Question

If a number of liquid droplets each of radius(r) coalesce to form a single drop of radius (R), then find the rise in temperature of the liquid in terms of the surface tension (T), density (ρ) and specific heat capacity (c) of the liquid.

Answer 1

$\Delta T = \frac{3T}{\rho c} \left(\frac{1}{r} - \frac{1}{R}\right)$

Answer 2

Let $n$ be the number of small droplets of radius $r$ coalescing into a single large droplet of radius $R$. By conservation of volume: $n \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow n = \left(\frac{R}{r}\right)^3$

The decrease in surface area is: $\Delta A = n \cdot 4\pi r^2 - 4\pi R^2 = \left(\frac{R}{r}\right)^3 \cdot 4\pi r^2 - 4\pi R^2 = \frac{4\pi R^3}{r} - 4\pi R^2 = 4\pi R^3 \left(\frac{1}{r} - \frac{1}{R}\right)$

The surface energy released is $W = T \cdot \Delta A = T \cdot 4\pi R^3 \left(\frac{1}{r} - \frac{1}{R}\right)$. This energy is converted into heat $Q = m \cdot c \cdot \Delta T$. The mass of the liquid is $m = \rho \cdot V_{final} = \rho \cdot \frac{4}{3}\pi R^3$. So, $Q = \rho \cdot \frac{4}{3}\pi R^3 \cdot c \cdot \Delta T$.

Equating $W$ and $Q$: $T \cdot 4\pi R^3 \left(\frac{1}{r} - \frac{1}{R}\right) = \rho \cdot \frac{4}{3}\pi R^3 \cdot c \cdot \Delta T$ $\Delta T = \frac{3T}{\rho c} \left(\frac{1}{r} - \frac{1}{R}\right)$