Q. A sphere at temperature $600K$ is placed in an environment of temperature is $200K$. Its cooling rate is $H$. If its temperature reduced to $400K$ then cooling rate in same environment will become
A) $(3/16)H$
B) $(16/3)H$
C) $(9/27)H$
D) $(1/16)H$
Explanation
(a) Rate of cooling $\propto (T^{4} - T_{0}^{4})$
$\Rightarrow \frac{H}{H^{'}} = \frac{(T_{1}^{4} - T_{0}^{4})}{(T_{2}^{4} - T_{0}^{4})} = \frac{600^{4} - 200^{4}}{400^{4} - 200^{4}}$
Or $H^{'} = \frac{(16 + 4)(16 - 4)H}{(36 + 4)(36 - 4)} = \frac{3}{16}H$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.