A vessel of 0.25 m3 capacity is filled with saturated steam at 1500 kPa. If the vessel is cooled until 25% of the steam has condensed, how much heat is transferred, and what is the final pressure?

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