Answer
a) $-0.238\ kJ/kg.K$
b) No.
Work Step by Step
a) From table A-6:
Initial ($P_1=0.6\ MPa, x_1=1$): $s_1=6.7593\ kJ/kg.K,\ u_1=2566.8\ kJ/kg$
From the energy balance:
$-W_s=\Delta U=m(u_2-u_1)$
Given $W_s=700\ kJ,\ m=2\ kg$:
$u_2=2216.8\ kJ/kg$
Final ($P_2=2\ MPa, u_2$): $x_2=0.8617,\ s_2=6.5215\ kJ/kg.K$
Hence:
$\Delta s = s_2-s_1=-0.238\ kJ/kg.K$
b) For an adiabatic process, entropy should at best remain constant, so the process is not realistic.
