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A black body at temperature $127^{\circ} \mathrm{C}$ radiates heat at the rate of $5 \mathrm{cal} / \mathrm{cm}^2 \mathrm{~s}$. At a temperature $927^{\circ} \mathrm{C}$, its rate of emission in units of $\mathrm{cal} / \mathrm{cm}^2 \mathrm{~s}$ will be
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The correct answer is:
405
From Stefan - Boltzmann's Law
$$
\begin{aligned}
& \mathrm{E}=\sigma \mathrm{T}^4 \\
& \Rightarrow \mathrm{E} \propto \mathrm{T}^4
\end{aligned}
$$
$\begin{aligned} \therefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4=\left(\frac{400}{1200}\right)^4 \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{1}{81} \\ \mathrm{E}_2 & =81 \times \mathrm{E}_1 \\ & =81 \times 5 \\ & =405 \mathrm{cal} / \mathrm{cm}^2 \mathrm{~s}\end{aligned}$
$$
\begin{aligned}
& \mathrm{E}=\sigma \mathrm{T}^4 \\
& \Rightarrow \mathrm{E} \propto \mathrm{T}^4
\end{aligned}
$$
$\begin{aligned} \therefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4=\left(\frac{400}{1200}\right)^4 \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{1}{81} \\ \mathrm{E}_2 & =81 \times \mathrm{E}_1 \\ & =81 \times 5 \\ & =405 \mathrm{cal} / \mathrm{cm}^2 \mathrm{~s}\end{aligned}$
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