The heat capacity of liquid water is 75.6 J/K-mol, while the enthalpy of fusion of ice is 6.0 kJ/mol — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Heat lost by water = $n C_p \Delta T = (500/18) \times 75.6 \times 20 = 42,000 \text{ J} = 42 \text{ kJ}$. Heat absorbed by melting ice = $n_{ice} \times 6.0 \text{ kJ/mol}$. Thus, $n_{ice} = 42 / 6.0 = 7 \text{ moles}$. Mass of ice = $7 \times 18 = 126 \text{ g}$. Number of 9.0 g cubes = $126 / 9.0 = 14$.