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.

If a drop of liquid breaks into smaller droplets, it results in loweri

1.	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.
Answer» Solution :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.