Animated Solution for Physics - Properties of Solids and Liquids: 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

Welcome, future engineers and physicists! Today, we are going to dive into a fascinating problem that beautifully combines the concepts of thermal radiation and steady-state equilibrium. This is a classic scenario that tests your understanding of the Stefan-Boltzmann Law and your ability to set up a heat balance equation.

Imagine you are standing in a room with three massive, parallel plates kept very close to each other. These aren't just any plates; they are ideal black surfaces, meaning they are perfect absorbers and perfect emitters of radiation. The outer plates are maintained at constant temperatures of $2T$ and $3T$. Our mission is to find the temperature of the middle plate once the system settles into a steady state.

Let's visualize the system. We have Plate 1 on the left at a temperature of $2T$, Plate 3 on the right at a hotter temperature of $3T$, and Plate 2 sandwiched in the middle. Let's assume the middle plate reaches an unknown steady temperature, which we will call $T_0$.

What exactly does steady state mean in this context? It means that Plate 2 is in thermal equilibrium. It is no longer heating up or cooling down. For this to happen, the rate at which it absorbs heat must perfectly balance the rate at which it emits heat. Since Plate 3 is the hottest ($3T$), heat will flow from Plate 3 to Plate 2. Let's call this heat flow $Q_1$. Simultaneously, since Plate 2 is hotter than Plate 1 ($2T$), heat will flow from Plate 2 to Plate 1. Let's call this heat flow $Q_2$.

In steady state, the heat gained equals the heat lost:

To quantify these heat flows, we turn to one of the most elegant equations in thermodynamics: the Stefan-Boltzmann Law. It states that the net heat transfer rate between two parallel black surfaces is proportional to the difference of the fourth powers of their absolute temperatures.

For the heat flowing from Plate 3 to Plate 2, we can write:

Similarly, for the heat flowing from Plate 2 to Plate 1, we write:

Here, $\sigma$ is the Stefan-Boltzmann constant, and $A$ is the surface area of the plates.

Now, we substitute our expressions for $Q_1$ and $Q_2$ into our steady-state condition:

Notice how the constants $\sigma$ and $A$ are present on both sides of the equation. Because the plates have the same area and are made of the same ideal material, these constants beautifully cancel out, leaving us with a pure temperature relationship:

Let's carefully expand the powers. Remember that $(3T{)}^4=3^4\cdot T^4=81T^4$, and $(2T{)}^4=2^4\cdot T^4=16T^4$. Substituting these back in, we get:

We are almost there! Let's group the $T_0$ terms on one side and the $T$ terms on the other. By adding $T_0^4$ to both sides and adding $16T^4$ to both sides, we obtain:

Now, we isolate $T_0^4$ by dividing by 2:

To find our final temperature $T_0$, we simply take the fourth root of both sides:

And there we have it! The steady-state temperature of the middle plate is exactly ${\left(\frac{97}{2}\right)}^{\frac{1}{4}}T$. This perfectly matches option (c).

This problem is a fantastic reminder of how powerful conservation laws are. By simply stating that 'energy in equals energy out', we were able to navigate through the physics of thermal radiation and arrive at a precise mathematical conclusion. Keep practicing, and always look for the underlying balance in physical systems!