Answer
a) $T_3 = 662.3°C$
b) $W= 240\ kJ$
c) $Q = 1214.61\ kJ/kg$
Work Step by Step
The steam undegoes two processes: i) 250 kPa -> 300 kPa at constant volume, ii) 0.8 m³ -> 1.6 m³ at constant pressure.
At the initial state:
$v_1=0.71873\ m³/kg,\ u_1 = 2536.8\ kJ/kg$
Since the initial volume is 0.8 m³, the mass is:
$m=1.113\ kg$
At the second state ($P_2=300\ kPa,\ v_2=v_1$):
$u_2 = 2653.2\ kJ/kg,\ h_2=2868.8\ kJ/kg$
At the final state ($P_3=300 kPa, v_3=\frac{1.6\ m³}{1.113\ kg}=1.4376\ m³/kg$):
$T_3 = 662.3°C,\ h_3 = 3843.70\ kJ/kg$
a) $T_3 = 662.3°C$
b) In the process 1-2 (constant volume) no work is done, in the process 2-3:
$W_{2-3}=P_2(V_3-V_2) = 240\ kJ$,
thus $W= 240\ kJ$
c) Energy balance for the 1-2 process:
$Q_{1-2} = \Delta U_{1-2} = 129.55\ kJ$
Energy balance for the 2-3 process:
$Q_{2-3}=\Delta H_{2-3} = 1085.06$
Therefore: $Q = 1214.61\ kJ/kg$
