Temperature of surrounding, T_{0} = 10 + 273 = 283 K

Specific heat of water, c_{p} = 4.18 kJ/kg K

The available energy of a system of mass m, specific heat c_{p}, 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 log_{e} \(\left(\cfrac{327}{283}\right)\)]

= 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.