Q. The rate of cooling at 600 K, if surrounding temperature is 300 K is$R$. The rate of cooling at 900 K is
A) $\frac{16}{3}R$
B) 2 $R$
C) 3 $R$
D) $\frac{2}{3}R$
Explanation
(a) Rate of cooling proportional to $\left( T^{4} - T_{0}^{4} \right),$ as per Stefan’s Law.
$\therefore\ \ \ \frac{R^{'}}{R} = \ \frac{(900)^{4} - (300)^{4}}{(600)^{4} - (300)^{4}}$
$= \ \frac{9^{4} - 3^{4}}{6^{4} - 3^{4}} = \ \frac{3^{4}\left( 3^{4} - 1 \right)}{3^{4}\left( 2^{4} - 1 \right)}\ $
= $\frac{80}{15} = \frac{16}{3}\ $
$R^{'}\ = \frac{16}{3}\ R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.