Question

Three very large plates of the 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) ${(\dfrac{{65}}{2})^{1/4}}T$
B) ${(\dfrac{{97}}{4})^{1/4}}T$
C) ${(\dfrac{{97}}{2})^{1/4}}T$
D) ${(97)^{1/4}}T$

Answer 1

Hint: The amount of heat energy released by the inner surfaces of first and third plates will be equal to the amount of heat energy absorbed by the second plate. Use the formula of emission power under steady state conditions to calculate the energy absorbed by the second plate. Emissivity of all the plates will be 1 as the plates have ideal black surfaces.

Formula Used:
Under steady state conditions, emission power, $P = \varepsilon \sigma A{T^4}$ where, $\varepsilon $ is the emissivity of the surface, $\sigma $ is the Stefan-Boltzmann constant, $A$ is the area of surface, $T$ is the temperature of the surface.

Complete step by step solution:
There are 3 plates, having the same area, two of them have temperatures $2T$ and $3T$ respectively. Between them, a third plate is there. Let its temperature be $t$ . Now, we need to find the temperature of the middle plate.

The amount of heat energy released by the inner surfaces of the plates whose temperature is given will be equal to the amount of heat energy absorbed by the middle plate. We know that under steady state conditions, emission power of a surface, $P = \varepsilon \sigma A{T^4}$ where, $\varepsilon $ is the emissivity of the surface, $\sigma $ is the Stefan-Boltzmann constant, $A$ is the area of surface, $T$ is the temperature of the surface.

Therefore, the temperature of the middle plate will be calculated by adding the energy absorbed by the inner surfaces of the other two plates.
Emission power of middle plate ${P_m} = \varepsilon \sigma 2A{t^4} = \sigma 2A{t^4}$
(since emissivity of ideal black surfaces is 1)
(both sides of this plate will be considered as it is absorbing from both sides. Hence, area $ = 2A$ )
Emission power of first plate ${P_f} = \varepsilon \sigma A{(2T)^4} = \sigma A{(2T)^4}$
(since emissivity of ideal black surfaces is 1)
(only 1 side of this plate will be considered as the middle plate can absorb heat from its 1 side only)
Emission power of third plate ${P_t} = \varepsilon \sigma A{(3T)^4} = \sigma A{(3T)^4}$
(since emissivity of ideal black surfaces is 1)
(only 1 side of this plate will be considered as the middle plate can absorb heat from its 1 side only)
As explained above, ${P_m} = {P_f} + {P_t}$
$ \Rightarrow \sigma 2A{t^4} = \sigma A{(2T)^4} + \sigma A{(3T)^4} = \sigma A[{(2T)^4} + {(3T)^4}]$
Cancelling like terms, we get $2{t^4} = {(2T)^4} + {(3T)^4} = 16{T^4} + 81{T^4} = 97{T^4}$
This implies, $t = {(\dfrac{{97}}{2})^{\dfrac{1}{4}}}T$
This will be the temperature of the middle plate.

Hence, C is the correct answer.

Note: Emissivity is defined as the measure of infrared energy that is radiated from an object. It can vary between the range of 0 to 1 depending on the type of object. An ideal black surface always has an emissivity of 1. When emissivity is 1, the object is said to be black body and when it is 0, the object is said to be a shiny mirror.