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(C) For the middle plate to be in a steady state,the net power received by the plate must equal the net power emitted by the plate.Let the temperature

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For the middle plate to be in a steady state,the net power received by the plate must equal the net power emitted by the plate.Let the temperature of the middle plate be $	heta$.The power received by the middle plate from the first plate (at $2T$) and the third plate (at $3T$) is given by the Stefan-Boltzmann law for black bodies: $P_{	ext{received}} = \sigma A (2T)^4 + \sigma A (3T)^4$.The middle plate emits radiation from both its sides,so the total power emitted is: $P_{	ext{emitted}} = 2 	imes \sigma A 	heta^4$.Equating the power received and emitted:$\sigma A (2T)^4 + \sigma A (3T)^4 = 2 \sigma A 	heta^4$$16 T^4 + 81 T^4 = 2 	heta^4$$97 T^4 = 2 	heta^4$$	heta^4 = \frac{97}{2} T^4$$	heta = \left( \frac{97}{2} ight)^{1/4} T$

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