Question

If a drop of liquid breaks into smaller droplets, it result in lowering of temperature of the droplets. Let a drop of radius R, breaks into N small droplets each of radius r. Estimate the drop in temperature.

Answer 1

When a big drop of radius R, breaks into n droplets each of radius r, the volume remains constant.
`therefore` Volume of big drop `=N xx ` Volume of each small drop
`(4)/(3)piR^(3)=N xx (4)/(3)pir^(3)`
or `R^(3)=Nr^(3)`
or `N=(R^(3))/(r^(3))`
Now, change in surface area `4piR^(2)-N4pir^(2)`
`=4pi(R^(2)-Nr^(2))`
Energy released `=T-DeltaA=S4pi(R^(2)-Nr^(2))" "`[T=Surface tension]
Due to releasing of this energy, the temperature is lowered.
If `rho` is the density and s is specific heat of liquid and its temperature is lowered by `Deltatheta` , then energy released `=ms Deltatheta" "` [s=specific heat `Deltatheta`=change in temperature]
`Txx4pi(R^(2)-Nr^(2))=((4)/(3)xxR^(3)xxrho)s Deltatheta" "[therefore m=vrho=(4)/(3)piR^(3)rho]`
`implies Deltatheta=(Txx4pi(R^(2)-nr^(2)))/((4)/(3)piR^(3)rhoxxs)`
`=(3T)/(rhos)[(R^(2))/(R^(3))-(Nr^(2))/(R^(3))]`
`=(3T)/(rhos)[(1)/(R)-((R^(3)//r^(3))xxr^(2))/(R^(3))]`
`=(3T)/(rhos)[(1)/(R)-(1)/(r)]`