Answer
$-1.94\times10^3J/K$
Work Step by Step
Heat leaves the water, so the change in entropy is negative. The heat transfer is the steam’s mass multiplied by the latent heat of vaporization.
$$\Delta S=\frac{Q}{T}=\frac{mL_v}{T}$$
$$=-\frac{(0.320kg)(2.26\times10^6J/kg)}{(100+273)K}$$
$$=-1.94\times10^3J/K$$
