Question

Calculate the amount of energy needed to break a drop of water \[2mm\] in diameter into ${10^9}$ droplets of equal size, taking the surface tension of water $7.3 \times {10^{ - 2}}N{m^{ - 1}}$.

Answer 1

Hint We will calculate the energy difference between the final and the initial drops and this difference will be the required energy needed to break the single big drop into ${10^9}$ droplets of equal size. The required energy of a single drop due to the surface tension is given by $E = TA$.

Complete Step by step solution
The radius of the drop is given. The initial area of the liquid drop is $\pi {r^2}$ where the diameter $d = 2mm$. Therefore, the area is
$\pi {r^2} = \pi {(1mm)^2} = \pi {(1 \times {10^{ - 3}}m)^2} = \pi \times {10^{ - 6}}{m^2}$.
Now the drop of water is made to break into ${10^9}$ droplets of equal size. This process would keep the volume of water as constant.
$\dfrac{4}{3}\pi {r_{initial}}^3 = {10^9} \times \dfrac{4}{3}\pi {r_{final}}^3$
Let ${r_{final}}$ be $R$, so that the above equation becomes ${r^3} = {10^9}{R^3}$.
$ \Rightarrow r = {10^3}R$, or
$ \Rightarrow R = {10^{ - 3}}r = {10^{ - 3}}(1mm)$
$\therefore R = {10^{ - 6}}m$
The final surface area of all the droplets combined is ${10^9} \times \pi {R^2}$.
$\therefore {10^9} \times \pi {({10^{ - 6}}m)^2} = \pi \times {10^{ - 3}}{m^2}$.
We see that the surface area has increased considerably from its initial value when there was only one big drop.
Now the energy $E$, stored in a drop due to the surface tension is given as
$E = TA$,
where $A$ is the surface area of the drop and $T$ is its surface tension.
Applying this formula in the initial and final situations, we get,
${E_{initial}} = T{A_{initial}}$ and ${E_{final}} = T{A_{final}}$.
$ \Rightarrow {E_{initial}} = T{A_{initial}} = (7.3 \times {10^{ - 2}}N{m^{ - 1}})(\pi \times {10^{ - 6}}{m^2}) = 7.3\pi \times {10^{ - 8}}Nm$
And,
$ \Rightarrow {E_{final}} = T{A_{final}} = (7.3 \times {10^{ - 2}}N{m^{ - 1}})(\pi \times {10^{ - 3}}{m^2}) = 7.3\pi \times {10^{ - 5}}Nm$
The required energy difference is the amount of energy needed to break the initial big drop into many small equal droplets.
This is found by ${E_{final}} - {E_{initial}}$
$ \Rightarrow {E_{final}} - {E_{initial}} = 7.3\pi \times {10^{ - 5}}Nm - 7.3\pi \times {10^{ - 8}}Nm = 22.91 \times {10^{ - 5}}Nm$
$\therefore \Delta E = 2.291 \times {10^{ - 4}}Nm$.
This is the extra energy required to break the drop.

Note We could have approached the problem in a different way. The energy of each individual drop of the final droplet formation could be calculated by the formula $E = TA$ and then it could have been multiplied by the number of droplets which were finally formed, i.e. ${10^9}$. This would give the same required energy difference.