In an irreversible process, 4000 J of heat transfer occurs from a hot reservoir at 600 K to a cold reservoir at 250 K, with no temperature change in either reservoir. What is the total change in entropy of the two reservoirs?
A6.67 J/K
B16.0 J/K
C9.33 J/K
D22.7 J/K
Answer & Solution
Correct answer: C. 9.33 J/K
1. The hot reservoir loses entropy as heat leaves it: ΔSh = -Qh/Th = -(4000 J)/(600 K) = -6.67 J/K.
2. The cold reservoir gains entropy as heat enters it: ΔSc = Qc/Tc = (4000 J)/(250 K) = 16.0 J/K.
3. Adding the two changes gives the total: ΔS_tot = ΔSh + ΔSc = -6.67 J/K + 16.0 J/K.
4. Carrying out the addition: -6.67 + 16.0 = 9.33 J/K.
5. Option A, 6.67 J/K, reports only the size of the hot reservoir's loss and drops its negative sign as well as the cold reservoir's gain.
6. Option B, 16.0 J/K, reports only the cold reservoir's gain and omits the hot reservoir's entropy loss entirely.
7. Option D, 22.7 J/K, does not come from correctly adding -6.67 J/K and 16.0 J/K, so it does not match the required combination.
_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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