Eight identical drops of mercury each of radius ‘r’ coalesce to form a single drop. If surface tension of mercury is ‘T’, the heat generated is 

$$\begin{gathered} \frac{4}{3}{\mathrm{\pi R}}^3=8\times \frac{4}{3}{\mathrm{\pi r}}^3\Rightarrow \mathrm{R}=2\mathrm{r} \\ \begin{array}{l}\therefore \mathrm{heat}\mathrm{generated}=\mathrm{Lost}\mathrm{energy}\\ =U_{\mathrm{i}}-U_{\mathrm{f}}\\ =8\times 4{\mathrm{\pi r}}^2\mathrm{T}-4{\mathrm{\pi R}}^2\mathrm{T}\\ =16{\mathrm{\pi r}}^2\mathrm{T}\end{array} \end{gathered}$$