A solid sphere of radius R and a hollow spher | vedclass

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(A) According to Stefan-Boltzmann law,the rate of loss of heat is given by $\frac{dQ}{dt} = \sigma e A (T^4 - T_s^4)$. Since both spheres have the sam

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Hollow sphere cools faster

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(A) According to Stefan-Boltzmann law,the rate of loss of heat is given by $\frac{dQ}{dt} = \sigma e A (T^4 - T_s^4)$. Since both spheres have the same outer radius $R$,their surface areas $A$ are equal. Thus,the rate of heat loss is the same for both.The rate of cooling (rate of change of temperature) is given by $\frac{dT}{dt} = \frac{dQ/dt}{ms} = \frac{\sigma e A (T^4 - T_s^4)}{ms}$,where $m$ is the mass and $s$ is the specific heat capacity.Since the hollow sphere has less mass $(m_{hollow} Therefore,the rate of cooling $\frac{dT}{dt}$ is greater for the hollow sphere. Hence,the hollow sphere cools faster.

$A$ solid sphere of radius $R$ and a hollow sphere of inner radius $r$ and outer radius $R$ made of copper are heated to the same temperature and are allowed to cool in the same environment. Then,choose the $CORRECT$ statement.

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