A large number of water drops, each of radius $$r$$, combine to have a drop of radius $$R$$. If the surface tension is $$T$$ and mechanical equivalent of heat is $$J$$, the rise in heat energy per unit volume will be:

When multiple drops coalesce, the total surface area decreases, releasing surface energy which is converted into heat energy.

Using conservation of volume for $$n$$ drops: 

$$V = n \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies n = \frac{R^3}{r^3}$$

Finding change in surface area: 

$$\Delta A = A_i - A_f = n(4\pi r^2) - 4\pi R^2 = \left(\frac{R^3}{r^3}\right)4\pi r^2 - 4\pi R^2 = 4\pi R^3 \left(\frac{1}{r} - \frac{1}{R}\right)$$

Finding rise in heat energy per unit volume:

$$\Delta U = T \Delta A = 4\pi R^3 T \left(\frac{1}{r} - \frac{1}{R}\right)$$

$$H = \frac{\Delta U}{J} = \frac{4\pi R^3 T}{J} \left(\frac{1}{r} - \frac{1}{R}\right)$$

$$\frac{H}{V} = \frac{\frac{4\pi R^3 T}{J} \left(\frac{1}{r} - \frac{1}{R}\right)}{\frac{4}{3}\pi R^3} = \frac{3T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$$