Answer
(a) $\Delta S = -210 J/K$
(b) $\Delta S_{net} = +22 J/K$
Work Step by Step
(a) Entropy change while it cools
$\Delta S = mc ln(\frac{T_2}{T_1})$
$\Delta S = (0.25kg ) (4190 J/kg.K) ln(\frac{293. 2 K}{358 .2 K})$
$\Delta S = -210 J/K$ The negative sign indicates that the heat comes out of the water to the surrounding.
(b) The heat from the water is
$Q = mc ΔT$
$Q = (0.250 kg)(4190 J/kg ⋅K)(−65.0 K)$
$Q = -6.809 \times 10^4 J$
The heat absorbed to the air is $Q = +6.809 \times 10^4 J$
Entropy of the air when the water is cooling is
$\Delta S = \frac{Q}{T} $
$\Delta S = \frac{+6.809 \times 10^4 J}{293.1 K} $
$\Delta S = +232 J/K$
Now the net entropy is
$\Delta S_{net} = -210 J/K + +232 J/K$
$\Delta S_{net} = +22 J/K$ The entropy is positive means the process is irreversible.
