Carnot cycle efficiency relationship For a heat engine operating on the Carnot cycle between a highest temperature T1 and a lowest temperature T2 (absolute units), which formula gives the thermal efficiency?

Introduction / Context:
The Carnot cycle sets the theoretical upper limit of efficiency for any heat engine operating between two thermal reservoirs. Its efficiency depends only on the reservoir temperatures when expressed in absolute units (Kelvin).

Given Data / Assumptions:

• Reversible cycle between hot reservoir at T1 and cold reservoir at T2.
• No internal irreversibilities; quasi-static processes.
• Temperatures are absolute (Kelvin) and T1 > T2 > 0.

Concept / Approach:

From entropy balance for a reversible cycle, Q1/T1 = Q2/T2. Efficiency η = 1 − (Q2/Q1) = 1 − (T2/T1). This simple ratio highlights why reducing T2 or increasing T1 raises the maximum possible efficiency, albeit material limits constrain T1 in practice.

Step-by-Step Solution:

Verification / Alternative check:

Plugging T2 → 0 gives η → 1, and T2 → T1 gives η → 0, matching physical expectations and limiting cases.

Why Other Options Are Wrong:

Forms with 1 − T1/T2 or (T2 − T1)/T2 invert temperatures incorrectly.η = T2/T1 is not an efficiency expression.(T1 − T2)/T2 overestimates efficiency and can exceed 1.

Common Pitfalls:

Using Celsius instead of Kelvin; only absolute temperatures are valid in the formula.

Final Answer:

η = 1 − T2/T1