Question

A solid sphere of radius $R,$ density $\rho$ and specific heat $s$ is initially at temperature $T_0$ Kelvin. When suspended in a surrounding of temperature $0K$, the time required to decrease the temperature of the sphere from $T_0$ to $\frac{T_0}{2}$ Kelvin is
(Assume that the sphere behaves like a black body)

• A. $\frac{5\rho sR}{9\sigma T_0^3}$
• B. $\frac{\rho sR}{9\sigma T_0^3}$
• C. $\frac{7\rho sR}{9\sigma T_0^3}$
• D. $\frac{8\rho sR}{9\sigma T_0^3}$

Answer 1

The correct option is C $\frac{7\rho sR}{9\sigma T_0^3}$

Since surrounding temp is $0K$
$\frac{dQ}{dt}=\sigma A(T^4-T_S^4)[\because e=1,T_S=0]$
$ms(-\frac{dT}{dt})=\sigma A{T}^4$
$\Rightarrow \rho \frac{4}{3}\pi R^{3}s(-\frac{dT}{dt})=\sigma 4\pi R^2{T}^4$
$\Rightarrow -\underset{T_0}{\overset{\frac{T_0}{2}}{\int }}\frac{dT}{T^4}=\frac{3\sigma }{\rho sR}\underset{0}{\overset{t}{\int }}dt$
$\Rightarrow +{\left[\frac{1}{3T^3}\right]}_{T_0}^{\frac{T_0}{2}}=+\frac{3\sigma }{\rho sR}t$
$\Rightarrow \frac{8}{T_0^3}-\frac{1}{T_0^3}=\frac{9\sigma t}{\rho sR}$

$\Rightarrow t=\frac{7\rho sR}{9\sigma T_0^3}$