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Phenomena on Liquid Surfaces. Help Me Solve This Question?
What work has to be performed to break up 1kg of pure water at 20 Degree Celcius into drops of one micron in diameter at the same temperature? The initial surface of the water is small as compared with the total surface of all the drops and it may be neglected.
1 Answer
- Andrew SmithLv 71 decade agoFavorite Answer
A drop of 1 micrometre diameter has a volume of 4/3 Pi() r^3 = 4/3* PI()* (0.5 * 10 ^ -6)^3
It has an area of 4* pi()*(0.5 * 10 ^ -6)^2
And this takes an amount of work = 0.04 * 4 * pi() * (0.5 * 10 ^ -6) ^2
(Using your value of 0.04 J / m^2 from a previous question)
So find the number of drops required to make 1Kg of water ( 10 ^ -3 m^3)
N= 10 ^ -3 / (4/3* PI()* (0.5 * 10 ^ -6)^3) = 1.9 * 10 ^ 15
Now multiply this by the energy per drop
= N * 0.04 * 4 * pi() * (0.5 * 10 ^ -6) ^2
= 240 J
