Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 1 / Section B / 10 / i / a / Solution

For identical spherical drops, total oil volume and interfacial area satisfyTaking $10\mathrm{vol}\%$ to mean the oil fraction of the final emulsion, $V_o/(V_o+V_w)=0.10$ gives $V_o=V_w/9=1.111\times 10^{-4}{\mathrm{m}}^3$. Thus $A=333.3{\mathrm{m}}^2$ and $E_s=16.7\mathrm{J}$. This corresponds to about $2.65\times 10^{13}$ drops.

There is a wording convention worth making explicit: if the oil addition instead means $0.10$ times the initial water volume, then $V_o=100\mathrm{mL}$, $A=300{\mathrm{m}}^2$ and $E_s=15.0\mathrm{J}$. These differ only in the stated volume-fraction convention. The calculation uses the oil–water interface and neglects the container and air interfaces.