JEE Advanced · Physics · 12. Thermal Properties
Two identical plates P and Q , radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures \(\mathrm{T}_{\mathrm{P}}\) and \(\mathrm{T}_{\mathrm{Q}}\), respectively, with \(\mathrm{T}_{\mathrm{Q}}<\mathrm{T}_{\mathrm{P}}\), 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 ________.

- A 3
- B 4
- C 5
- D 6
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
Initially :
\(\mathrm{W}_0=\sigma\left(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_{\mathrm{Q}}^4\right)\)

Finally :
Putting heat currents equal in steady state :

\(\begin{aligned} & \sigma\left(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4\right)=\sigma\left(\mathrm{T}_1^4-\mathrm{T}_2^4\right) \\ & \sigma\left(\mathrm{T}_1^4-\mathrm{T}_2^4\right)=\sigma\left(\mathrm{T}_2^4-\mathrm{T}_{\mathrm{Q}}^4\right)\end{aligned}\)
Adding :
\(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4=\mathrm{T}_2^4-\mathrm{T}_{\mathrm{Q}}^4\)
\(\Rightarrow \mathrm{T}_1^4+\mathrm{T}_2^4=\mathrm{T}_{\mathrm{P}}^4+\mathrm{T}_{\mathrm{Q}}^4\)
and \(\quad \Rightarrow \mathrm{T}_1^4-\mathrm{T}_2^4=\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4\)
Adding : \(\mathrm{T}_1^4=\frac{2 \mathrm{~T}_{\mathrm{P}}^4+\mathrm{T}_{\mathrm{Q}}^4}{3}\)
So \(\mathrm{W}_{\mathrm{S}}=\sigma\left(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4\right)\)
\(=\sigma\left(\mathrm{T}_{\mathrm{P}}^4-\left(\frac{2 \mathrm{~T}_{\mathrm{P}}^4+\mathrm{T}_{\mathrm{Q}}^4}{3}\right)\right)=\sigma\left(\frac{\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_{\mathrm{Q}}^4}{3}\right)\)
hence \(\frac{\mathrm{W}_{\mathrm{S}}}{\mathrm{W}_0}=3\)
\(\mathrm{W}_0=\sigma\left(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_{\mathrm{Q}}^4\right)\)

Finally :
Putting heat currents equal in steady state :

\(\begin{aligned} & \sigma\left(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4\right)=\sigma\left(\mathrm{T}_1^4-\mathrm{T}_2^4\right) \\ & \sigma\left(\mathrm{T}_1^4-\mathrm{T}_2^4\right)=\sigma\left(\mathrm{T}_2^4-\mathrm{T}_{\mathrm{Q}}^4\right)\end{aligned}\)
Adding :
\(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4=\mathrm{T}_2^4-\mathrm{T}_{\mathrm{Q}}^4\)
\(\Rightarrow \mathrm{T}_1^4+\mathrm{T}_2^4=\mathrm{T}_{\mathrm{P}}^4+\mathrm{T}_{\mathrm{Q}}^4\)
and \(\quad \Rightarrow \mathrm{T}_1^4-\mathrm{T}_2^4=\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4\)
Adding : \(\mathrm{T}_1^4=\frac{2 \mathrm{~T}_{\mathrm{P}}^4+\mathrm{T}_{\mathrm{Q}}^4}{3}\)
So \(\mathrm{W}_{\mathrm{S}}=\sigma\left(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4\right)\)
\(=\sigma\left(\mathrm{T}_{\mathrm{P}}^4-\left(\frac{2 \mathrm{~T}_{\mathrm{P}}^4+\mathrm{T}_{\mathrm{Q}}^4}{3}\right)\right)=\sigma\left(\frac{\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_{\mathrm{Q}}^4}{3}\right)\)
hence \(\frac{\mathrm{W}_{\mathrm{S}}}{\mathrm{W}_0}=3\)
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