Answer
a) $m = 7.567\ kg $
b) $T_2 = 179.88°C$
c) $Q = -4499.85\ kJ $
Work Step by Step
At the initial state (1MPa, 450°C):
$v_1=0.3304\ m³/kg,\ h_1=3371.8\ kJ/kg$
At the final state(saturated steam, 1MPa):
$T_2 = 179.88°C,\ h_2=2777.1\ kJ/kg$
a) With an initial volume of 2.5 m³:
$m = 7.567\ kg $
b)$T_2 = 179.88°C$
c)For a stationary constant-pressure closed system, the energy balance reduces to:
$Q=\Delta H$
$Q = -4499.85\ kJ $
