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Suppose a liquid drop can evaporate using the energy released by the reduction of its own surface energy, so that its temperature stays constant. What is the minimum radius of the drop for which this is possible? Surface tension is T, liquid density is rho, and latent heat of vaporization is L.

1. rho*L/T
2. sqrt(T/(rho*L))
3. T/(rho*L)
4. 2T/(rho*L)

Correct answer: 2T/(rho*L)

Solution

For a thin shell dr, change in surface area dA = 8*pi*r*dr and change in volume dV = 4*pi*r²*dr. Energy from surface = T*8*pi*r*dr; energy needed = L*rho*4*pi*r²*dr. For evaporation to be sustained, T*8*pi*r >= L*rho*4*pi*r², giving r <= 2T/(rho*L). The minimum radius at which this just works (boundary) is r = 2T/(rho*L).

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