Question

Twenty seven droplets of water, each

of radius 0.1 mm coalesce into a single

drop. Find the change in surface energy.

Surface tension of water is 0.072 N/m.​

Answer 1

The radius of each water droplet $\sf{=0.1\ mm=0.1\times10^{-3}\ m.}$

Let the radius of the new single drop be R.

Even after coalescing, the volume of the 27 water droplets remain unchanged. Thus,

$\longrightarrow\sf{27\times\dfrac{4}{3}\,\pi\left(0.1\times10^{-3}\right)^3=\dfrac{4}{3}\,\pi R^3}$

$\longrightarrow\sf{R^3=27\left(0.1\times10^{-3}\right)^3\ m^3}$

$\longrightarrow\sf{R=3\times0.1\times10^{-3}\ m}$

$\longrightarrow\sf{R=0.3\times10^{-3}\ m}$

The relation between surface tension and surface energy is,

$\longrightarrow\sf{Surface\ Energy=Surface\ Tension\times Surface\ Area}$

Since the drop is assumed to be spherical, surface energy,

$\longrightarrow\sf{W=4\pi\sigma r^2}$

where $\sf{\sigma=0.072\ N\,m^{-1}}$ is surface tension and r is the radius of the drop.

Total surface energy of the 27 water droplets is,

$\longrightarrow\sf{W_1=27\times0.072\times4\pi\left(0.1\times10^{-3}\right)^2}$

$\longrightarrow\sf{W_1=2.44\times10^{-7}\ J}$

Surface energy of the single drop formed after coalescing is,

$\longrightarrow\sf{W_2=0.072\times4\pi\left(0.3\times10^{-3}\right)^2}$

$\longrightarrow\sf{W_2=0.81\times10^{-7}\ J}$

Hence the change in surface energy is,

$\longrightarrow\sf{\Delta W=W_1-W_2}$

$\longrightarrow\sf{\Delta W=2.44\times10^{-7}\ J-0.81\times10^{-7}\ J}$

$\longrightarrow\underline{\underline{\sf{\Delta W=1.63\times10^{-7}\ J}}}$