Question

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 by $\frac{3T}{J}(\frac{1}{r}-\frac{1}{R})$ where T is the surface tension of water and J is the mechanical equivalent of heat

Answer 1

Let n be the number of little droplets Since volume will remain constant hence volume of n little droplets = volume of single drop
$\therefore$ $n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$ or $n{r}^3=R^3$
Decrease in surface area = n x $4\pi r^2-4\pi R^2$
or $\Delta A=4\pi [n{r}^2-R^2]=4\pi [\frac{n{r}^3}{r}-R^2]=4\pi [\frac{R^3}{r}-R^2]=4\pi R^3[\frac{1}{r}-\frac{1}{R}]$
Energy evolved W = T x decrease in surface are =T x $4\pi R^3[\frac{1}{r}-\frac{1}{R}]$
Heat produced, $Q=\frac{W}{J}=\frac{4\pi T{R}^3}{J}[\frac{1}{r}-\frac{1}{R}]$ But $Q=msd\theta$
where m is the mass of big drop s is the specific heat of water and $d\theta$ is the rise in temperature
$\therefore$ $\frac{4\pi T{R}^3}{J}[\frac{1}{r}-\frac{1}{R}]=$ volume of big drop x density of water x sp. heat of water x de
or, $\frac{4}{3}\pi R^3\times 1\times 1\times d\theta =\frac{4\pi T{R}^3}{J}(\frac{1}{r}-\frac{1}{R})$ or, $d\theta =\frac{3T}{J}[\frac{1}{r}-\frac{1}{R}]$