Question

Let $n$ droplets each of radius $r$ coalesce to form a large drop of radius $R$. Assuming that the droplets are incompressible and $S$ is the surface tension. Calculate the rise in temperature, if $C$ is the specific heat and $\rho$ is the density.

• A. $\frac{3(n-1)}{\rho RC}S$
• B. $\frac{3(n^{1/3}-1)}{\rho RC}S$
• C. $\frac{3(n^{2/3}-1)}{\rho RC}S$
• D. $\frac{3}{\rho RC}(1-\frac{1}{n^{1/3}})S$

Answer 1

The correct option is B $\frac{3(n^{1/3}-1)}{\rho RC}S$

When $n$ droplets coalesce to form one large drop, surface area decreases and energy is released but volume of drops does not change.
Hence,
$\frac{4}{3}\pi r^{3}n=\frac{4}{3}\pi R^3$
$R=n^{1/3}r$
Energy loss $=S\times (4\pi r^{2}n-4\pi R^2)$
This loss increases the temperature of the water.
$mC\Delta T=S\times (4\pi r^{2}n-4\pi R^2)$
$\rho \times \frac{4}{3}\pi R^3\times C\times \Delta T=S\times (4\pi r^{2}n-4\pi R^2)$

$\Delta T=\frac{3[n^{1/3}-1]S}{\rho RC}$