If a drop of liquid breaks into smaller droplets,it results in the lowering of the temperature of the droplets. Let a drop of radius $R$ break into $N$ small droplets,each of radius $r$. Estimate the drop in temperature.

(N/A) The volume of the liquid remains constant during the process.
Volume of large drop = $N \times$ Volume of small drop
$\frac{4}{3} \pi R^3 = N \times \frac{4}{3} \pi r^3 \implies R^3 = N r^3 \implies N = \frac{R^3}{r^3}$.
The change in surface area $\Delta A$ is given by the final surface area minus the initial surface area:
$\Delta A = N(4 \pi r^2) - 4 \pi R^2 = 4 \pi (N r^2 - R^2)$.
Since $N = R^3/r^3$,we have $\Delta A = 4 \pi (\frac{R^3}{r^3} r^2 - R^2) = 4 \pi R^2 (\frac{R}{r} - 1)$.
The energy released due to the increase in surface area is $E = S \Delta A = 4 \pi S R^2 (\frac{R}{r} - 1)$,where $S$ is the surface tension.
This energy is absorbed from the internal energy of the liquid,causing a temperature drop $\Delta \theta$.
$E = m C \Delta \theta$,where $m = \rho V = \rho (\frac{4}{3} \pi R^3)$ and $C$ is the specific heat capacity.
Equating the energy: $4 \pi S R^2 (\frac{R}{r} - 1) = \rho (\frac{4}{3} \pi R^3) C \Delta \theta$.
Solving for $\Delta \theta$: $\Delta \theta = \frac{3 S}{\rho C R} (\frac{R}{r} - 1) = \frac{3 S}{\rho C} (\frac{1}{r} - \frac{1}{R})$.