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 the mechanical equivalent of heat is J, then the rise in temperature will be

Conserving volume,n×43πr3=43πR3⇒nr3=R3 Decrease in surface area =n×4πr2−4πR2⇒ΔA=4πR3(1r−1R) Energy released=W=T× decrease in surface area=T×4πR3(1r−1R) Heat produced, Q=WJ=4πTR3J(1r−1R)Also,Q=msΔT=43πR3×1×1×ΔT∴43πR3×1×1×ΔT=4πTR3 J(1r−1R)⇒ΔT=3 T J(1r−1R)