A large number of water drops, each of radius r , combine to have a...

$n \times \dfrac{4}{3}\pi{r}^{3} = \dfrac{4}{3}\pi{R}^{3}$

$∴ {n}^{1 / 3}r = R$

$∴$ Total change in surface energy

$= \left(n\left(4\pi{r}^{2}\right) - 4\pi{R}^{2}\right)T$

$⇒ 4\piT\left(n{r}^{2} - {R}^{2}\right)$

$∴$ Heat energy

$= \dfrac{4\piT\left(n{r}^{2} - {R}^{2}\right)}{J \times \dfrac{4}{3}\pi{R}^{3}} = \dfrac{3T}{J}\left(\dfrac{n{r}^{2}}{{R}^{3}} - \dfrac{1}{R}\right)$

Put $n{r}^{3} = {R}^{3}$

$∴ \dfrac{3T}{J}\left(\dfrac{1}{r} - \dfrac{1}{R}\right)$