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.

Explanation

Volume of big single drop =Amount of energy released
`i.e, 4/3 pi R^(3)=4/3 pir^(3) N`
`or N=R^(3)/r^(3)`
Amount of energy released
`triangleU=" Change in surface area"`
`or triangleU=4pi (R^(2)-Nr^(2)) xx T`
Let all the energy released is used in lowering the temperature. If s is the specific heat of the fluid of drops then the change in temperature will be
`triangle theta=(triangleU)/(ms)=(4pi T(R^(2)-Nr^(2)))/(4/3 pi R^(3)ps)`
where p is the density of the fluid
`therefore triangle theta=(3T)/(ps) (1/R-r^(2)/R^(3)N)`
`or triangle theta=(3T)/(ps) (1/R-(r^(2)R^(3))/(R^(3)r^(3)) =(3T)/(ps) (1/R-1/r)`