If a drop of liquid breaks into smaller droplets,

Volume of the liquid of drop of radius R = (N)(volume of liquid droplet of radius r)

$\frac{4}{3} \pi R^{3}=N \times \frac{4}{3} \pi r^{3}$

$N=\frac{R^{3}}{r^{3}}$

Energy released is:

$\Delta U=T \times \Delta A=T\left[4 \pi R^{2}-N\left(4 \pi r^{2}\right)\right]=4 \pi T\left(R^{2}-N r^{2}\right)$

When the energy is released, the temperature reduces.

$\Delta T=\frac{\Delta U}{m c}=\frac{4 \pi T\left(R^{2}-N r^{2}\right)}{\frac{4}{3} R^{3} \rho c}$

Where c is the specific heat of the liquid.

Solving the above equation, it can be said that ∆T will be negative and therefore, the temperature of the droplet falls.