Question

A solid copper sphere of density $\rho$, specific heat capacity $C$ and radius $r$ is initially at $200K$. It is suspended inside a chamber whose walls are at $0K$. The time required (in (is) for the temperature of the sphere to drop to $100K$ 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, $T=200K$ and $T_0=0K$
As the rate of fall of temperature,
$\frac{\Delta T}{\Delta t}=\frac{\sigma Ae\left(T^4-T_0^4\right)}{ms}$
where, $\sigma =$ Stefan's constant,
$A=$ area of sphere,
and $e=$ emissivity $=1$ and
$S=$ specific heat capacity.
So, $t=\frac{ms\Delta T}{\sigma A\left(T^4-T_0^4\right)}\left(\because T_0=0K\right)$
$\Rightarrow t=\frac{(\rho V)C(200K-100K)}{\sigma (A)\left(200^4-0^4\right)}$
$\Rightarrow t=\frac{\rho \frac{4}{3}\pi r^{3}C\times 100}{\sigma 4\pi r^2\times (200{)}^4}$
$\Rightarrow t=\frac{1}{48}\frac{r\rho C}{\sigma }\times 10^{-6}s$
$=\frac{1}{48}\frac{\rho rC}{\sigma }\mu s$