Question

Two balls of same material and same surface finish have their diameters in the ratio $1:2$. They are heated to the same temperature and are left in a room to cool by radiation, then the initial rate of loss of heat:

• A. Will be same for the balls
• B. For larger ball is half that of other ball
• C. For larger ball is twice that of other ball
• D. For larger ball is four times that of the other ball

Answer 1

The correct option is D For larger ball is four times that of the other ball

As we know,
By Stefan - Boltzmann's law
$\frac{\Delta Q}{\Delta t}=ϵ\sigma A(T^4-T_0^4)$
Given, balls are of same material and both are at same temperature.
Hence, $\sigma ,ϵ$ & $T$ will be same for both balls.
Thus,
$\frac{\Delta Q}{\Delta t}\propto A$
$\frac{\Delta Q}{\Delta t}\propto r^2$
Let, $r_1$ is the radius of small ball.
$r_2$ is the radius of larger ball.
Given: $\{\frac{r_1}{r_2}=\frac{1}{2}\}$
So, $\frac{\frac{\Delta Q}{\Delta t_1}}{\frac{\Delta Q}{\Delta t_2}}={\left(\frac{r_1}{r_2}\right)}^2={\left(\frac{r_1}{2r_1}\right)}^2=\frac{1}{4}$
Hence, initial rate of loss of heat for larger ball will be four times that of the smaller ball.