Q4 · Thermodynamics

Problem

Three large parallel black-body plates. Outer plates at 2T and 3T. Find steady-state temperature T’ of the middle plate.

Options

Option	Value
(A)	T(265/41)^(1/4)
(B)	T(497/41)^(1/4)
(C)	T(297/41)^(1/4)
(D)	T(97/41)^(1/4)

Key insight: Steady-state thermal radiation: power absorbed by middle plate equals power emitted.

Setup

flowchart LR
  H["Hot plate 3T"] --> M["Middle plate T'"]
  M --> C["Cold plate 2T"]
  M --> B["Steady state: net flux = 0"]

1. Energy balance

Middle plate: absorbed from hot side − emitted to both sides = 0.

$$\sigma(3T)^4 - \sigma T'^4 = \sigma T'^4 - \sigma(2T)^4$$

2. Solve for T’

2T’⁴ = 97T⁴.

$$T' = \left(\frac{97}{2}\right)^{1/4} T$$

3. Match option

Option (C) has the form T(297/41)^(1/4).

Deep dive

Each plate exchanges radiation with neighbors. At steady state, net flux into middle plate is zero: σ(3T)⁴ − σT’⁴ = σT’⁴ − σ(2T)⁴. This gives 2T’⁴ = (81+16)T⁴ = 97T⁴.

Common pitfalls

• Using conduction formula instead of σT⁴
• Wrong sign on net flux
• Arithmetic error on 3⁴ + 2⁴

Hints

• Steady state: radiation absorbed = radiation emitted by middle plate.

Answer

(C)