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)

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{{M\alpha }}{{16\pi {R^2}\sigma }}\,\ln \left( {\frac{16}{3}} \right)\)

Step-by-step Solution

Detailed explanation

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 \(2{T_{{0^,}}}\)

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

Same subject

Explore more questions on app

From JEE Mains

Explore more questions on app