Question

8 spherical droplets, each of radiua 'r' of a liquid of density 'p' and surface tension 'T' coalesce to form a big drop. If 's' be the specific heat capacity of the liquid, find the rise in temperature of the liquid in this process.

Answer 1

Volume of 8 spherical droplets after merging = 8 * 4π/3 * r³
                  = 4π/3 * (2r)³
So when the droplets merge the radius becomes 2r.

density of liquid = d,     specific heat = s
Surface tension = T = surface energy / area.
Surface energy = T * area

Surface energy of the 8 spherical droplets = 8 * 4πr² * T
Surface energy of the big droplet = 4π (2r)² * T

The additional Surface Energy is used in increasing the big droplet's temperature      = 16π r² T
Mass of the drop = m = 32π/3 r³ * d

Increase in temperature = energy/m*s = 16πr² T/ [32π/3  r³ d s]
         = 3T / [2 r d s]

We find that the gravitational potential energy before merging is not exactly equal to the gravitational PE after merging into the big drop.
   gravitational PE before merging = 8 * (4π/3 * r³ * d) g * r = 32 π/3 * r⁴dg
   gravitational PE after merging = (4π/3 * (2r)³ * d) * g * (2r) = 64 π/3* r⁴dg
We ignored this assuming that it is not significant.

In case we consider that too,
   increase in temperature = [16πr² T - 32 π/3 r⁴ dg]/[32 π/3 r³ d s]
             = [ 3/2 T  -  r² d g ] / [ r d s]