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

now,

(final area - initial area) of surface

As per the law of conservation of mass,

Final volume =initial volume of one drop which has a radius R which is then split into N drops r

So, $\frac{4}{3}\pi R^{3}=N.\frac{4}{3}\pi r^3$

$\Delta t=\frac{4\pi \sigma \times 3}{N.4\pi r^{3}ps}\left [ Nr^{2}-R^{2} \right ]$

$\Delta t=\frac{3\sigma}{Nps}\left [ \frac{Nr^{2}}{r^{3}}-\frac{R^{2}}{r^{3}} \right ]$

$\Delta t=\frac{3\sigma N}{Nps}\left [ \frac{1}{r}-\frac{1}{R} \right ]$

as R>r

Hence, we can say that  will be positive and therefore,

So, the temperature increases as the formation smaller drops take place, and this change takes place through absorption of energy through the surroundings.

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