Question

An insulated piston–cylinder device contains 5 l of saturated liquid water at a constant pressure of 150 kpa. an electric resistance heater inside the cylinder is now turned on, and 2200 kj of energy is transferred to the steam. determine the entropy change of the water during this process.

Answer 1

At the initial state: v1 = vf = 0.001053 m 3 /kg, h1 = hf = 467.11 kJ/kg, and s1 = sf = 1.4336 kJ/kgK.
 The mass of the water is: m = V/v1 = 0.005/0.001053 = 4.7483 kg.
 To find the final state, we will use the First Law:
 Q12 = m(h2 - h1) for closed system undergoing a constant pressure process.
 h2 = 1Q2/m + h1 = 2200/4.7483 + 467.11 = 930.43 kJ/kg.
 At P2 = P1 = 150 kPa, this is a saturated mixture.
 hf = 467.11 kJ/kg, hfg = 2226.5 kJ/kg, sf = 1.4336 kJ/kgK, and sfg = 5.7897  kJ/kgK
 s2 = sf + sfg (h2 – hf )/hfg = 1.4336 + 5.7897(930.43 – 467.11)/2226.5 =   2.6384 kJ/kgK.
 The entropy change of water is:
 Delta Ssys= m(s2 – s1) = 4.7483(2.6384 – 1.4336) = 5.72 kJ/K.