Mass of ice, m_{ice} = 1 kg

Temperature of ice, T_{ice} = 0 + 273 = 273 K

Mass of water, m_{wate}r = 12 kg

Temperature of water, T_{water} = 27 + 273 = 300 K

Surrounding temperature, T_{0} = 15 + 273 = 288 K

Specific heat of water = 4.18 kJ/kg K

Specific heat of ice = 2.1 kJ/kg K

Latent heat of ice = 333.5 kJ/kg

Let T_{c} = common temperature when heat flows between ice and water stops.

Heat lost by water = Heat gained by ice i.e.,

12 × 4.18(300 – T_{c} ) = 4.18(T_{c} – 273) + 333.5

or 15048 – 50.16Tc = 4.18T_{c} – 1141.14 + 333.5

or 54.34 T_{c} = 15855.64

∴ T_{c} = 291.8 K or 18.8°C

Change of entropy of water

= 12 × 4.18 loge \(\left(\cfrac{291.8}{300}\right)\) = – 1.39 kJ/K\(\cfrac{333.5}{273}\) =

Change of entropy of ice = 1 × 4.18 loge \(\left(\cfrac{291.8}{300}\right)\) = 1.499 kJ/K

Net change of entropy, ∆S = – 1.39 + 1.499 = 0.109 kJ/K

Hence, net increase in entropy = 0.109 kJ/K.

Increase in unavailable energy = T_{0}∆S = 288 × 0.109 = 31.39 kJ.