Answer
$\Delta S=30.24\frac{J}{K}$
Work Step by Step
We can find the entropy change as;
$\Delta S=\frac{Q}{T}$......eq(1)
Also,
$Q=mL_v$ where $L_v$ is water heat of vaporization and its value is $2256000\frac{J}{kg}$.
Thus,
$Q=(0.012)(2256000)=11280J$
We plug in the known values in eq(1) to obtain:
$\Delta S=\frac{11280}{373}=30.24\frac{J}{K}$