Q.The ratio of emissive power of a perfect blackbody at 1327 ºC and 527 ºC is (A) 4:1
(B) 16:1
(C) 2:1
(D) 8:1

Stefan's law gives the total radiant energy from a black body as $\sigma A T^4$ per second; multiply by time for the total energy.

By Stefan's law, the power radiated by a perfectly black body of surface area $A$ at absolute temperature $T$ is

$P = \sigma A T^{4}$

Given $A = 200\text{ cm}^2 = 200\times10^{-4}\text{ m}^2 = 0.02\text{ m}^2$, $T = 127^\circ\text{C} = 400\text{ K}$, and $\sigma = 5.67\times10^{-8}\text{ W m}^{-2}\text{K}^{-4}$:

$T^4 = (400)^4 = 2.56\times10^{10}\text{ K}^4$

…

Rate of loss of heat by radiation follows Stefan's law with a surrounding-temperature correction.

By Stefan's law (with surroundings at temperature $T_0$), the net rate of loss of heat is $R \propto (T^4-T_0^4)$.

Convert to kelvin: $T_1=827+273=1100$ K, $T_2=427+273=700$ K, $T_0=27+273=300$ K.

$\frac{R_1}{R_2} = \frac{T_1^4-T_0^4}{T_2^4-T_0^4} = \frac{(1100)^4-(300)^4}{(700)^4-(300)^4}$

Computing: $1100^4 = 1.4641\times10^{12}$, $700^4=2.401\times10^{11}$, $300^4=8.1\times10^{9}$. …