Answer
$Q=-2170\ kJ$
Work Step by Step
For the two tanks as the system, the energy balance reduces to:
$Q=\Delta U=mu_2-m_Au_{1,A}-m_Bu_{1,B}$
From steam tables:
Initial state in tank A (0.2m³, 400 kPa, 80% quality):
$v_{1,A}=0.37015\ m³/kg,\ u_{1,A}=2163.3\ kJ/kg$
Hence $m_A=0.5403\ kg$
Initial state in tank B (0.5m³, 200 kPa, 250°C):
$v_{1,B}=1.1989\ m³/kg,\ u_{1,B}=2731.4\ kJ/kg$
Hence $m_B=0.4170\ kg$
And $m=0.9573\ kg$
At the final state:
$v_2=\frac{V_A+V_B}{m}=0.73117\ m³/kg$
$T_2=25°C$, so: $P_2=3.17\ kPa,\ v_{2,L}=0.001003\ m³/kg,\ v_{2,G}=43.340\ m³/kg$,
$u_{2,L}=104.83\ kJ/kg,\ u_{2,G}=2304.3\ kJ/kg$
And $x_2=\frac{v_2-v_{2,L}}{v_{2,G}}=0.01685$
therefore: $u_2=143.65\ kJ/kg$
Back to the energy balance:
$Q=-2170\ kJ$
