Steady State Temperature of Middle Plate Between Ideal Black Surfaces

To find the steady-state temperature of the intermediate plate, let:

• $T_1 = 2T$ be the temperature of the first plate,
• $T_3 = 3T$ be the temperature of the third plate,
• $T'$ be the steady-state temperature of the intermediate (middle) plate,
• $A$ be the surface area of each plate,
• $\sigma$ be the Stefan-Boltzmann constant.

Since the plates are very large, parallel, close to each other, and behave as ideal blackbodies with emissivity $\varepsilon = 1$ and absorptivity $\alpha = 1$, the radiation exchange between adjacent surfaces can be treated as direct transfer with a view factor of approximately $1$.

The intermediate plate has very high thermal conductivity, so both of its surfaces are maintained at the same uniform temperature $T'$.

1. Radiation absorbed by the middle plate:

   • From the first plate (at temperature $T_1$): $P_{\text{abs, 1}} = \sigma A T_1^4 = \sigma A (2T)^4$
   • From the third plate (at temperature $T_3$): $P_{\text{abs, 3}} = \sigma A T_3^4 = \sigma A (3T)^4$
   • Total power absorbed by the middle plate: $P_{\text{abs}} = \sigma A \left[(2T)^4 + (3T)^4\right]$

2. Radiation emitted by the middle plate: The middle plate emits thermal radiation from both of its surfaces (one facing the first plate and one facing the third plate): $P_{\text{emit}} = 2 \sigma A (T')^4$

3. Steady-state condition: In the steady-state, the net rate of heat transfer to the middle plate must be zero, meaning the total power absorbed must equal the total power emitted: $P_{\text{abs}} = P_{\text{emit}}$ $\sigma A \left[(2T)^4 + (3T)^4\right] = 2\sigma A (T')^4$

   Canceling $\sigma A$ from both sides gives: $2(T')^4 = (2T)^4 + (3T)^4$ $2(T')^4 = (16 + 81)T^4 = 97T^4$ $(T')^4 = \frac{97}{2} T^4$

   Taking the fourth root on both sides: $T' = \left(\frac{97}{2}\right)^{\frac{1}{4}} T$

Therefore, the correct option is C.