Question

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.

Answer 1

Mass of ice, m_ice = 1 kg 

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

Mass of water, m_water = 12 kg 

Temperature of water, T_water = 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 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 = T₀∆S = 288 × 0.109 = 31.39 kJ.