JEE Advanced 2025 Paper 1 · Q10 · Heat Radiation / Stefan-Boltzmann

Question

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 $T_Q < T_P$. 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 $Q$, so that there are now four plates in the order $P, A, B, Q$. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from $P$ to $Q$ in the steady state is $W_S$, then the ratio $\dfrac{W_0}{W_S}$ is _____.

SolveKar AI · Accepted Answer

For two parallel black-body plates at temperatures $T_P$ and $T_Q$, the net radiative flux from $P$ to $Q$ is, by Stefan-Boltzmann, $W_0 = \sigma (T_P^4 - T_Q^4)$. With two extra plates inserted, label intermediate steady-state temperatures $T_1, T_2$ for plates $A, B$. In the steady state every gap carries the same net flux $W_S$ (no heat accumulates on any plate). Hence $W_S = \sigma (T_P^4 - T_1^4) = \sigma (T_1^4 - T_2^4) = \sigma (T_2^4 - T_Q^4)$. Adding the three equations: $3 W_S = \sigma (T_P^4 - T_Q^4) = W_0$. Therefore $\dfrac{W_0}{W_S} = 3$. (In general, inserting $n$ identical black-body shields between two black plates reduces the net flux by a factor of $n+1$.) Answer: $3$.

JEE Advanced 2025 Paper 1 · Q10 · Heat Radiation / Stefan-Boltzmann

Solution