InterviewSolution
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InterviewSolution
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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. |
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Answer» Volume of a big drop of radius R = N xx volume of small drop let of radius r `:. (4)/(3) pi R^3 = N xx(4)/(3)pi r^3 or R^3 = N r^3 or N = R^3//r^3` Change in surface area `Delta U = ST xx` change in surface area `= Txx 4pip[R^2 - N r^2]` All this energy relased is at the cost of loweiring the temperature. Mass of the big drop of liquid, `m =(4)/(3) pi R^3 rho` Let s be the specific heat of liquid drop, `Delta theta` the decrease in temperature. Then `Delta theta = (Delta U)/(ms) = (Txx4pi(R^2 - Nr^2))/((4)/(3) pi R^3 rho)s = (3T)/(rho s) [(1)/(R) - (N r^2)/(R^3)] = (3T)/(rho s)[(1)/(R) - (R^3)/(r^3) xx(r^2)/(R^3)]` `= (3T)/(rho)[(1)/(R) - (1)/(r)]` |
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