- (D) 5.00 g ice to water = 5.00 × 80 cal = 400 cal. 1 g at 4° can go to 100°C
as water and absorb 1 cal/°C. Then 400 cal − (100° − 4°) = 400 − 96 =
304 cal. 304 cal can change or 0.56 g of water to steam. There
obviously is not enough heat to vaporize all the water. If done in joules, 5.00 g
of ice to water = 5.00 × 3.34 × 10^2 g/J = 1.67 × 10^3 J.
1 g at 4°C to 100°C = Δtemp = 96°
96° × 4.18 J/g/°C = 17.3 J/g = 1.73 × 10^1 J = 0.0173 × 10^3 J
1.67 × 10^3 J − 0.0173 × 10^3 J = 1.65 × 10^3 J left for vaporization. Since
1 g requires 2.26 × 10^3 J for vaporization, (D) is the answer.
