Answer
(a) $Q = 1.17 \times 10^5 J$
$\Delta S = 428.6 J/K$
(b) $\Delta S = -392.6 J/K$
(c) $\Delta S_{net} = 36 J/K$
Work Step by Step
(a) The heat flow into the ice is
$Q = mL_f$
$Q = (0.350 kg)(3.34×10^5 J/kg) $
$Q = 1.17 \times 10^5 J$
The entropy of the water
$\Delta S = \frac{Q}{T} $
$\Delta S = \frac{ 1.17 \times 10^5 J}{273 K} $
$\Delta S = 428.6 J/K$
(b) The change of entropy in the body at $T = 298 K$ is
$\Delta S = \frac{Q}{T} $
$\Delta S = \frac{-1.17 \times 10^5 J}{298 K} $
$\Delta S = -392.6 J/K$
(c) $\Delta S_{net} = 428.6 J/K + (-392.6 J/K)$
$\Delta S_{net} = 36 J/K$
