Answer
a) $T_2 = 605.7°C$
b) $W=45\ kJ$
c)$Q = 288.31\ kJ$
Work Step by Step
At the initial state (200 kPa, 200°C):
$v_1=1.08049\ m³/kg,\ u_1=2654.6\ kJ/kg$
With an initial volume of 0.4 m³, the mass is:
$m = 0.37\ kg$
Since the final volume is 0.6 m³:
$v_2 = 1.6207\ m³/kg$
Therefore at the final state ($P_2=250\ kPa$):
$T_2 = 605.7°C,\ u_2=3312.2\ kJ/kg$
a) $T_2 = 605.7°C$
b)The boundary work is the area under the trapezoid shape defined by $(P_1,V_1),(P_2,V_2)$:
$W = (V_2-V_1)\frac{P_1+P_2}{2}$
$W=45\ kJ$
c)The energy balance for the stationary closed-system:
$Q-W=\Delta U$
$Q = 288.31\ kJ$
On the P-v diagram, both the initial and final states are to the right of the saturation curve, connected by a straight line.
