$5.9.2.$ Find the entropy increment of $1\text{ kg}$ of ice as it melts.
The ice melts at the constant temperature. Using the expression for the entropy:$$\Delta S = \int_{1}^{2}\frac{\delta Q}{T}$$
Considering the constant melting temperature $T$
$$\Delta S = \frac{1}{T}\int_{1}^{2}\delta Q = \frac{Q}{T}$$
For melting, $Q = \lambda m$, where $\lambda$ is a heat of fusion of ice, $Q$ is the heat the ice absorbs. This yields:
$$\boxed{\Delta S = \frac{\lambda m}{T}}$$
Substituting $m = 1 \text{ kg}$, $T = 273 \text{ K}$, $\lambda = 3.33\cdot 10^5~\mathrm{\frac{J}{kg}}$
$$\Delta S = 1219~\mathrm{\frac{J}{K}}\approx 1.2~\mathrm{\frac{kJ}{K}}$$
$$\Delta S = 1.2~\mathrm{\frac{kJ}{K}}$$