If a number of identical droplets of water, each of radius r, coalesce to form a single drop of radius R, the resulting rise in the temperature of water is given by (here $\rho$ is the density of water, s its specific heat and $\sigma$ its surface tension)

$\mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}=\frac{4\mathrm{\pi }{\mathrm{TR}}^3}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ ;$\text{ But }\mathrm{Q}=\mathrm{msd\theta }$,where T surface tension

where m is the mass of big drop s is the specific heat of water and dθ is the rise in temperature.

$\therefore \frac{4\mathrm{\pi }{\mathrm{TR}}^3}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]=\text{ volume of big drop }\mathrm{x}\text{ density of water(ρ) }\mathrm{x}\text{ sp. heat of }$$\mathrm{water}\left(s\right)\times ∆\mathrm{\theta }$

$\text{ or, }\frac{4}{3}{\mathrm{\pi R}}^3\times \rho \times s\times \mathrm{d\theta }=\frac{4{\mathrm{\pi TR}}^3}{\mathrm{J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)\text{ or, }\mathrm{d\theta }=\frac{3\mathrm{T}}{J\rho s}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$;  Where J is mechanical equivalent of heat.