Answer
$\Delta T=-42.3°C$
$v_2=0.0345\ m³/kg$
Work Step by Step
From tables A-11 to A-13:
Inlet ($P_1=0.7\ MPa$, sat. liquid): $T_1=26.69°C,\ h_1=88.2\ kJ/kg$
Outlet ($P_2=0.16\ MPa, h_2=h_1$): $T_2=-15.60°C,\ x_2=0.2745$
The temperature drop is: $\Delta T=-42.3°C$
The specific volume is:
$v_2=v_{2,L}+x_2v_{2,G}$
$v_2=0.0345\ m³/kg$
