Answer
a) $\Delta s = 0.240\ kJ/kg.K$
b) Yes.
Work Step by Step
a) From the energy balance:
$-W=\Delta U=mc_v(T_2-T_1)$
Given $W=600\ kJ,\ m=5\ kg,\ c_v=0.718\ kJ/kg.K,\ T_1=427\ °C=700\ K$
$T_2=532.9\ K$
$\Delta s = c_p\ln\left(\dfrac{T_2}{T_1}\right)-R\ln\left(\dfrac{P_2}{P_1}\right)$
With $P_2=100\ kPa,\ P_1=600\ kPa,\ R=0.287\ kJ/kg.K,\ c_p=1.005\ kJ/kg.K$
$\Delta s = 0.240\ kJ/kg.K$
b) Since the entropy increases, this is a realistic process.
