The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol · K. The conversion of 50.0 g of ice at 0.00°C to liquid water at 22.0°C requires kJ of heat.

Question

Quizplus · Accepted Answer

First, we need to determine the heat required to melt the ice:
q1 = nΔHfus = (50.0 g ÷ 18.0153 g/mol) × 6.01 kJ/mol = 16.7 kJ
Next, we need to determine the heat required to raise the temperature of the resulting water from 0.00°C to 22.0°C:
q2 = nCΔT = (50.0 g ÷ 18.0153 g/mol) × 75.3 J/mol · K × (22.0°C - 0.00°C) = 21.3 kJ
The total heat required is the sum of q1 and q2:
qtotal = q1 + q2 = 16.7 kJ + 21.3 kJ = 38.0 kJ, which is closest to choice D, 21.3.

The Heat of Fusion of Water Is 6