Answer
a) $T_2=261.7°C$
b) $m_i=0.0176\ kg$
Work Step by Step
From tables A-4 to A-6:
Initial ($P_1=0.5\ MPa, T_1=200°C$): $v_1=0.42503\ m³/kg,\ u_1=2643.3\ kJ/kg$
Inlet ($P_i=1\ MPa, T_2=350°C$): $h_i=3158.2\ kJ/kg$
$m=V/v$
Initial $V_1=0.01\ m³$: $m_1=0.0235\ kg$
Final $V_2=0.02\ m³$: $m_2=0.02/v_2$
From the material balance:
$m_i=m_2-m_1$
From the energy balance:
$-W_{b,e}+m_ih_i=m_2u_2-m_1u_1$
Since $W_b=-P(V_2-V_1)=5\ kJ$
$-W_b+m_ih_i=m_2u_2-m_1u_1$
$m_2(h_i-u_2)=m_1(h_i-u_1)+5$
By trial and error:
$T_2=261.7°C,\ v_2=0.4858\ m³/kg,\ u_2=2742.4\ kJ/kg$
Hence the final mass is: $m_2=0.0412\ kg$
$m_i=0.0176\ kg$
