One kilogram of ice at 0°C melts into water at 0°C. The latent heat of fusion for water is 334 kJ/kg. What is the increase in entropy of the ice during this melting?
A3.34 x 10^2 J/K
B3.34 x 10^3 J/K
C6.11 x 10^2 J/K
D1.22 x 10^3 J/K
Answer & Solution
Correct answer: D. 1.22 x 10^3 J/K
1. First find the heat needed to melt the ice: Q = m x Lf = (1.00 kg)(334 kJ/kg) = 334 kJ = 3.34 x 10^5 J.
2. Convert the melting temperature to kelvins: T = 0°C + 273 = 273 K.
3. Apply the entropy definition: ΔS = Q/T = (3.34 x 10^5 J)/(273 K).
4. Carrying out the division: 3.34 x 10^5 / 273 = 1.22 x 10^3 J/K.
5. Option A, 3.34 x 10^2 J/K, and option B, 3.34 x 10^3 J/K, both just rescale the heat Q itself rather than dividing it by the temperature.
6. Option C, 6.11 x 10^2 J/K, does not follow from dividing 3.34 x 10^5 J by 273 K, so it does not match the required substitution.
_Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy_
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