Question

Calculate the energy evolved when 8 droplets of water (surface tension `0.072 Nm^(-1)`) of radius `1.2 mm ` each combine into one.

Answer 1

Here, `S=0.072 Nm^(-1)`
`r=0.5 mm = 0.5 xx 10^(-3)m`.
Let R be the radius of big drop formed.
As, volume of the big drop = ` "volume of 8 small drops"`
`4/3 pi R^(3) = 8xx4/3 pi r^(3)`
or `R=2 r = 2 xx 0.5 xx 10^(-3) = 10^(-3)m`,
Surface area of big drop = `4 pi r^(2)`
`= 4 pi xx (10^(-3))^(2)`
`= 4 pi xx 10^(-6) m^(2)`
Surface area of 8 small drops = `8 xx 4 pi r^(2)`
`= 8 xx 4 pi (0.5 xx 10^(-3))^(2)`
`= 8 pi xx 10^(-6)m^(2)`
Decrease in surface area
` 8 pi xx 10^(-6) - 4 pi xx 10^(-6)`
`=4 pi xx 10^(-6)m^(2)`
Energy evolved = S.T = decrease in area
`=0.072 xx 4 pi xx 10^(-6)`
`=9.05 xx 10^(-7) J`