How many grams of water can be cooled from 41 ∘c to 19 ∘c by the evaporation of 62 g 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.) ?

62 g of water are vaporized and the energy required is 2.4 kJ/g

So 62g x 2.4 kJ/g = 148.8 kJ or 148,800 Joules

Q = mCΔT

Q is energy in joules, m is mass of water, C is the specific heat, delta T is change in temp

148,800 = m (4.18) (41 - 19) = 1618g or 1.6 kg of water