Theory (for any two spherical drops)

When two spherical drops of radii \(r_1\) and \(r_2\) coalesce to form a single spherical drop of radius \(R\), the total volume is conserved but the surface area decreases. The decrease in surface energy equals the energy released:

\[ \textbf{Volume conservation:}\quad \frac{4}{3}\pi R^3=\frac{4}{3}\pi\left(r_1^3+r_2^3\right)\Rightarrow R=\left(r_1^3+r_2^3\right)^{1/3} \] \[ \textbf{Surface areas:}\quad A_i=4\pi\left(r_1^2+r_2^2\right),\qquad A_f=4\pi R^2 \] \[ \textbf{Energy released:}\quad \Delta E=S\,(A_i-A_f)=4\pi S\left(r_1^2+r_2^2-R^2\right) \]

All radii must be in meters for \(\Delta E\) in joules.

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