8 identical drops of water, each of radius 1 mm coalesce to form a single drop. Then heat generated in the process is [surface tension of water is 0.072 N/m]

$\frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi r^3\Rightarrow R=2r$

$U_f=\text{Final surface energy = 4}\pi {\text{R}}^2.T=4\pi {\left(2r\right)}^{2}T=16\pi r^{2}T$

$U_i=\text{Initial surface energy = 8}\times \text{4}\pi {\text{r}}^{2}T=32\pi r^{2}T$

$\therefore$ Heat generated = Lost energy $=U_i-U_f=32\pi r^{2}T-16\pi r^{2}T$

$=16\pi \times {\left({10}^{-3}\right)}^2\times 0.072J=1.152\pi \mu J$