The enthalpy of fusion of ice is 6.020 kJ/mol. The heat capacity of liquid water is 75.4 J/mol °C. What is the smallest number of ice cubes at 0 °C, each containing 1 mol of water, necessary to cool 500. g of liquid water initially at 20 °C to 0 °C?

Question

Quizplus · Accepted Answer

To cool 500 g of water (27.8 mol) from 20 °C to 0 °C, the heat removed is q = m * C * ΔT = 27.8 mol * 75.4 J/mol°C * 20°C = 41,832 J = 41.832 kJ. Each ice cube absorbs 6.020 kJ to melt, so 41.832 kJ / 6.020 kJ/mol = 6.95, rounding up gives 7 ice cubes.

The Enthalpy of Fusion of Ice Is 6