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.

Answers (1)

now, \Delta E=\sigma

(final area - initial area) of surface

\Delta E=m s \Delta t

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

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

\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 ]

\Delta t=\frac{3\sigma}{ps}\left [ \frac{1}{r}-\frac{1}{R} \right ] as R>r

Hence, we can say that \Delta t will be positive and therefore, \frac{1}{R} < \frac{1}{r}

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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