8 identical drops of water, each of radius 1 mm coalesce to form a single drop. Then heat generated in the process is [surface tension of water is 0.072 N/m]

43πR3=8×43πr3⇒R=2rUf=Final surface energy = 4πR2.T=4π2r2T=16πr2TUi=Initial surface energy = 8×4πr2T=32πr2T∴ Heat generated = Lost energy =Ui−Uf=32πr2T−16πr2T=16π×10−32×0.072J=1.152πμJ