JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Heat & Thermodynamics

Two identical plates P and Q, radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures

T_{P}

and

T_{Q}

, respectively, with

T_{Q} < T_{P}

, as shown in Fig. 1. The radiated power transferred per unit area from P to Q is

W_{0}

. Subsequently, two more plates, identical to P and Q, are introduced between P and Q, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from P to Q (Fig. 2) in the steady state is

W_{S}

, then the ratio

\frac{W _{0}}{W _{S}}

is ____.

Answer: 3

Solution

Initially (Fig. 1):

W_{0} = σ (T_{P}^{4} - T_{Q}^{4})

Finally (Fig. 2): With two intermediate plates at temperatures

T_{1}

and

T_{2}

, in steady state:

σ (T_{P}^{4} - T_{1}^{4}) = σ (T_{1}^{4} - T_{2}^{4}) = σ (T_{2}^{4} - T_{Q}^{4})

From these equations:

T_{P}^{4} - T_{1}^{4} = T_{2}^{4} - T_{Q}^{4}

T_{1}^{4} + T_{2}^{4} = T_{P}^{4} + T_{Q}^{4}

Solving:

T_{1}^{4} = \frac{2 T _{P}^{4} + T _{Q}^{4}}{3}

Power transferred:

W_{S} = σ (T_{P}^{4} - T_{1}^{4}) = σ (T_{P}^{4} - \frac{2 T _{P}^{4} + T _{Q}^{4}}{3}) = σ \frac{T _{P}^{4} - T _{Q}^{4}}{3}

Therefore:

\frac{W _{0}}{W _{S}} = 3

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer

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JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Heat & Thermodynamics

Solution

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