Q.Assuming the water vapour to be a perfect gas, calculate the internal energy change when 1 mol of water at 100 °C and 1 bar pressure is converted to ice at 0 °C. Given the enthalpy of fusion of ice is 6.00 kJ mol$^{-1}$ and heat capacity of water is 4.2 J/g °C.

Concept: State functions and step-wise enthalpy — break the overall change into steps and use the state-function property of $H$ and $U$.

Step 1: Cool water from 100 °C to 0 °C

Mass of 1 mol water = 18 g.

$q_1 = m C \Delta T = 18 \times 4.2 \times (-100) = -7560\ \text{J} = -7.56\ \text{kJ}$

At constant pressure, $q_p = \Delta H$, so $\Delta H_1 = -7.56\ \text{kJ}$.

Step 2: Freeze water at 0 °C

$\Delta H_2 = -\Delta H_{\text{fusion}} = -6.00\ \text{kJ}$ (exothermic).

Step 3: Total enthalpy change

$\Delta H = \Delta H_1 + \Delta H_2 = -7.56 - 6.00 = -13.56\ \text{kJ}$. …

Internal energy is a state function, so the overall change equals the sum of the two steps. Because no gas is produced or consumed at any stage, $\Delta n_g = 0$ throughout and the $\Delta(PV)$ term is negligible, giving $\Delta U \approx \Delta H$ for each step.

Step 1 — Cool water from $100\,^\circ\text{C}$ to $0\,^\circ\text{C}$

Molar heat capacity of water:

Step 2 — Freeze water at $0\,^\circ\text{C}$

Freezing is the reverse of fusion, so

Total change

$\Delta H_\text{total} = \Delta H_1 + \Delta H_2 = -7.56 + (-6.00) = -13.56\ \text{kJ mol}^{-1}.$ …