Given, Mass of ice, $m = 1 \text{ kg}$ Latent heat of fusion, $L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$ Specific heat of water, $s = 4.2 \times 10^3 \text{ J kg}^{-1} \text{ K}^{-1}$ Initial temperature of ice, $T_0 = 0^\circ \text{C} = 273 \text{ K}$ Final temperature of water, $T_f = 100^\circ \text{C} = 373 \text{ K}$ The total change in entropy ($\Delta S$) consists of two parts: the entropy change during the phase change (melting) and the entropy change during the temperature rise of the water. 1. Entropy change for melting ice at $273 \text{ K}$: $$ \begin{aligned} \Delta S_1 &= \frac{m L_f}{T_0} \\ &= \frac{1 \times 3.36 \times 10^5}{273} \\ &= 1230.769 \text{ J K}^{-1} \end{aligned} $$ 2. Entropy change for heating water from $273 \text{ K}$ to $373 \text{ K}$: $$ \begin{aligned} \Delta S_2 &= \int_{T_0}^{T_f} \frac{m s \text{ d}T}{T} \\ &= m s \ln \left( \frac{T_f}{T_0} \right) \\ &= 1 \times 4200 \times \ln \left( \frac{373}{273} \right) \\ &= 4200 \times 0.3110 \\ &= 1306.200 \text{ J K}^{-1} \end{aligned} $$ Total entropy change: $$ \begin{aligned} \Delta S &= \Delta S_1 + \Delta S_2 \\ &= 1230.769 + 1306.200 \\ &= 2536.969 \text{ J K}^{-1} \end{aligned} $$ The total entropy increase is $2536.969 \text{ J K}^{-1}$.