Question

If a number of little droplets of water, each of radius $r$ , coalesce to form a single drop of radius $R$ , show that the rise in temperature will be given by $\dfrac{{3T}}{J}\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)$ where $T$ is the surface tension of water and $J$ is the mechanical equivalent of heat.

Answer 1

Hint : 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 due to lower energy held by the larger droplets than multiple little droplets due to the lower smaller surface area. This energy will be utilized in increasing the temperature of the water.

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
Rise in temperature due to heat $Q$ : $Q = mC\Delta T$ where $m$ is the mass of water, $C$ is its specific heat, $\Delta T$ is the change in temperature.

Complete step by step answer
We’ve been given that a number of little droplets of water, each of radius $r$ , coalesce to form a single drop of the radius $R$ . When this happens, the water will lose its energy that is stored in the form of surface tension as the total surface area of the water decreases. We can find the change in energy of the water due to surface tension as
$\Rightarrow W = T\Delta A$
The change in the area can be calculated as the difference between the surface area of the bigger drop and the total surface area of $n$ the number of smaller drops.
$\Rightarrow \Delta A = n4\pi {r^2} - 4\pi {R^2}$
Since the volume of water will remain constant, we can also write
$\Rightarrow \dfrac{4}{3}\pi {R^3} = n\dfrac{4}{3}\pi {r^3}$ which on simplifying gives us
$\Rightarrow {R^3} = n{r^3}$
Coming back to the change in the area, we can multiply and divide the first term on the right side by $r$ , we can write
$\Rightarrow \Delta A = \dfrac{{n4\pi {r^3}}}{r} - 4\pi {R^2}$
Since ${R^3} = n{r^3}$ , we can write
$\Rightarrow \Delta A = \dfrac{{4\pi {R^3}}}{r} - 4\pi {R^2}$
$\Rightarrow \Delta A = 4\pi {R^3}\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)$
Then the energy released in mixing these drops will be
$\Rightarrow W = T4\pi {R^3}\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)$
This energy will be released for the mixing of $n$ drops into one drop. Since $J$ is the mechanical equivalent of heat, the heat released in the process will be
$\Rightarrow Q = \dfrac{{T4\pi {R^3}}}{J}\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)$
This heat will be utilized in raising the temperature of the water according to the relation $Q = mC\Delta T$
So, we can calculate the change in temperature as
$\Rightarrow \Delta T = \dfrac{Q}{{mC}}$
$\Rightarrow \Delta T = \dfrac{{\dfrac{{T4\pi {R^3}}}{J}\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)}}{{mC}}$
To find the mass of the water in these droplets, we will use the relation of density and volume of the water as
$\Rightarrow m = \rho V$ where $\rho = 1$ for water and $V = \dfrac{4}{3}\pi {R^3}$ , so we can calculate the change in temperature after substituting the value of the mass of water and specific heat capacity of water $C = 1$ as
$\Rightarrow \Delta T = \dfrac{{3T}}{J}\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)$
Hence we have proved that the temperature rise when a number of little droplets of water, each of radius $r$ , coalesce to form a single drop of the radius $R$ will be $\Delta T = \dfrac{{3T}}{J}\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)$ .

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.