Answer
$T_2=202.5°C$
$W_b=3125\ kJ$
Work Step by Step
From table A-6:
Initial ($P_1=0.1\ MPa, T_1=150°C$): $v_1=1.9367\ m³/kg,\ u_1=2582.8\ kJ/kg$
Since the volume increases linearly with the pressure:
$V_2=\dfrac{P_2}{P_1}V_1$
Given $V_1=50\ m³,\ P_2=0.15\ MPa$:
$V_2=75\ m³$
Hence since $m=V/v$:
$m_1=25.82\ kg$
and since the mass doubles: $m_2=51.64\ kg$, therefore:
$v_2=1.4525\ m³/kg$
Form table A-6 ($P_2,v_2)$: $T_2=202.5°C$
Since the volume increases linearly with the pressure:
$W_b=\dfrac{P_1+P_2}{2}(V_2-V_1)$
$W_b=3125\ kJ$
