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Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. First and third plates are maintained at absolute temperatures $2 T$ and $3 T$ respectively.
Temperature of the middle plate in steady state is
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Temperature of the middle plate in steady state is
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Verified Answer
The correct answer is:
$\left(\frac{97}{2}\right)^{\frac{1}{4}} \mathrm{~T}$
Let assume temperature of middle plate is $T_0$.
As temperatures are constant,
$\therefore$ Heat gained by middle surface $=$ Heat lost by middle surface.
So, $\sigma A\left[(3 T)^4-T_0^4\right]=\sigma A\left[T_0^4-(2 T)^4\right]$
$$
\Rightarrow \quad T_0^4=\frac{97}{2} T^4 \quad \Rightarrow T_0=\left(\frac{97}{2}\right)^{1 / 4} T
$$
As temperatures are constant,
$\therefore$ Heat gained by middle surface $=$ Heat lost by middle surface.
So, $\sigma A\left[(3 T)^4-T_0^4\right]=\sigma A\left[T_0^4-(2 T)^4\right]$
$$
\Rightarrow \quad T_0^4=\frac{97}{2} T^4 \quad \Rightarrow T_0=\left(\frac{97}{2}\right)^{1 / 4} T
$$
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