Answer:
the final pressure is P=353.5 kPa and the heat lost by the system (heat transferred) is Q= -3614.7327 kJ ( negative sign means heat outflow)
Explanation:
since the steam is at saturated state at 1500 kPa , from tables of saturated steam
at P= 1500 kPa → specific volume v= 0.13225 m³/kg , ug=2593.3 kJ/kg
thus
m = V/v = 0.25 m³ / 0.13225 m³/kg = 1.89 kg
for condensation until 25% of steam, then mass of steam is
mg₂ = 0.25* 1.89 kg = 0.4725 kg
if we neglect the volume occupied by the liquid , then the steam occupies
v₂ = V/m = 0.25 m³ / 0.4725 kg =0.529 m³/kg
returning to the saturated steam table
at v₂=0.529 m³/kg → 353.5 kPa , uL₂= 584.1585 kJ/kg , ug₂= 2548.4965 kJ/kg
then from the first law of thermodynamics
ΔU= Q - W , but since V=constant , dV=0 and W=∫PdV =0
Q= ΔU
Q= (mg₂*ug₂+ml₂*uL₂) - (m*ug) = 1.89 kg* (0.25*2548.4965 kJ/kg + 0.75*584.1585 kJ/kg - 2593.3 kJ/kg) = -3614.7327 kJ
