As water droplets combine to form a larger drop,

When $n$ small droplets of radius $r$ combine to form a single large drop of radius $R$, the total volume remains constant.

$\dfrac{4}{3}\pi R^3 = n \times \dfrac{4}{3}\pi r^3 \Rightarrow R = n^{1/3}r$

The initial total surface area is $A_i = n \times 4\pi r^2$.

The final surface area is $A_f = 4\pi R^2 = 4\pi n^{2/3}r^2$.

Since $n > n^{2/3}$ for $n > 1$, the total surface area decreases ($A_f < A_i$).

Surface energy is directly proportional to the surface area ($U = T \times A$, where $T$ is surface tension). As the surface area decreases, the total surface energy of the system decreases.

This loss in surface energy is liberated to the surroundings.