Solution

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 $m = 1 \text{ kg}$, $T = 273 \text{ K}$, $\lambda = 3.33\cdot 10^5~\mathrm{\frac{J}{kg}}$

$$\Delta S = 1219~\mathrm{\frac{J}{K}}\approx 1.2~\mathrm{\frac{kJ}{K}}$$