Two solid spheres A and B made of the same ma | vedclass

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(B) According to the Stefan-Boltzmann law,the rate of heat loss $dQ/dt$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$.For a sphere,$A = 4 \pi r^2$ an

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$r_B/r_A$

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(B) According to the Stefan-Boltzmann law,the rate of heat loss $dQ/dt$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$.For a sphere,$A = 4 \pi r^2$ and $dQ = mc(dT) = (\rho V c) dT = \rho (4/3 \pi r^3) c dT$.Thus,the rate of change of temperature is given by: $\rho (4/3 \pi r^3) c (-dT/dt) = \sigma (4 \pi r^2) (T^4 - T_0^4)$.Simplifying this,we get: $(-dT/dt) = [3 \sigma / (\rho c r)] (T^4 - T_0^4)$.Since the material is the same,$\rho$ and $c$ are constant. For the same temperature difference,$(-dT/dt) \propto 1/r$.Therefore,the ratio of the rate of change of temperature of $A$ and $B$ is: $(dT/dt)_A / (dT/dt)_B = r_B / r_A$.

Two solid spheres $A$ and $B$ made of the same material have radii $r_A$ and $r_B$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature of $A$ and $B$ is :

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