Given, radius of small drop =r Radius of big drop =R Surface tension =T and mechanical equivalent of heat =J As, small drops combine to form big drop. ∴ Volume of big drop (VB)=n× Volume of small drop (VS) ⇒
4
3
πR3=n⋅
4
3
πr3 ⇒nr3=R3 ⇒r=
R
n1∕3
. . . (i) Surface energy of small drop (ES)= Surface tension (T)× Area (A) ⇒ES=n×4πr2T and EB=4πR2T Now, change in energy will be ∆E=EB−ES=4πT(nr2−R2) ∴ Heat energy per unit volume =