A solid copper sphere of density $\rho$, specific heat capacity $C$ and radius $r$ is initially at $200…
A solid copper sphere of density $\rho$, specific heat capacity $C$ and radius $r$ is initially at $200 \mathrm{~K}$. It is suspended inside a chamber whose walls are at $0 \mathrm{~K}$. The time required (in $\mu \mathrm{s})$ for the temperature of the sphere to drop to $100 \mathrm{~K}$ is
( $\sigma$ is Stefan's constant and all the quantities are in SI units.)
$48 \frac{r \rho C}{\sigma}$
$\frac{1}{48} \frac{r \rho C}{\sigma}$
$\frac{27}{7} \frac{r \rho C}{\sigma}$
$\frac{7}{27} \frac{r \rho C}{\sigma}$
Solution
Here, $T=200 \mathrm{~K}$ and $T_0=0 \mathrm{~K}$
As the rate of fall of temperature,
$
\frac{\Delta T}{\Delta t}=\frac{\sigma A e\left(T^4-T_0^4\right)}{m s}
$
where, $\sigma=$ Stefan's constant, $A=$ area of sphere, and $e=$ emissivity $=1$ and $S=$ specific heat capacity.
$
\begin{aligned}
& \text { So, } \quad t=\frac{m s \Delta T}{\sigma A\left(T^4-T_0^4\right)} \\
& \left(\because T_0=0 \mathrm{~K}\right) \\
& \Rightarrow \quad t=\frac{(\rho V) C(200 \mathrm{~K}-100 \mathrm{~K})}{\sigma(A)\left(200^4-0^4\right)} \\
& \Rightarrow \quad t=\frac{\rho \frac{4}{3} \pi r^3 C \times 100}{\sigma 4 \pi r^2 \times(200)^4} \\
& \Rightarrow \quad t=\frac{1}{48} \frac{r \rho C}{\sigma} \times 10^{-6} \mathrm{~s}=\frac{1}{48} \frac{\rho r C}{\sigma} \mu \mathrm{s} \\
&
\end{aligned}
$
Hence, the correct option is (b)