Two ice pieces of identical mass move towant each other with identical speeds, c

Question

Two ice pieces of identical mass move towant each other with identical speeds, collide absolutely plastically and evaporate. What was the speed of each piece of ice just before collision if their temperature before impact was 0 degree C, while thermal (root mean squared) speed of water molecule after collission was equal to the escape speed from planet. Specific heat of water 4.086 J/g degree C Specific heat of vapor 1.996 J/g degree C, Latent heat water fusion 334 J/g, latent heat of water vaporization 2260 J/g. Radius of earth 6400 km, free fall acceleration near Earth surface g = 9.81 m/s. Present your answer in miles per hour (li/h) with correct amount of significant numbers.

Explanation / Answer

Initial KE of both icecubes = m*v^2
Finally, assume that the KE was used up in evaporating and heating the water
vrms = sqroot(3RT/M) = 11.2*10^3 m/s
now molar mass per kg of water = M
m kg water = m litre water = m1/22.4 mol = m18/22.4*1000
T = m4043.3212 K ( steam)

so, m*33400 + m*2260000 + m*4186*100 + m*19960*(m4043.3212 - 100) = mv^2
v = 1514575.693 miles/s = 420.71547 miles/hr

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