To determine the surface energy released when two identical water drops coalesce to form a larger drop, let's go through the process step by step:

Volume Conservation:

When two droplets, each with a radius $ R $, merge to form a larger droplet, the total volume is conserved. Therefore:

$ 2 \times \frac{4}{3} \pi R^3 = \frac{4}{3} \pi r^3 $

Solving for $ r $, the radius of the larger drop, we find:

$ r = 2^{1/3} R $

Initial Surface Energy ($ U_i $):

The initial surface energy of the two smaller droplets is given by the surface area of each multiplied by the surface tension $ T $:

$ U_i = 2 \times 4 \pi R^2 T $

Final Surface Energy ($ U_f $):

The surface energy of the larger droplet is given by:

$ U_f = 4 \pi r^2 T = 4 \pi R^2 T (2^{2/3}) $

Energy Released:

The energy released in the process is the difference between the initial and final surface energy:

$ \text{Heat lost} = U_i - U_f = 4 \pi R^2 T \left[2 - 2^{2/3}\right] $

This expression gives us the amount of surface energy that is released when two identical water drops combine to form a single larger drop.