# 1 kg of ice at 0°C is mixed with 12 kg of water at 27°C.

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1 kg of ice at 0°C is mixed with 12 kg of water at 27°C. Assuming the surrounding temperature as 15°C, calculate the net increase in entropy and unavailable energy when the system reaches common temperature :

Given : Specific heat of water = 4.18 kJ/kg K ; specific heat of ice = 2.1 kJ/kg K and enthalpy of fusion of ice (latent heat) = 333.5 kJ/kg.

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Mass of ice, mice = 1 kg

Temperature of ice, Tice = 0 + 273 = 273 K

Mass of water, mwater = 12 kg

Temperature of water, Twater = 27 + 273 = 300 K

Surrounding temperature, T0 = 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 Tc = common temperature when heat flows between ice and water stops.

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

12 × 4.18(300 – Tc ) = 4.18(Tc – 273) + 333.5

or 15048 – 50.16Tc = 4.18Tc – 1141.14 + 333.5

or 54.34 Tc = 15855.64

∴ Tc = 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 = T0∆S = 288 × 0.109 = 31.39 kJ.