Question

A number of water droplets each of radius $r$ coalesce to form a droplet of radius $R.$ The rise in temperature is (Surface tension of water =S )

• A. $\frac{2S}{rJ}$
• B. $\frac{3S}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$
• C. $\frac{3S}{rJ}$
• D. $\frac{3S}{J}\left(\frac{1}{r}+\frac{1}{R}\right)$

Answer 1

Answer: (B)

Let $n$ drops coalesce to form big drop. Then
$n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
or $n=\frac{R^3}{r^3}$
As heat produeed $=\frac{\text{ work done }}{J}$
$\therefore ms\Delta T=\frac{\text{ surface tension }\times \text{ decrease in area }}{J}$
or $ms\Delta T=\frac{S}{J}\times \left[n\times 4\pi r^2-4\pi R^2\right]$
or $\frac{4}{3}\pi R^3\times 1\times 1\times \Delta T=\frac{S}{J}\times \left[n\times 4\pi r^2-4\pi R^2\right]$
or $\Delta T=\frac{3S\times 4\pi }{J4\pi R^3}\left[n{r}^2-R^2\right]=\frac{3S}{J{R}^3}\left[\frac{R^3}{r^3}{r}^2-R^2\right]$
$=\frac{3S{R}^3}{J{R}^3}\left[\frac{1}{r}-\frac{1}{R}\right]=\frac{3S}{J}\left[\frac{1}{r}-\frac{1}{R}\right]$