Question

How much energy must be supplied to change 36 g of ice at 0°C to water at room temperature 25°C

Data for water, $H_2O \Delta H_{fusion}^{\circ}=6.01 kJ mol^{-1}$

$C_{p,liquid}=4.18 J.K^{-1}g^{-1}$

• A. 12 kJ
• B. 16 kJ
• C. 19 kJ
• D. 22 kJ

Answer 1

Answer: (B)

16 kJ

Answer 2

The total energy required is the sum of the energy for melting and the energy for heating.

1. Molar mass of water is approximately 18 g/mol.
2. Moles of water: $n = \frac{36 \text{ g}}{18 \text{ g/mol}} = 2 \text{ mol}$.
3. Energy for melting: $Q_{fusion} = n \times \Delta H_{fusion}^{\circ} = 2 \text{ mol} \times 6.01 \text{ kJ/mol} = 12.02 \text{ kJ}$.
4. Energy for heating water: $Q_{heating} = \text{mass} \times C_{p,liquid} \times \Delta T = 36 \text{ g} \times 4.18 \text{ J/K/g} \times (25-0) \text{ K} = 3762 \text{ J} = 3.762 \text{ kJ}$.
5. Total energy: $Q_{total} = Q_{fusion} + Q_{heating} = 12.02 \text{ kJ} + 3.762 \text{ kJ} = 15.782 \text{ kJ}$. This is closest to 16 kJ.