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

Explanation

Radius of larger drop =R
Radius of smaller drop =r
Volume of large drop =Volume of small drops N
`(4)/(3)piR^(3)=N((4)/(3)pir^(3))`
`thereforeR^(3)=Nr^(3)`
`N=(R^(3))/(r^(3))` ….(1)
Change in area ,
`DeltaA` = area of large drop - area of small drops N
`=4piR^(2)-N(4pir^(2))`
`=4Deltapi(R^(2)-Nr^(2))`
`therefore` Energy relased `=SDeltaA`
`=S4pi(R^(2)-Nr^(2))`....(2)
(Where S =surface tension)
Due to this energy relased ,temperature decrease in temperature is `DeltaQ`.
Then energy released,
`Q=CmDeltatheta`
`=C((4)/(3)piR^(3)rho)Deltatheta` ...(3)
(Where m = volume V `xx` density `rho=(4)/(3)piR^(3)rho)`
Equating equation (1) and (2), `Sxx4pi(R^(2)-Nr^(2))=((4)/(3)piR^(3)rho)CxxDeltatheta`
`thereforeDeltatheta=(Sxx4pi(R^(2)-Nr^(2)))/(((4)/(3)piR^(3)rho)C)`
`Deltatheta=(3S)/(rhoC)[(R^(2)-Nr^(2))/(R^(3))]`
`Deltatheta=(3S)/(rhoC)[(1)/(R)-(Nr^(2))/(R^(3))]`
Putting the value of N from equation (1),
`Deltatheta=(3S)/(rhoC)[(1)/(R)-(R^(3))/(r^(3))((r^(2))/(R^(3)))]`
`=(3S)/(5C)[(1)/(R)-(1)/(r)]` ....(4)
decrease in temperature.