3.00 kg of ice at 0 degrees Celsius (273 K) melts into water at 0 degrees Celsius. Using Lf(water) = 334 kJ/kg, what is the increase in entropy?

1. First find the heat needed to melt the ice: Q = mLf = (3.00 kg)(334,000 J/kg) = 1.002 x 10^6 J. 2. Apply ΔS = Q divided by T with T = 273 K, the melting temperature of ice. 3. Dividing gives ΔS = 3670 J/K, which is 3.67 x 10^3 J/K. 4. 1.22 x 10^3 J/K is the value for only 1.00 kg of ice, and 3.67 x 10^4 J/K shifts the decimal point by one place. 5. 1.10 x 10^6 J/K confuses the heat Q itself with the entropy change. 6. So 3.67 x 10^3 J/K is correct. _Source: OpenStax Physics (CC BY 4.0), Ch 12 "Thermodynamics", section 12.3 Second Law of Thermodynamics: Entropy_