Question

A water drop of radius 1 mm is broken into ${10^6}$ identical drops, surface tension of water is 72 dynes/cm find the energy spent in this process.

Answer 1

In this solution, we will use the law of conservation of energy to find the rise in temperature. The decrease in energy of the droplets will due to the smaller droplets having a larger surface area when summed up.

Formula used: In this solution, we will use the following formula
Change in energy due to change in the area of water droplets: $W = T\Delta A$ where $T$ is the tension of the water surface and $\Delta A$ is the change in the area.

Complete step by step answer
We’ve been given that droplets of water of radius 1 mm, break to form multiple smaller droplets. When this happens, the water will spend its energy that is stored in the form of surface tension as the total surface area of the water increases. We can find the change in energy of the water due to surface tension as
$\Rightarrow W = T\Delta A$
To find the surface area, we need to find the radius of the smaller drops. Let $R = 1\,mm$ be the radius of the larger drop and $r$ be the radius of the smaller droplets. Since the volume of water will remain constant, we can write that
$\Rightarrow \dfrac{4}{3}\pi {R^3} = n\dfrac{4}{3}\pi {r^3}$ where $n = {10^6}$ is the number of smaller droplets.
$\Rightarrow {R^3} = \,{10^6}\,{r^3}$
Solving for $r$, we get
$\Rightarrow r = \,\dfrac{R}{{{{10}^2}}}$
Then the change in surface area will be
$\Rightarrow \Delta A = n4\pi {r^2} - 4\pi {R^2}$
Substituting $r = \,\dfrac{R}{{{{10}^2}}}$ in the above equation, we get
$\Rightarrow \Delta A = 4\pi {R^2}({10^2} - 1)$ and as $R = 1\,mm = 0.1\,cm$, we get
$\Rightarrow \Delta A = 4\pi {10^{ - 6}} \times (99)$
Now we can calculate the work done as
$\Rightarrow W = T\Delta A$
$\Rightarrow W = 72 \times 4\pi {10^{ - 6}} \times (99)$
Hence the work done will be
$\Rightarrow W = 895.2768\,{\text{dyne}}$.

Note
Here we have assumed that all the energy released when the drops merge will go in raising the temperature of the water however, in reality, some other internal and external barriers have to be overcome which will utilize the energy released while mixing. Also due to turbulence in the motion of water, the volume of water does not remain constant when mixing of drops due to spraying of water but for an ideal case, we can assume that the volume remains constant.