Question

A black coloured solid sphere of radius $R$ and mass $M$ is inside a cavity with vacuum inside. The walls of the cavity are maintained at temperature $T_0$. The initial temperature of the sphere is $3T_0$. If the specific heat of the material of the sphere varies as $\alpha T^3$ per unit mass with the temperature $T$ of the sphere, where $\alpha$ is a constant, then the time taken for the sphere to cool down to temperature $2T_0$ will be
($\sigma$ is Stefan Boltzmann constant)

• A. $\frac{M\alpha }{4\pi R^2\sigma }ln\left(\frac{3}{2}\right)$
• B. $\frac{M\alpha }{4\pi R^2\sigma }ln\left(\frac{16}{3}\right)$
• C. $\frac{M\alpha }{16\pi R^2\sigma }ln\left(\frac{16}{3}\right)$
• D. $\frac{M\alpha }{16\pi R^2\sigma }ln\left(\frac{3}{2}\right)$

Answer 1

Answer: (A)

$\frac{M\alpha }{4\pi R^2\sigma }ln\left(\frac{3}{2}\right)$

Answer is A.
For a small change in temperature, heat loss is $dQ=MSdT=M\alpha T^{3}dT$
Differentiating w.r.t dt above, we get

$dQ/dt=M\alpha \frac{T^{3}dT}{dt}$

We know from stefan's law that energy per unit of time is $\frac{dQ}{dt}=\sigma A{T}^4$
Therefore from above, we write

$M\alpha \frac{T^{3}dT}{dt}=\sigma A{T}^4$

Integrating above from $3T_o$ to $2T_o$ and $A=4\pi R^2$ we get

${\int }_o^{t}dt=\frac{M\alpha }{\sigma 4\pi R^2}{\int }_{2T_o}^{3T_o}\frac{dT}{T}$

Or $t=\frac{M\alpha }{\sigma 4\pi R^2}\times \ln T{|}_{2T_o}^{3T_o}$

Or $t=\frac{M\alpha }{\sigma 4\pi R^2}\times \ln \frac{3T_o}{2T_o}$

$t=\frac{M\alpha }{4\pi R^2\sigma }log\left(3/2\right)$