If a number of little droplets of water, each of radius r, coalesce to form a single drop of radius R, show that the rise in temperature will be given

$\frac{3\mathrm{T}}{\mathrm{J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$

where, T is the surface tension of water and J is the mechanical equivalent of heat.

Let n be the number of little droplets.

Since, volume will remain constant, hence volume of n little droplets = volume of single drop

$\therefore \mathrm{n}\times \frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}{\mathrm{\pi R}}^3$ or ${\mathrm{nr}}^3={\mathrm{R}}^3$

Decrease in surface area $=\mathrm{n}\times 4{\mathrm{\pi r}}^2-4{\mathrm{\pi R}}^2$ or $\mathrm{\Delta A}=4\mathrm{\pi }\left[{\mathrm{nr}}^2-{\mathrm{R}}^2\right]$

$$\begin{gathered} =4\mathrm{\pi }\left[\frac{{\mathrm{nr}}^3}{\mathrm{r}}-{\mathrm{R}}^2\right]=4\mathrm{\pi }\left[\frac{{\mathrm{R}}^3}{\mathrm{r}}-{\mathrm{R}}^2\right] \\ =4{\mathrm{\pi R}}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \end{gathered}$$

Energy evolved $\mathrm{W}=\mathrm{T}\times$ decrease in surface area

$=\mathrm{T}\times 4{\mathrm{\pi R}}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$

Heat produced, $\mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}=\frac{4{\mathrm{\pi TR}}^3}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$

But $\mathrm{Q}=\mathrm{ms}△\theta$

where, m is the mass of big drop, s is the specific heat of water and $d \theta$ is the rise in temperature.

$\therefore \frac{4{\mathrm{\pi TR}}^3}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]=$ Volume of big drop

$\times$ density of water $\times$ specific heat of water $\times △\theta$

or 

$$\begin{gathered} \frac{4}{3}{\mathrm{\pi R}}^3\times 1\times 1\times \mathrm{d\theta }=\frac{4{\mathrm{\pi TR}}^3}{\mathrm{J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right) \\ \text{ or } △\theta =\frac{3\mathrm{T}}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \end{gathered}$$