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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)
$\frac{{M\alpha }}{{4\pi {R^2}\sigma }}\,\ln \left( {\frac{3}{2}} \right)$
$\frac{{M\alpha }}{{4\pi {R^2}\sigma }}\,\ln \left( {\frac{16}{3}} \right)$
$\frac{{M\alpha }}{{16\pi {R^2}\sigma }}\,\ln \left( {\frac{16}{3}} \right)$
$\frac{{M\alpha }}{{16\pi {R^2}\sigma }}\,\ln \left( {\frac{3}{2}} \right)$
Solution
In the given problem, fall in temperature of sphere,
$dT = \left( {3{T_0} – 2{T_0}} \right) = {T_0}$
Tmperature of surrounding, ${T_{surr}} = {T_0}$
Initial temperature of sphere, ${T_{initial}} = 3{T_0}$
Specific heat of the material of the sphere varies as,
$c = \alpha {T^3}\,per\,unit\,mass\,\left( {\alpha = a\,constant} \right)$
Applying formula,
$\frac{{dT}}{{dt}} = \frac{{\sigma A}}{{McJ}}\left( {{T^4} – T_{surr}^4} \right)$
$ \Rightarrow \frac{{{T_0}}}{{dt}} = \frac{{\sigma 4\pi {R^2}}}{{M\alpha {{\left( {3{T_0}} \right)}^3}\,J}}\left[ {{{\left( {3{T_0}} \right)}^4} – {{\left( {{T_0}} \right)}^4}} \right]$
$ \Rightarrow dt = \frac{{M\alpha 27T_0^4J}}{{\alpha 4\pi {R^2} \times 80T_0^4}}$
Solving we get,
Time taken for the sphere to cool down temperature $2{T_{{0^,}}}$
$t = \frac{{M\alpha }}{{16\pi {R^2}\sigma }}In\left( {\frac{{16}}{3}} \right)$
