A Carnot engine operates between a hot reservoir at 800 K and a cold reservoir at 200 K. If the working substance is changed from an ideal gas to steam, what happens to the efficiency?
AEfficiency increases because steam has a higher heat capacity and can absorb more heat
BEfficiency decreases because steam is harder to compress at low temperatures
CEfficiency stays the same — Carnot efficiency depends only on the two reservoir temperatures
DEfficiency changes because reversibility depends on the properties of the working substance
The Carnot efficiency η_C = 1 − T_C/T_H depends only on the temperatures of the two reservoirs, not on the working substance. For 800 K and 200 K, η_C = 1 − 200/800 = 75%, regardless of whether the working fluid is an ideal gas, steam, or any other substance. This follows from the derivation: because every step is reversible, zero entropy is generated, and the entropy bookkeeping gives Q_C/Q_H = T_C/T_H — a relation involving only temperatures. The identity of the working substance never enters.
Question 2 Multiple Choice
An inventor claims to have built a heat engine that operates between reservoirs at 1000 K and 300 K and achieves 80% efficiency. The maximum Carnot efficiency between these temperatures is 70%. What does thermodynamics say about this claim?
AThe claim is plausible if the working substance is chosen carefully enough
BThe claim is impossible — it would allow construction of a device that transfers heat from cold to hot with no other effect, violating the second law
CThe claim is fine as long as the engine operates slowly enough to be quasi-static
DThe claim could work if the engine uses an irreversible process that generates negative entropy
No engine operating between two temperatures can exceed Carnot efficiency. The proof is by contradiction: if such an engine existed, you could run a Carnot refrigerator (powered by this engine) that pumps more heat from cold to hot than the engine requires. The net result would be spontaneous heat flow from cold to hot — the Clausius violation. Entropy cannot be negative, and real engines generate additional entropy through irreversibility, pushing them further below the Carnot limit.
Question 3 True / False
The Carnot cycle is the most efficient practical engine design and should be used whenever maximum efficiency is required.
TTrue
FFalse
Answer: False
The Carnot cycle is a theoretical ideal, not a practical design. Its steps must be quasi-static (infinitely slow) to remain reversible. In the limit of reversibility, the power output — work per unit time — approaches zero. A perfectly efficient engine that produces no power is useless for engineering. Real engines sacrifice some thermodynamic efficiency to operate at finite speed and deliver actual power. Carnot efficiency is a theoretical upper bound, not an engineering blueprint.
Question 4 True / False
In a Carnot cycle, because every step is reversible, the total entropy of the universe does not change — the entropy gained from the hot reservoir exactly equals the entropy delivered to the cold reservoir.
TTrue
FFalse
Answer: True
Reversibility is precisely defined as producing no net entropy. In the Carnot cycle: the hot reservoir loses entropy ΔS_H = −Q_H/T_H (heat flows out), and the cold reservoir gains entropy ΔS_C = +Q_C/T_C (heat flows in). Since no entropy is generated internally, total entropy change is zero, which requires Q_H/T_H = Q_C/T_C. This relationship is the key step in deriving the Carnot efficiency formula η_C = 1 − T_C/T_H. Real engines generate entropy internally (due to friction, heat flow across finite temperature differences, etc.), so Q_C/Q_H > T_C/T_H, and their efficiency falls below Carnot.
Question 5 Short Answer
Why does the efficiency of a Carnot cycle depend only on the two reservoir temperatures and not on the nature of the working substance or the details of the cycle steps?
Think about your answer, then reveal below.
Model answer: Because the Carnot cycle is reversible at every step, no entropy is generated anywhere in the cycle. The only entropy changes are the heat exchanges with the reservoirs: entropy flows in from the hot reservoir (Q_H/T_H) and flows out to the cold reservoir (Q_C/T_C). Zero net entropy change requires these to be equal: Q_H/T_H = Q_C/T_C. This relation — derived purely from the constraint of reversibility — gives Q_C/Q_H = T_C/T_H, and therefore efficiency η = 1 − Q_C/Q_H = 1 − T_C/T_H. The working substance never appears in this argument.
The T-S diagram makes this especially vivid: the Carnot cycle is a perfect rectangle on T-S axes. The heat absorbed is the area under the top edge (Q_H = T_H × ΔS), the heat rejected is the area under the bottom edge (Q_C = T_C × ΔS), and the net work is the enclosed area. The ratio Q_C/Q_H = T_C/T_H follows immediately from the geometry of the rectangle.