Answer
(a) $\Delta S = 189. 1 J/K$
(b) $\Delta S = -154.5 J/K$
(c) $\Delta S_{net} = 34.6 J/K$
(d) The process is irreversible. See explanation.
Work Step by Step
(a) The change of entropy of water is
$\Delta S = mcln(\frac{T_2}{T_1})$
$\Delta S = (250 \times 10^{-3} kg)(4190 J/kg.K) ln (\frac{351.15 K}{293.15K})$
$\Delta S = 189. 1 J/K$
(b) The change of entropy in the heating element is
$\Delta S = \frac{-mc\Delta T}{T_{element}} = \frac{-(250 \times 10^{-3} kg)(4190 J/kg.K)(351.15K - 293.15K)}{393.15K} $
$\Delta S = -154.5 J/K$
(c) The system of water and heating element
$\Delta S_{net} = 189.1 J/K - 154.5 J/K$
$\Delta S_{net} = 34.6 J/K$
(d) The process is irreversible. This is because it involves a process of heating a liquid from the heating element that follows the second law of thermodynamics. Heat is being delivered at a higher temperature than water, and the magnitude of entropy loss of the source will be less than the entropy gain of the water. Hence the net entropy change is positive.
