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. \(\frac{M\alpha}{4\pi R^2\sigma}\ln\left(\frac32\right)\)
• 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. \(\frac{M\alpha}{16\pi R^2\sigma}\ln\left(\frac32\right)\)

Hint:

Net radiated power \(=\sigma A(T^4-T_0^4)\) equals \(-Mc\,dT/dt\) with \(c=\alpha T^3\); integrate.

Answer 1

The Correct Option is C

The sphere is black, so the net power lost is \(\sigma\,4\pi R^2\,(T^4 - T_0^4)\).
Heat balance: \(-M\alpha T^3\,\frac{dT}{dt} = 4\pi R^2\sigma\,(T^4 - T_0^4)\).
\(dt = -\frac{M\alpha}{4\pi R^2\sigma}\,\frac{T^3\,dT}{T^4 - T_0^4}\).
\(t = \frac{M\alpha}{4\pi R^2\sigma}\int_{2T_0}^{3T_0}\frac{T^3\,dT}{T^4-T_0^4} = \frac{M\alpha}{4\pi R^2\sigma}\cdot\frac14\ln\left(T^4 - T_0^4\right)\Big|_{2T_0}^{3T_0}\).
\(= \frac{M\alpha}{16\pi R^2\sigma}\ln\frac{81-1}{16-1} = \frac{M\alpha}{16\pi R^2\sigma}\ln\frac{16}{3}\).
Answer: (3) \(\frac{M\alpha}{16\pi R^2\sigma}\ln\left(\frac{16}{3}\right)\)