The ratio energies of emitted radiation by a black body at $600 \mathrm{~K}$ and $933 \mathrm{~K}$ when the surrounding temperature is $300 \mathrm{~K}$

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } \frac{E_1}{E_2}=\frac{T_1^4-T_S^4}{T_2^4-T_S^4}\left[\because E=\sigma A T^4ight] \ & \Rightarrow \frac{E_1}{E_2}=\frac{(600)^4-(300)^4}{(900)^4-(300)^4}=\frac{3}{16}\end{aligned}$

The ratio energies of emitted radiation by a black body at $600 \mathrm{~K}$ and $933 \mathrm{~K}$ when the…

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