the power transferred per unit area is given by Stefan-Boltzmann law as:$W_0 = \sigma(T_P^4 - T_Q^4)$For Fig. $2$,let the temperatures of the two intermediate plates be $T_1$ and $T_2$. In the steady state,the heat flux $W_S$ through each gap must be the same:$W_S = \sigma(T_P^4 - T_1^4) = \sigma(T_1^4 - T_2^4) = \sigma(T_2^4 - T_Q^4)$From these equations,we have:$T_P^4 - T_1^4 = T_1^4 - T_2^4 = T_2^4 - T_Q^4 = W_S / \sigma$Adding these three equations:$(T_P^4 - T_1^4) + (T_1^4 - T_2^4) + (T_2^4 - T_Q^4) = 3(W_S / \sigma)$$T_P^4 - T_Q^4 = 3(W_S / \sigma)$Substituting $W_0 = \sigma(T_P^4 - T_Q^4)$:$W_0 / \sigma = 3(W_S / \sigma)$$W_0 = 3W_S$Therefore,the ratio $\frac{W_0}{W_S} = 3$.

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