Question

Three very large plates of same area are kept parallel (in vertical plane) 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. \((97)^{1/4}T\)
• B. \(\left(\frac{97}{2}\right)^{1/4}T\)
• C. \(\left(\frac{97}{4}\right)^{1/4}T\)
• D. \(\left(\frac{65}{2}\right)^{1/4}T\)

Hint:

In steady state the middle plate absorbs as much radiation as it emits from both faces.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Black plates absorb all radiation falling on them and radiate according to Stefan's law, \(\sigma T^{4}\) per unit area.

Step 2: Energy balance
Let the middle plate be at temperature \(T_2\). It receives from both neighbours and radiates from both faces:
\(\sigma(2T)^{4}+\sigma(3T)^{4}=2\sigma T_2^{4}\)

Step 3: Calculation
\(16T^{4}+81T^{4}=2T_2^{4}\), so \(T_2^{4}=\dfrac{97}{2}T^{4}\).

Step 4: Result
\(T_2=\left(\dfrac{97}{2}\right)^{1/4}T\), option (B).

Final Answer:
The middle plate settles at \(\left(\dfrac{97}{2}\right)^{1/4}T\). \[ \boxed{\text{(B) }\left(\dfrac{97}{2}\right)^{1/4}T} \]