A large number of water drops, each of radius \(r\), combine to have a drop of radius \(R\). If the surface tension is \(T\) and mechanical equivalent of heat is \(J ,\) the rise in heat energy per unit volume will be

\(\therefore n ^{1 / 3} r = R\)

\(\therefore\) Total change in surface energy

\(=\left(n\left(4 \pi r^{2}\right)-4 \pi R^{2}\right) T\)

\(\Rightarrow 4 \pi T \left( nr ^{2}- R ^{2}\right)\)

\(\therefore\) Heat energy

\(=\frac{4 \pi T \left( nr ^{2}- R ^{2}\right)}{ J \times \frac{4}{3} \pi R ^{3}}=\frac{3 T }{ J }\left(\frac{ nr ^{2}}{ R ^{3}}-\frac{1}{ R }\right)\)

Put \(nr ^{3}= R ^{3}\)

\(\therefore \frac{3 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)\)