n droplets of water each of radius r coalesce to form a big drop of radius R. If the energy released during coalescence goes into heating the drop, then the rise in temperature will be (take T as surface tension of water and J as mechanical equivalent of heat)

 

Gain in surface energy, $\Delta U=T\Delta A$  

from volume centenary,   $\frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$  

$\Rightarrow r=\frac{R}{4}$  

Initial surface area, Ai = 4pR²

final surface area,   $A_f=64\times 4\pi r^2$  

  $\therefore \Delta U=12\pi R^2.T=0.075\times 12\times 3.14\times 10^{-4}=2.8\times 10^{-4}J$  

Fact. Due to decrease in intermolecular forces

P? - P? = 4T/a

P? - P? = 4T/b

P? - P? = 4T (1/a - 1/b)

Also, P? - P? = 4T/r

4T/r = 4T (1/a - 1/b)

1/r = (b-a)/ab

r = ab / (b-a)

From volume conservation

$n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R^3=n{r}^3.........\left(1\right)$

Decrease in surface area = $n\times 4\pi r^2-4\pi R^2$

$\left(\Delta A\right)=4\pi \left[n{r}^2-R^2\right]=4\pi \left[\frac{n\times r^3}{r}-R^2\right]=4\pi \left[\frac{R^3}{r}-R^2\right]=4\pi R^3\left[\frac{1}{r}-\frac{1}{R}\right]$

Energy released (W) = $T\times \Delta A=4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]$

Heat produced (Q) = $\frac{W}{J}=\frac{4\pi T{R}^3}{J}\left[\frac{1}{r}-\frac{1}{R}\right]$

$Now,Q=ms\Delta \theta$

$\Rightarrow \frac{4\pi R^{3}T}{J}\left[\frac{1}{r}-\frac{1}{R}\right]=\left(\frac{4}{3}\pi R^3\right)\times \left|x\right|\times \Delta \theta =\frac{3T}{J}\left[\frac{1}{r}-\frac{1}{R}\right]$

 $\frac{4}{3}\pi R^3=729\frac{4}{3}\pi r^3$

R³ = 729r³

$R=\left(729\right)\frac{1}{3}\left(r\right)\left(\frac{1}{3}\right)$

R = 9r

$\Delta u=T\left(4\pi r^2\right)\times 729-T\times 4\pi R^2$

$=T\times 4\pi \times 8R^2$

$\begin{array}{l}=75.39\times 10^{-5}J\\ \Delta u=7.5\times 10^{-4}J\end{array}$