Question

A single drop of water with radius $R$ is formed from coalesce of many identical little droplets of radius $r$ having serface tension $T$ . The rise in temperature in this process is given by $\frac{XT}{\rho J}\left(\frac{1}{r} - \frac{1}{R}\right),$ where $J$ is the mechanical equivalent of heat and $\rho $ is density of water.
Find $X$ .

Answer 1

Let there be $'n'$ small drops each of radius $r$ which coalesce to form a single drop of radius $R$ .
Then
$\frac{4}{3}\pi r^{3}\times n=\frac{4}{3}\pi R^{3}$
$\Rightarrow n=\left(\frac{R}{r}\right)^{3}$
Initial surface area of the 'n' small drops
$=4\pi r^{2}\times n$
$=4\pi r^{2}\times \left(\frac{R}{r}\right)^{3}=\frac{4 \pi R^{3}}{r}$
Final surface area of the big drop formed $=4\pi R^{2}$
Decrease in surface area,
$\Delta A=\frac{4 \pi R^{3}}{r}-4\pi R^{2}=4\pi R^{3}\left(\frac{1}{r} - \frac{1}{R}\right)$
Due to surface tension of the liquid, decrease in area of the liquid surface results in the release of
energy.
The energy released,
$W=T\times \Delta A=4\pi R^{3}T\left(\frac{1}{r} - \frac{1}{R}\right)$
If the energy released gets converted into heat, then heat produced,
$Q=\frac{W}{J}=\frac{4 \pi R^{3} T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$ $....\left(\right.i\left.\right)$
Let $\theta $ be the rise in temperature of the drop.
$\therefore Q=$ mass $\times $ specific heat $\times $ rise in temperature
$=\frac{4}{3}\pi R^{3}\times \rho \times c\times \theta $
For water $c=1calg^{- 1}^\circ C^{- 1}$
$\therefore Q=\frac{4}{3}\pi R^{3}\rho \times 1\times \theta $ $....\left(\right.ii\left.\right)$
From equations $\left(\right.i\left.\right)$ and $\left(\right.ii\left.\right)$ ,
$\frac{4}{3}\pi R^{3}\rho \theta =\frac{4 \pi R^{3} T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$
$\therefore \theta =\frac{3 T}{\rho J}\left(\frac{1}{r} - \frac{1}{R}\right)\Rightarrow X=3$