Two water drops each of radius 'r' coalesce to from a bigger drop. If ' T ' is the surface tension, the surface energy released in this process is :

\(\begin{aligned} & 2 \times \frac{4}{3} \pi \mathrm{R}^3=\frac{4}{3} \pi \mathrm{r}^3 \Rightarrow \mathrm{r}=2^{1 / 3} \mathrm{R} \\ & \mathrm{U}_{\mathrm{i}}=2 \times 4 \pi \mathrm{R}^2 \mathrm{~T} \\ & \mathrm{U}_{\mathrm{f}}=4 \pi \mathrm{r}^2 \mathrm{~T}=4 \pi \mathrm{R}^2 \mathrm{~T} 2^{2 / 3} \\ & \therefore \text { Heat lost }=\mathrm{u}_{\mathrm{i}}-\mathrm{u}_{\mathrm{f}}=4 \pi \mathrm{R}^2 \mathrm{~T}\left[2-2^{2 / 3}\right]\end{aligned}\)