alleryStart learning →1.00 kg of ice at 0 degrees Celsius (273 K) absorbs 3.34 x 10^5 J of heat while melting into water at 0 degrees Celsius. What is the increase in entropy?
A1.22 x 10^1 J/K
B1.22 x 10^2 J/K
C1.22 x 10^3 J/K
D1.22 x 10^4 J/K
Answer & Solution
Correct answer: C. 1.22 x 10^3 J/K
1. Change in entropy is given by ΔS = Q divided by T.
2. Substitute Q = 3.34 x 10^5 J and T = 273 K, the melting temperature of ice.
3. Dividing gives ΔS = 1223 J/K, which rounds to 1.22 x 10^3 J/K.
4. 1.22 x 10^4 J/K is ten times too large, and 1.22 x 10^1 or 1.22 x 10^2 J/K are a hundred and ten times too small.
5. So 1.22 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_
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