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(C) In the steady state,the energy absorbed by the middle plate is equal to the energy released by the middle plate.Let the temperature of the middle

Vedclass · Accepted Answer

$\left(\frac{97}{2}\right)^{\frac{1}{4}} \ T$

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(C) In the steady state,the energy absorbed by the middle plate is equal to the energy released by the middle plate.Let the temperature of the middle plate be $T'$.The radiation energy incident on the middle plate from the first plate (at $3T$) is $\sigma A (3T)^4$.The radiation energy emitted by the middle plate towards the first plate is $\sigma A (T')^4$.The net energy absorbed from the first plate is $\sigma A (3T)^4 - \sigma A (T')^4$.The radiation energy incident on the middle plate from the third plate (at $2T$) is $\sigma A (2T)^4$.The radiation energy emitted by the middle plate towards the third plate is $\sigma A (T')^4$.The net energy released to the third plate is $\sigma A (T')^4 - \sigma A (2T)^4$.Equating the net energy absorbed and released:$\sigma A (3T)^4 - \sigma A (T')^4 = \sigma A (T')^4 - \sigma A (2T)^4$$(3T)^4 - (T')^4 = (T')^4 - (2T)^4$$81T^4 + 16T^4 = 2(T')^4$$97T^4 = 2(T')^4$$(T')^4 = \frac{97}{2} T^4$$T' = \left(\frac{97}{2}\right)^{\frac{1}{4}} T$

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