Answer
$\Delta S=-0.132\ kJ/K$
Work Step by Step
From table A-6:
Initial ($P_1=0.2\ MPa, T_1=150°C$): $v_1=0.95986\ m³/kg,\ s_1=7.2810\ kJ/kg.K$
Final ($v_2=v_1, T_2=40°C$): $x_2=0.04914,\ s_2=0.9499\ kJ/kg.K$
$m=V/v,\ V=0.02\ m^3$
$m=0.02084\ kg$
$\Delta S=m(s_2-s_1)=-0.132\ kJ/K$
