A 4 L jug of milk at 25°C is placed in refrigerator where it is cooled down to a temperature of 5°C. Assuming the milk has the property of liquid water, find the entropy generated in the cooling process.
Answer: C
v1 = vf = 0.001003 m3/kg, h = hf = 104.87 kJ/kg; sf = 0.3673 kJ/kg K
h = hf = 20.98 kJ/kg, s = sf = 0.0761 kJ/kg K
P = constant = 101 kPa => 1W2 = mP(v2 – v1);
m = V/v1 = 0.004 / 0.001003 = 3.988 kg
1Q2 = m(h2 − h1) = 3.988 (20.98 – 104.87) = -3.988 × 83.89 = -334.55 kJ
1S2(gen) = m(s2 − s1) − 1Q2/T(refrig)
= 3.988 (0.0761 − 0.3673) − (−334.55 / 278.15) = − 1.1613 + 1.2028
= 0.0415 kJ/K.