Asked by Anonymous

How many grams of water can be cooled from 31 C to 20 C by the evaporation of 37 grams of water? (The heat of vaporization of water in this temperature range is 2.4 kJ/g. The specific heat of water is 4.18 J/g*K.
Answered by DrBob222
heat water cooled = heat water lost.
mass x specific heat water x delta T = delta H x 37 g H2O.
mass x 4.18 x (31-20) = 2.4 x 10^3 x 37
solve for mass.
Check my thinking.
Answered by Anonymous
1931
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