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Question

Chemistry Question on Thermodynamics

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

(652)4T\bigg(\frac{65}{2}\bigg)^4 T

B

(974)14T\bigg(\frac{97}{4}\bigg)^{\frac{1}{4}} T

C

(972)14T\bigg(\frac{97}{2}\bigg)^{\frac{1}{4}} T

D

(97)4T(97)^4 T

Answer

(972)14T\bigg(\frac{97}{2}\bigg)^{\frac{1}{4}} T

Explanation

Solution

Let temperature of middle plate in steady state is T0_0 Q1=Q2 \, \, \, Q_1 =Q_2 Q= \, \, \, \, \, \, Q = net rate of heat flow σA(3T)4σAT04=σAT04σAT04σA(2T)4\therefore \, \, \, \, \sigma A (3 T)^4 - \sigma A T_0^4 =\sigma A T_0^4 - \sigma AT_0^4 -\sigma A(2T)^4 Solving this equation, we get T0=(972)1/4T \, \, \, \, \, \, \, \, \, \, \, T_0 =\bigg(\frac{97}{2}\bigg)^{1/4}T