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 :

R is the radius of bigger drop.<br><br>r is the radius of n water drops.<br><br>Water drops are combined to make bigger drop.<br><br>So,<br><br>Volume of n drops = volume of bigger drop<br><br>$n\left( {{4 \over 3}\pi {r^3}} \right) = {4 \over 3}\pi {R^3}$<br><br>$\Rightarrow$ $R = r{n^{1/3}} \Rightarrow n = {\left( {{R \over r}} \right)^3}$<br><br>Loss in surface energy, $\Delta$U = T $\times$ (Change in surface area)<br><br>$\Delta$U = T (n4$\pi$r² $-$ 4$\pi$R²)<br><br>$$\Delta U = 4\pi T\left[ {{{\left( {{R \over r}} \right)}^3}{r^2} - {R^2}} \right] = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over J}$$<br><br>$\therefore$ $${{\Delta U} \over V} = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over {J \times {4 \over 3}\pi {R^3}}} = {{3T} \over J}\left[ {{1 \over r} - {1 \over R}} \right]$$