Answer
$Q = 46.85\ kJ$
Work Step by Step
At the initial state (100°C, saturated, 12.3% quality), from steam tables:
$v_{1,L} = 0.001043 m³/kg,\ v_{1,G} = 1.6720\ m³/kg$
$u_{1,L} = 419.06 kJ/kg,\ u_{1,G} = 2506.0\ kJ/kg$
$v_1=v_{1,L}+x_1(v_{1,G}-v_{1,L})$
$v_1 = 0.20657\ m³/kg$
Therefore, with a volume of 10L, the mass is:
$m = 0.04841\ kg$
$u_1=u_{1,L}+x_1(u_{1,G}-u_{1,L})$
$u_1 = 675.76\ kJ/kg$
The process occurs at a constant volume, therefore, the final state is $T_2=150°C,\ v_2=v_1$:
$v_{2,L} = 0.001091 m³/kg,\ v_{2,G} = 0.39248\ m³/kg$
$u_{2,L} = 631.66 kJ/kg,\ u_{2,G} = 2559.1\ kJ/kg$
$x_2 = \frac{(v_2-v_{1,L})}{(v_{1,G}-v_{1,L})}$
$x_2 = 0.525 $
$u_2 = 1643.54\ kJ/kg $
For a stationary, constant-volume, closed system:
$Q = \Delta U$
$Q = 46.85\ kJ$
