Question

Calculate the energy released when 1000 small water drops each of same radius `10^(-7)m` coalesce to form one large drop. The surface tension of water is `7.0xx10^(-2)N//m`.

Answer 1

Let r be the radius of smaller drops and R of bigger one. Equating the initial and final volumes, we have
`(4)/(3) pi R^(3) =(1000)((4)/(3) pi r^(3))`
or `R=10 r=(10)(10^(-7)) m`
or `R=10^(-6) m`
Further , the water drops have only one free surface.
Therefore,
`DeltaA=4pi R^(2)-(1000)(4pi r^(2))`
`=4pi [(10^(-6))^(-2)-(10^(3))(10^(-7))^(2)]`
`= -36 pi(10^(-12))m^(2)`
Here, negative sign implies that surface area is decreasing.
Hence, emergy released in the process.
`U=T|DeltaA|`
`=(7xx10^(-2))(36 pi xx10^(-12))J=7.9xx10^(-12) J`