MHT CET · Physics · Thermal Properties of Matter
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
- A 405
- B 35
- C 45
- D 350
Answer & Solution
Correct Answer
(A) 405
Step-by-step Solution
Detailed explanation
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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