If 2.00 m3 of water at 0°C is frozen and cooled to -15°C by being in contact wit

Question

If 2.00 m3 of water at 0°C is frozen and cooled to -15°C by being in contact with a great deal of ice at -15°C, what would be the total change in entropy of the process?
J/K

Explanation / Answer

The mass of the water to freeze is: m = ?·V = 1000kg/m³ · 2.0m³ = 2000kg The heat exchanged by the freezing water Q = m·( - L + c·?T) with latent heat of fusion of water L = 333.55kJ/kg heat capacity of ice c = 2.05kJ/kgK => Q1 = 2000kg · ( - 333.55kJ/kg + 2.05kJ/kgK·(-25K) = 2000kg · ( - 384.8kJ/kg) = -769.600kJ The same amount is absorbed by the surrounding excess of ice. Q2 = -Q1 = 769.600kJ To get the change of entropy of the freezing ice and the surrounding ice, use relation: dS = dQ/T For the freezing ice portion: dS1= m·( - L/Tm + c·dT/T) (Tm is the absolute melting/freezing temperature,i.e. Tm = 273.15K) => ?S1 = m·( - L/Tm + ? Tm?Tf c/T dT)= m·( - L/Tm + c·ln(Tf/Tm) ) (Tf is the final absolute temperature, i.e. Tf =(273.165 -25)K=248.15K ) So ?S1 = 2000kg · ( - 333.55kJ/kg/273.15K + 2.05kJ/kgK·ln(248.15/273.15) ) = -2836kJ/K Because surrounding ice is in large absorbs the heat at (almost) constant temperature. Therefore ?S2 = Q2/T2 = 769.600kJ/248.15K = 3101kJ/K The total change of entropy is the sum of the partial changes: ?S = ?S1 + ?S2 = -2836kJ/K + 3101kJ/K = 265kJ/K

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