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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. The first and third plates are maintained at temperatures $2 T$ and $3 T$ respectively. The temperature of the middle (i.e. second) plate under steady state condition is
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Verified Answer
The correct answer is:
$\left(\frac{97}{2}\right)^{1 / 4} T$
Under steady conditions, the heat gained per second by a plate is equal to the heat released per second by the plate.
$\begin{array}{l}
\frac{\text { Heat gained }}{\text { Second }}[\text { by }(2) \text { from }(1)]+\frac{\text { Heat gained }}{\text { Second }} \\
{\left[\left(\text { by }((2) \text { from }(3)]=\frac{\text { Heat gained }}{\text { Second }} \text { (by } 2\right)\right.}
\end{array}$
$\begin{array}{ll}\therefore \quad & \sigma A(2 T)^{4}+\sigma A(3 T)^{4} \\ & =\sigma(2 A)(T)^{4} \\ \therefore \quad & T=\left[\frac{97}{2}\right]^{1 / 4} T\end{array}$
$\begin{array}{l}
\frac{\text { Heat gained }}{\text { Second }}[\text { by }(2) \text { from }(1)]+\frac{\text { Heat gained }}{\text { Second }} \\
{\left[\left(\text { by }((2) \text { from }(3)]=\frac{\text { Heat gained }}{\text { Second }} \text { (by } 2\right)\right.}
\end{array}$
$\begin{array}{ll}\therefore \quad & \sigma A(2 T)^{4}+\sigma A(3 T)^{4} \\ & =\sigma(2 A)(T)^{4} \\ \therefore \quad & T=\left[\frac{97}{2}\right]^{1 / 4} T\end{array}$
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