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 sphere is $3T_0$. If the specific heat of the material of the sphere varies as $\alpha T^3$ with the 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

The correct option is C $\frac{M\alpha }{16\pi R^2\sigma }\ln \left(\frac{16}{3}\right)$

$\text{Solution}:$

In the given problem, fall in temperature of sphere,

$dT=(3T_0-2T_0)=T_0$

Temperature of surrounding, $T_{surr}=T_0$

Initial temperature of sphere, $T_{initial}=3T_0$

Specific heat of the material of the sphere varies as,

$c=a{T}^3$ per unit mass ($\alpha$ = a constant) Applying formula,

$\frac{dT}{dt}=\frac{\sigma A}{McJ}(T^4-T_{surr}^4)$

$⟹\frac{T_0}{dt}=\frac{\sigma 4\pi R^2}{M\alpha {\left(3T_0\right)}^{3}J}[{\left(3T_0\right)}^4-{\left(T_0\right)}^4]$

$⟹dt=\frac{M\alpha 27T_0^{4}J}{\sigma 4\pi R^2\times 80T_0^4}$

Solving we get,

Time taken for the sphere to cool down temperature $2T_0$,

$t=\frac{M\alpha }{16\pi R^2\sigma }ln\left(\frac{16}{3}\right)$

$\text{Hence C is the correct option}$