## Conceptual Physics (12th Edition)

Calculate the heat required to vaporize 2 kg of ethyl alcohol. Q = mL = (2000 g)(200 cal/g) = 400,000 calories. We know that 1 g of $0^{\circ}C$ ice requires 80 calories to form from 1 g of $0^{\circ}C$ water, so we can form (400000/80) = 5000 grams of ice, or 5 kg. Another way to solve the problem is to note that ethyl alcohol's heat of vaporization (200 cal/g) is 2.5 times water's heat of fusion (80 cal/g). If they both change phase in an isolated system, 2.5 times more ice will change phase. (2.5)(2 kg) = 5 kg.