Q. A solid copper sphere (density $\rho$ and specific heat capacity $c$) of radius $r$ at an initial temperature $200K$ is suspended inside a chamber whose walls are at almost $0K$. The time required (in $\mu\ s$) for the temperature of the sphere to drop to $100\ K$ is
A) $\frac{72}{2}\ \frac{r\rho c}{\sigma}$
B) $\frac{7}{72}\ \frac{r\rho c}{\sigma}$
C) $\frac{27}{7}\ \frac{r\rho c}{\sigma}$
D) $\frac{7}{27}\ \frac{r\rho c}{\sigma}$
Explanation
(b) $\frac{dT}{dt} = \frac{\sigma A}{mcJ}(T^{4} - T_{0}^{4})$ [In the given problem fall in temperature of body $dT = (200 - 100) = 100K$, temp. of surrounding $T_{0} = 0K$, Initial temperature of body $T = 200K$]
$\frac{100}{dt} = \frac{\sigma 4\pi r^{2}}{\frac{4}{3}\pi r^{3}\rho cJ}(200^{4} - 0^{4})$
$\Rightarrow dt = \frac{r\rho cJ}{48\sigma} \times 10^{- 6}s = \frac{r\rho c}{\sigma}.\frac{4.2}{48} \times 10^{- 6}$
$= \frac{7}{80}\ \frac{r\rho c}{\sigma}\mu s \approx \frac{7}{72}\frac{r\rho c}{\sigma}\mu s\ \ \lbrack As\ J = 4.2\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.