Two bodies A and B at temperatures ' $\mathrm{T}_1$ ' $\mathrm{K}$ and ' $\mathrm{T}_2$ ' $\mathrm{K}$…
Two bodies A and B at temperatures ' $\mathrm{T}_1$ ' $\mathrm{K}$ and ' $\mathrm{T}_2$ ' $\mathrm{K}$ respectively have the same dimensions. Their emissivities are in the ratio $1: 3$. If they radiate the same amount of heat per unit area per unit time, then the ratio of their temperatures $\left(T_1: T_2\right)$ is
$1: 3$
$3^{1 / 4}: 1$
$9^{1 / 4}: 1$
$81: 1$
Solution
From Stefan - Boltzmann's law
$\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{e}\left(\sigma \mathrm{AT}^4\right)$
Given $\mathrm{A}$ and $\frac{\mathrm{dQ}}{\mathrm{dt}}$ are same for both the bodies
$\begin{gathered}
\Rightarrow \mathrm{e}_1 \mathrm{~T}_1^4=\mathrm{e}_2 \mathrm{~T}_2^4 \\
\therefore \quad\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4=\frac{\mathrm{e}_2}{\mathrm{e}_1}=\frac{3}{1} \\
\Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sqrt[4]{3}}{1}=\frac{3^{\frac{1}{4}}}{1}
\end{gathered}$
.