Q. Two bodies $A\ $and $B$ having temperatures 327$℃$ and 427$℃$ are radiating heat to the surrounding. The surrounding temperature is 27$℃$ . The ration of rate of heat radiation of $A$ to that of $B$ is
A) 0.52
B) 0.31
C) 0.81
D) 0.42
Explanation
(a) If temperature of surrounding is considered, then net loss of energy of a body by radiation
$\mathcal{Q =}Ae\sigma(T^{4} - T_{0}^{4})$
$\Rightarrow$ $\mathcal{Q \propto (}T^{4} - T_{0}^{4})$
$\therefore$ $\frac{\mathcal{Q}_{1}}{\mathcal{Q}_{2}} = \frac{T_{1}^{4} - T_{0}^{4}}{T_{2}^{4} - T_{0}^{4}}$
=$\frac{(273 + 327)^{4} - (273 + 27)^{4}}{(273 + 427)^{4} - (273 + 27)^{4}}$
$= \ \frac{(600)^{4} - (300)^{4}}{(700)^{4} - (300)^{4}} = 0.52$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.