Thank you for reporting, we will resolve it shortly
Q.
If temperature of a black body increases from $-73^{\circ} C$ to $327^{\circ} C$, then ratio of emissive power at these two temperatures is
Thermal Properties of Matter
Solution:
According to Stefan's law, the radiant energy emitted by a perfectly black body per unit area per second (i.e., emissive power or radiancy) is directly proportional to the fourth power of its absolute temperature.
i.e., $E \propto T^{4}$ or $E=\sigma T^{4}$
where $\sigma$ is Stefan's constant whose value is
$5.67 \times 10^{-8} W / m ^{2} K ^{4}$
As $T K=t^{\circ} C +273$
$\therefore T_{1}=-73^{\circ} C +273=200 \,K $;
$ T_{2}=327^{\circ} C +273=600 \,K$
$\therefore \frac{E_{1}}{E_{2}}=\left(\frac{T_{1}}{T_{2}}\right)^{4}$
$=\left(\frac{200 \,K }{600 \,K }\right)^{2}=\left(\frac{1}{3}\right)^{4}=\frac{1}{81}$