If a number of little droplets of water, each of radius `r`, coalesce to form a single drop of radius `R`, show that the rise in temperature will be given by `(3T)/J(1/r-1/R)` where `T` is the surface tension of water and `J` is the mechanical equivalent of heat.

Explanation

suppose `n` small water drop combine and form a big drop. During this process so valume will also be conserved
`("Volume")_("initial") = ("Volume")_("final")`
`((4)/(3)pir^(3)) xx n = (4)/(3)piR^(3) rArr n = (R^(3))/(r^(3))`
Loss in surface energy
`DeltaU_(loss) = TDeltaA_(ions) = T(4pir^(3)xxn - 4piR^(2))`
Put `n = (R^(3))/(r^(3))`
get `DeltaU_(loss) = T(4pir^(2) xx (R^(3))/(r^(3)) - 4piR^(2))`
`DeltaU_(loss) = T4piR^(3)((1)/(r) - (1)/(R))`
`T(4piR^(3))((1)/(r) - (1)/(R)) = ms Deltatheta` when `m = rho(vol)`
`T(4piR^(3))((1)/(r) - (1)/(R)) = (1 gm//cm^(3))((4)/(3)piR^(3))(1 cal//gm^(@)C)DeltaT`
get `Deltatheta = (3T)/(J)[(1)/(r) - (1)/(R)]`