Temperature of surrounding, T0 = 10 + 273 = 283 K
Specific heat of water, cp = 4.18 kJ/kg K
The available energy of a system of mass m, specific heat cp, and at temperature T, is given by
Available energy,
Now, available energy of 20 kg of water at 90°C,
Available energy of 30 kg of water at 30°C,
Total available energy,
(A.E.)total = (A.E.)20 kg + (A.E.)30 kg
= 798.38 + 85.27 = 883.65 kJ
If t°C is the final temperature after mixing, then
20 × 4.18 × (90 – t) = 30 × 4.18 (t – 30)
20(90 – t) = 30 (t – 30)
t = \(\cfrac{20\times90+30\times30}{20+30}\) = 54°C
Total mass after mixing = 20 + 30 = 50 kg
Available energy of 50 kg of water at 54°C
(A.E.)50 kg = 50 × 4.18 [ (327 - 283) - 283 loge \(\left(\cfrac{327}{283}\right)\)]
= 209 (44 – 40.89) = 649.99 kJ
∴ Decrease in available energy due to mixing
= Total energy before mixing – Total energy after mixing
= 883.65 – 649.99 = 233.66 kJ.