question_answer

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 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is

A) \[{{\left( \frac{65}{2} \right)}^{1/4}}T\]

B) \[{{\left( \frac{97}{4} \right)}^{1/4}}T\]

C) \[{{\left( \frac{97}{2} \right)}^{1/4}}T\]

D) \[{{(97)}^{1/4}}T\]

Correct Answer: C

Solution :

[c] : Let T' be the temperature of i n m the middle plate (II) and A be area of each plate. Under steady state, the rate of energy received by the middle plate is equal to 2T r 3T rate of energy emitted by it. i.e.,\[\sigma A{{(3T)}^{4}}-\sigma A{{(T')}^{4}}=\sigma A{{(T')}^{4}}-\sigma A{{(2T)}^{4}}\] or\[\sigma A[{{(3T)}^{4}}-{{(T')}^{4}}]=\sigma A[{{(T')}^{4}}-{{(2T)}^{4}}]\] or\[[{{(3T)}^{4}}-{{(T')}^{4}}={{(T')}^{4}}-{{(2T)}^{4}}\] or\[2{{(T')}^{4}}={{T}^{4}}({{3}^{4}}+{{2}^{4}})={{T}^{4}}(81+16)=97{{T}^{4}}\] or\[T{{'}^{4}}=\frac{97}{2}{{T}^{4}}orT'={{\left( \frac{97}{2} \right)}^{1/4}}T\]