Question

A black body emits heat at the rate of 20 W. When its temperature is 227${}^o$C. Another black body emits heat at the rate of 15W, when its temperature is 277${}^o$C. Compare the area of the surface of the two bodies, if the surrounding is at NTP :

• A. 16 : 1
• B. 1 : 4
• C. 12 : 1
• D. 1 : 12

Answer 1

Answer: (C)

12 : 1

According to the Stefan-Boltzmann law,

Power radiated $P=e\sigma A(T^4-T_0^4)$

where $e$ is emmissivity, $\sigma$ is Stefan's constant, $A$ is surface area of body, $T$ is the temperature of body and $T_0$ is surrounding temperature.

Hence comparing for the two surfaces,

$\frac{P_1}{P_2}=\frac{A_1(T_1^4-T_0^4)}{A_2(T_2^2-T_0^4)}$

Given : $P_1=20$ W, $P_2=15$W, $T_1={227}^o+273=500$ K and $T_2={277}^o+273=550$ K

Surrounding temperature $T_0=293$ K

$⟹\frac{A_1}{A_2}=\frac{P_1(T_2^4-T_0^4)}{P_2(T_1^4-T_0^4)}$

$=\frac{20\times ({550}^4-{293}^4)}{15\times ({500}^4-{293}^4)}$ $=12$