Solution

Resulting change of entropy can be represented as the sum of changes of the entropy in each stage: heating, then evaporation.

$$\Delta S = \Delta S_{1} + \Delta S_{2}$$

I. Heating the water: As $\delta Q=cmdT$:

$$\Delta S_{1} =\int_{T_{1}}^{T_{2}}\frac{\delta Q}{T} = cm\int_{T_{1}}^{T_{2}}\frac{dT}{T} = cm\cdot ln\frac{T_{2}}{T_{1}}$$

Where $T_{1} = 293~\text{K}$ and $T_{2} = 373~\text{K}$ and $c = 4.2~\mathrm{\frac{kJ}{kg\cdot \text{ K}}}$ is the specific heat of water.

II. When water evaporates the temperature is constant.

$$\Delta S_{2} = \int_{1}^{2}\frac{\delta Q}{T} = \frac{1}{T_{2}}\int_{1}^{2}\delta Q = \frac{Q}{T_{2}} = \frac{mL}{T_{2}}$$

Here $L$ is the heat of vaporization of the water.

The resulting change in entropy:

$$\boxed{\Delta S =cm\cdot ln\frac{T_{2}}{T_{1}} + \frac{mL}{T_{2}}}$$

$$\Delta S = 7189 ~\mathrm{\frac{J}{K}} \approx 7.2 ~\mathrm{\frac{kJ}{K}}$$