Question

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 $100K$ is

• A. $\frac{72}{7}\frac{r\rho c}{\sigma }$
• B. $\frac{7}{72}\frac{r\rho c}{\sigma }$
• C. $\frac{27}{7}\frac{r\rho c}{\sigma }$
• D. $\frac{7}{27}\frac{r\rho c}{\sigma }$

Answer 1

Answer: (B)

$\frac{dT}{dt}=\frac{\sigma A}{mcJ}\left(T^4-T_0^4\right)$ [In the given problem, fall in temperature of body $dT=(200-100)=100K$, temp. of surrounding $T_0=0K$, Initial temperature of body $T=200K$ ]
$\frac{100}{dt}=\frac{\sigma 4\pi r^2}{\frac{4}{3}\pi r^3\rho cJ}\left(200^4-0^4\right)$
$\Rightarrow dt=\frac{r\rho cJ}{48\sigma }\times 10^{-6}\cdot s=\frac{r\rho c}{\sigma }\cdot \frac{4.2}{48}\times 10^{-6}$
$=\frac{7}{80}\frac{r\rho c}{\sigma }\mu s\simeq \frac{7}{72}\frac{r\rho c}{\sigma }\mu s$[ As $J=4.2]$