A hollow copper sphere S and a hollow copper | vedclass

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(A) According to Newton's law of cooling,the rate of loss of heat is given by $\frac{dQ}{dt} = \varepsilon A \sigma (T^4 - T_0^4) \approx \varepsilon

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(A) According to Newton's law of cooling,the rate of loss of heat is given by $\frac{dQ}{dt} = \varepsilon A \sigma (T^4 - T_0^4) \approx \varepsilon A \sigma (4T_0^3) \Delta T$.Since the rate of loss of heat is also $\frac{dQ}{dt} = mc \frac{dT}{dt}$,we have $mc \frac{dT}{dt} = \varepsilon A \sigma (4T_0^3) \Delta T$.Thus,the rate of fall of temperature is $\frac{dT}{dt} = \frac{\varepsilon A \sigma 4T_0^3}{mc} \Delta T$.For both the sphere and the cube,the surface area $A$ is the same. However,the volume $V$ of a sphere is less than the volume of a cube for the same surface area. Since the mass $m = \rho V$,the mass of water in the sphere is less than the mass of water in the cube $(m_S Since $\frac{dT}{dt} \propto \frac{1}{m}$,the rate of cooling is faster for the sphere $(S)$ than for the cube $(C)$.Therefore,the temperature of the sphere drops faster than the temperature of the cube,which is represented by the graph where the curve for $S$ is steeper than the curve for $C$.

$A$ hollow copper sphere $S$ and a hollow copper cube $C$,both with negligible thin walls of the same surface area,are filled with water at $90^{\circ}C$ and allowed to cool in the same environment. The graph that correctly represents their cooling is:

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