Question

A black-colored solid sphere of radius $R$ and mass $M$ is inside a cavity with a vacuum inside. The walls of the cavity are maintained at temperature ${T_0}$. The initial temperature of the sphere is $3{T_0}$. 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 $2{T_0}$ will be
( $\sigma$ is Stefan Boltzmann constant)
A) $\dfrac{{M\alpha }}{{4\pi {R^2}\sigma }}\ln \left( {\dfrac{3}{2}} \right)$
B) $\dfrac{{M\alpha }}{{4\pi {R^2}\sigma }}\ln \left( {\dfrac{{16}}{3}} \right)$
C) $\dfrac{{M\alpha }}{{16\pi {R^2}\sigma }}\ln \left( {\dfrac{{16}}{3}} \right)$
D) $\dfrac{{M\alpha }}{{16\pi {R^2}\sigma }}\ln \left( {\dfrac{3}{2}} \right)$

Answer 1

Hint
Since the sphere is a black body, it will radiate away its energy and reduce its temperature. The rate of energy radiated can be calculated from the Stefan–Boltzmann law which can then be equated with the rate of heat loss in the blackbody
Formula used:
Stefan–Boltzmann law: $\dfrac{{dQ}}{{dt}} = \sigma A{T^4}$ where $\dfrac{{dQ}}{{dt}}$ is the energy radiated by a black body at temperature $T$ and having area $A$.

Complete step by step answer
Since the body in the question is a perfect black body at temperature $3{T_0}$, it will radiate away energy in the form of radiation according to Stefan–Boltzmann law and this will cause a decrease in the temperature of the body.
The small change in temperature of a body $dT$ due to energy loss $dQ$ can be calculated as:
$\Rightarrow dQ = McdT$
Substituting the value of specific heat, we get
$\Rightarrow dQ = M\alpha {T^3}dT$
Differentiating the above equation to get the rate of heat loss, we get
$\Rightarrow \dfrac{{dQ}}{{dt}} = \dfrac{{M\alpha {T^3}dT}}{{dt}}$
Comparing the rate of energy loss with the rate of energy radiated away from the black body as calculated from Stefan–Boltzmann law, we get:
$\Rightarrow \dfrac{{M\alpha {T^3}dT}}{{dt}} = \sigma A{T^4}$
On integrating both sides for the temperature $3{T_0}$ to $2{T_0}$ in time $t = 0$ to $t = t$, and substituting $A = 4\pi {R^2}$, we get
$\Rightarrow \int\limits_{T = 2{T_0}}^{3{T_0}} {M\alpha {T^3}dT} = \int\limits_{t = 0}^t {\sigma (4\pi {R^2}){T^4}} dt$
Taking out the constants from integration, we get
$\Rightarrow t = \dfrac{{M\alpha }}{{\sigma 4\pi {R^2}}}\int\limits_{T = 2{T_0}}^{3{T_0}} {\dfrac{{dT}}{T}}$
$\Rightarrow t = \dfrac{{M\alpha }}{{\sigma 4\pi {R^2}}}\left( {\ln T|_{T = 2{T_0}}^{T = 3{T_0}}} \right)$
On simplifying the natural logarithm term, we get
$\Rightarrow t = \dfrac{{M\alpha }}{{\sigma 4\pi {R^2}}}\ln \left( {\dfrac{3}{2}} \right)$ which corresponds to option A.

Note
Since the specific heat of the material is a function of the temperature itself, we cannot directly use the temperature difference of the body to determine the heat radiated away to cause that temperature difference in the body, but rather we need to integrate it. Unless mentioned otherwise, a black colored body should be taken to behave as a perfect black body.