JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Heat & Thermodynamics

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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

(A) $(\frac{65}{2})^{\frac{1}{4}} T$
(B) $(\frac{97}{4})^{\frac{1}{4}} T$
(C) $(\frac{97}{2})^{\frac{1}{4}} T$ ✓ Correct
(D) $(97)^{\frac{1}{4}} T$

Answer: (C)

Solution

Under steady state, the middle plate receives radiation from both outer plates and emits from both sides. Energy balance gives:

σ A (2 T)^{4} + σ A (3 T)^{4} = 2 σ A (T^{'})^{4}

, where

T^{'}

is the temperature of the middle plate. Simplifying:

16 T^{4} + 81 T^{4} = 2 (T^{'})^{4}

, so

97 T^{4} = 2 (T^{'})^{4}

. Therefore

T^{'} = (\frac{97}{2})^{1/4} T

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option

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JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Heat & Thermodynamics

Solution

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