Answer
See explanation.
Work Step by Step
(a) $\Delta S = \frac{Q}{T} $
$\Delta S = \frac{mL_V}{T} $
$\Delta S = \frac{(1.00 kg)(2256 \times 10^3 J/kg) }{(373. 15 K)}$
$\Delta S = 6045.8 J/K$
(b) The magnitude of the entropy change is roughly five times the value found in Example 20.5 This is because water is less ordered than ice and steam.
