1. **State the problem:** Calculate the enthalpy change when 50 g of ice at -10°C is heated to water at 30°C. 2. **Given data:** - Mass, $m = 50$ g - Initial temperature of ice, $T_i = -10^\circ C$ - Final temperature of water, $T_f = 30^\circ C$ - Specific heat capacity of ice, $c_{ice} = 2.09$ J/g$^\circ$C - Specific heat capacity of water, $c_{water} = 4.184$ J/g$^\circ$C - Heat of fusion of ice, $\Delta H_{fusion} = 6.01$ kJ/mol - Molar mass of water, $M = 18.015$ g/mol 3. **Process steps:** - Heat ice from -10°C to 0°C - Melt ice at 0°C to water - Heat water from 0°C to 30°C 4. **Calculate heat to warm ice from -10°C to 0°C:** $$q_1 = m \times c_{ice} \times \Delta T = 50 \times 2.09 \times (0 - (-10)) = 50 \times 2.09 \times 10 = 1045 \text{ J}$$ 5. **Calculate heat to melt ice at 0°C:** - Number of moles of ice: $$n = \frac{m}{M} = \frac{50}{18.015} \approx 2.776 \text{ mol}$$ - Heat for fusion: $$q_2 = n \times \Delta H_{fusion} = 2.776 \times 6.01 \times 1000 = 16679.76 \text{ J}$$ 6. **Calculate heat to warm water from 0°C to 30°C:** $$q_3 = m \times c_{water} \times \Delta T = 50 \times 4.184 \times (30 - 0) = 50 \times 4.184 \times 30 = 6276 \text{ J}$$ 7. **Total heat absorbed:** $$q_{total} = q_1 + q_2 + q_3 = 1045 + 16679.76 + 6276 = 24000.76 \text{ J}$$ 8. **Convert total heat to kilojoules:** $$q_{total} = \frac{24000.76}{1000} = 24.001 \text{ kJ}$$ **Final answer:** The enthalpy change is approximately **24.0 kJ**. **Heating curve description:** - From -10°C to 0°C: temperature of ice increases (solid phase) - At 0°C: phase change from ice to water (melting) - From 0°C to 30°C: temperature of water increases (liquid phase)