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 of ice, $m = 50$ 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. **Approach:** The heating process involves three steps: a. Heating ice from -10°C to 0°C b. Melting ice at 0°C to water c. Heating water from 0°C to 30°C 4. **Step 1: Heating ice from -10°C to 0°C** Use formula $q = m c \Delta T$ $$q_1 = 50 \times 2.09 \times (0 - (-10)) = 50 \times 2.09 \times 10 = 1045 \text{ J}$$ 5. **Step 2: Melting ice at 0°C** Calculate moles of ice: $$n = \frac{m}{M} = \frac{50}{18.015} \approx 2.776 \text{ mol}$$ Heat required for fusion: $$q_2 = n \times \Delta H_{fusion} = 2.776 \times 6.01 = 16.68 \text{ kJ} = 16680 \text{ J}$$ 6. **Step 3: Heating water from 0°C to 30°C** $$q_3 = m c_{water} \Delta T = 50 \times 4.184 \times (30 - 0) = 50 \times 4.184 \times 30 = 6276 \text{ J}$$ 7. **Total enthalpy change:** $$q_{total} = q_1 + q_2 + q_3 = 1045 + 16680 + 6276 = 24001 \text{ J} = 24.001 \text{ kJ}$$ **Final answer:** The enthalpy change is approximately **24.0 kJ**. 8. **Heating curve description:** - From -10°C to 0°C, temperature of ice rises (sensible heat) - At 0°C, phase change from ice to water occurs (latent heat) - From 0°C to 30°C, temperature of water rises (sensible heat)