10⁶ droplets of a liquid with the surface tension 0.077 j/m² merge to a single drop. the energy released is approximately what ?
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Let us say r is the radius of each single drop before these coalesce and R is the radius of the new bubble after these coalesce. Since volume of liquid remains same, we have
n×
3
4
πr
3
=
3
4
πR
3
⇒R=n
3
1
r .... (I)
Initial surface energy of single drop =E=(4πr
2
)×T
Initial surface energy of n drops =nE=n(4πr
2
)×T
final surface energy of big single drop =(4πR
2
)×T
=(4π(n
3
1
r)
2
)×T
=n
3
2
(4πr
2
×T)=n
3
2
E from (I)
Clearly
n
3
2
E<nE
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