To determine the energy released when 216 small liquid drops coalesce into a larger drop, follow these steps:
Number of small drops
=216Surface area of each small drop
=A=4πr2Surface tension of the liquid
=TCalculations:
Volumes:
Volume of one small drop:
πr3Total volume for 216 small drops:
216×πr3Volume of the large drop:
Since volume is conserved, the volume of the large drop is:
πR3=216×πr3R3=6r3⇒R=6rSurface Areas:
Total surface area of 216 small drops:
216×4πr2=864πr2Surface area of the large drop:
4π(6r)2=144πr2Decrease in Surface Area:
∆A=864πr2−144πr2=720πr2Energy Released:
Using the formula
E=T⋅∆A :
E=T⋅720πr2E=180⋅T⋅4πr2E=180ATThus, the energy released during the coalescence process is
180AT.
