A number of water droplets each of radius r coalesce to form a droplet of radius R. The rise in temperature is (S.T. of water =T, mechanical equivalent of heat =J).

 Let R be the radius of a single drop. Then
$$\begin{gathered} \frac{4}{3}{\mathrm{\pi R}}^3=\mathrm{n}\times \frac{4}{3}{\mathrm{\pi r}}^3\text{ or }{\mathrm{R}}^3={\mathrm{nr}}^3 \\ \mathrm{Decrease} \mathrm{in} \mathrm{surface} \mathrm{area} \\ =\left(4{\mathrm{\pi r}}^2\mathrm{n}-4{\mathrm{\pi R}}^2\right) \\ =4\mathrm{\pi }\left[{\mathrm{nr}}^2-{\mathrm{R}}^2\right]=4\mathrm{\pi }\left[\frac{{\mathrm{nr}}^3}{\mathrm{r}}-{\mathrm{R}}^2\right] \\ =4\mathrm{\pi }\left[\frac{{\mathrm{R}}^3}{\mathrm{r}}-{\mathrm{R}}^2\right]=4{\mathrm{\pi R}}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\ \text{ Heat produced }=\frac{4{\mathrm{\pi R}}^3\mathrm{T}}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\ \mathrm{Rise} \mathrm{in} \mathrm{temperature} \mathrm{d}0 \mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{4}{3}{\mathrm{\pi R}}^3\times 1\times \mathrm{d\theta }=\frac{4{\mathrm{\pi R}}^3\mathrm{T}}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\ \mathrm{d\theta }=\frac{3\mathrm{T}}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \end{gathered}$$