Question

A solid copper sphere of density $\rho ,$ specific heat $c$ and radius $r$ is at temperature $T_1$. It is suspended inside a chamber whose walls are at temperature $0K$. The time required for the temperature of sphere to drop to $T_2$ is $\frac{r\rho c}{xe\sigma }(\frac{1}{T_2^3}-\frac{1}{T_1^3})$. Find the value of $x$ ?
Take the emmissivity of the sphere to be equal to $e$.

Answer 1

Rate of heat loss $\frac{dQ}{dt}=eA\sigma T^4$
$\Rightarrow mc\left(\frac{-dT}{dt}\right)=eA\sigma T^4$
Temperature decreases with time
$m=\frac{4}{3}\pi r^3\rho ,A=4\pi r^2$

$\Rightarrow -\frac{dT}{dt}=\frac{3e\sigma }{\rho cr}{T}^4\Rightarrow -\underset{0}{\overset{t}{\int }}dt=\frac{r\rho c}{3e\sigma }\underset{T_1}{\overset{T_2}{\int }}\frac{dT}{T^4}\Rightarrow t=+\frac{r\rho c}{9e\sigma }[\frac{1}{T_2^3}-\frac{1}{T_1^3}]\Rightarrow x=9$