First slide

Heat transfer-Radiation

Question

There are two spherical balls A and B of the same material with same surface, but the diameter of A is half that of B. If A and B are heated to the same temperature and then allowed to cool, then

Moderate

A

rate of cooling is same in both
B

rate of cooling of A is four times that of B
C

rate of cooling of A is twice that of B
D

rate of cooling of A is $\frac{1}{4}$ times that of B

Solution

Rate of cooling, $R_{C} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc} = \frac{Aεσ (T^{4} - T_{0}^{4})}{VρC}$
$\Rightarrow R_{C} \propto \frac{A}{V} \propto \frac{1}{r} \propto \frac{1}{(Diameter)} (∵ m = ρV)$
Since diameter of A is half that of B, its rate of cooling will be twice that of B.

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