A solid sphere and a hollow sphere of the sam | vedclass

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(A) The rate of cooling is given by the formula: $\frac{d	heta}{dt} = \frac{P}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.Since both sphere

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The hollow sphere will cool at a faster rate for all values of $T$

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(A) The rate of cooling is given by the formula: $\frac{d	heta}{dt} = \frac{P}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.Since both spheres have the same material,size (surface area $A$),and are in the same surroundings,the rate of heat loss $P$ is the same for both.The rate of cooling $\frac{d	heta}{dt}$ is inversely proportional to the mass $m$ of the object: $\frac{d	heta}{dt} \propto \frac{1}{m}$.Since the solid sphere has a larger mass $(m_{solid} > m_{hollow})$,its rate of cooling will be lower.Therefore,the hollow sphere,having a smaller mass,will cool at a faster rate for all values of $T$.

$A$ solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in the same surroundings. If the temperature difference between each sphere and its surroundings is $T$,then

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