Answer
$\Delta S_{surr}\geq-0.8708\ kJ/K$
Work Step by Step
From table A-5:
Initial ($P_1=175\ kPa, x_1=0.10$): $v_1=0.1013\ m³/kg,\ s_1=2.0537\ kJ/kg.K$
Final ($P_2=150\ kPa, x_2=0.40$): $v_2=0.4644\ m³/kg,\ s_2=3.7494\ kJ/kg.K,\ s_{out}=s_G=7.2231\ kJ/kg.K$
Since $m=V/v,\ V_t=0.02\ m^3$
$m_1=0.1974\ kg$
$m_2=0.04307\ kg$
From the mass balance:
$\frac{dm}{dt}=-\dot{m}$
From the entropy balance and the second law:
$\dot{S}_{gen}=\frac{dS}{dt}-\dot{S}_{in}+\dot{S}_{out}\geq0$
$\frac{dS_{surr}}{dt}+\frac{d(ms)}{dt}+\dot{m}s\geq0$
Combining with the mass balance and integrating over time:
$\Delta S_{surr}+m_2s_2-m_1s_1-s_{out}(m_2-m_1)\geq0$
Hence:
$\Delta S_{surr}\geq-0.8708\ kJ/K$
