Question

A solid copper sphere of density $\rho$, specific heat capacity $C$ and radius $r$ is initially at $200\, K$. It is suspended inside a chamber whose walls are at $0\, K$. The time required (in (is) for the temperature of the sphere to drop to $100 \,K$ is
($\sigma$ is Stefan's constant and all the quantities are in SI units)

• A. $48 \frac{r\rho C}{\sigma}$
• B. $\frac{1}{48} \frac{r\rho C}{\sigma}$
• C. $\frac{27}{7} \frac{r\rho C}{\sigma}$
• D. $\frac{7}{27} \frac{r\rho C}{\sigma}$

Answer 1

Answer: (B)

Here, $T=200 \,K$ and $T_{0}=0 \,K$
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.
So, $ t=\frac{m s \Delta T}{\sigma A\left(T^{4}-T_{0}^{4}\right)} \left(\because T_{0}=0\, K \right)$
$ \Rightarrow t=\frac{(\rho V) C(200\, K -100\, K )}{\sigma(A)\left(200^{4}-0^{4}\right)}$
$\Rightarrow t=\frac{\rho \frac{4}{3} \pi r^{3} C \times 100}{\sigma 4 \pi r^{2} \times(200)^{4}} $
$ \Rightarrow t=\frac{1}{48} \frac{r \rho C}{\sigma} \times 10^{-6} s$
$ =\frac{1}{48} \frac{\rho r C}{\sigma} \,\mu s $