๐ŸŒก๏ธ Full Lesson ยท Thermodynamics
ฮท = 1 โˆ’ (Tc/Th)
Carnot Efficiency

No real engine can beat this theoretical ceiling โ€” efficiency depends only on the hot and cold reservoir temperatures.

The Memory Trick
๐Ÿ’ก ฮท = 1 โˆ’ (Tc/Th)

Carnot efficiency represents the absolute theoretical maximum efficiency any heat engine could ever achieve, operating between a hot reservoir at temperature Th and a cold reservoir at temperature Tc (both in Kelvin): ฮท = 1 โˆ’ (Tc/Th). No real engine โ€” no matter how well-engineered โ€” can exceed this limit.

Why It Works
This formula depends ONLY on the two reservoir temperatures โ€” not on what fuel is used, what the engine is made of, or any other engineering detail. That's precisely what makes it a fundamental physical limit rather than just a practical engineering benchmark: it emerges directly from the Second Law of Thermodynamics itself.
Step by Step
Working With Carnot Efficiency
1
Bigger temperature difference means higher efficiency
The larger the gap between Th and Tc, the closer Tc/Th gets to zero, and the closer efficiency gets to 1 (100%) โ€” but it can never actually reach 100% unless Tc reaches absolute zero.
A power plant with a hot reservoir at 800 K and cold reservoir at 300 K has Carnot efficiency = 1 โˆ’ (300/800) = 0.625, or 62.5% โ€” far higher than one with a smaller temperature gap.
2
Real engines always fall short
Real-world engines never actually achieve Carnot efficiency, due to unavoidable practical factors like friction, turbulence, and other irreversible processes that the idealized Carnot cycle assumes away.
A car engine's actual efficiency is typically well below 40%, far short of its theoretical Carnot maximum, due to friction, heat loss, and incomplete combustion.
3
Temperatures MUST be in Kelvin
Since this formula involves a ratio of temperatures, using Celsius (which has an arbitrary zero point) would give a meaningless result โ€” Kelvin's absolute zero starting point is essential here.
Using 27ยฐC instead of 300 K for a cold reservoir would produce a nonsensical efficiency calculation entirely disconnected from the physical reality.
๐Ÿฅ Worked Example
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at 350 K. What is its maximum possible (Carnot) efficiency?
1
Apply the Carnot efficiency formula: ฮท = 1 โˆ’ (Tc/Th).
2
Plug in values: ฮท = 1 โˆ’ (350/500) = 1 โˆ’ 0.7 = 0.3.
3
Interpret: the theoretical maximum efficiency is 30% โ€” no engine operating between these two specific temperatures could ever exceed 30% efficiency, no matter how well engineered.
๐Ÿ“Œ Exam Application
Exams test correctly applying ฮท = 1 โˆ’ (Tc/Th) with temperatures in Kelvin, and understanding that this represents an unreachable theoretical ceiling that real engines always fall short of.
โš ๏ธ Most Common Carnot Efficiency Mistakes
The most common trap is plugging Celsius temperatures directly into the Carnot efficiency formula instead of converting to Kelvin first โ€” since the formula is a ratio of absolute temperatures, using Celsius produces a completely incorrect result.
โœ“ Quick Self-Test
1) Write the formula for Carnot efficiency. ฮท = 1 โˆ’ (Tc/Th), with temperatures in Kelvin. 2) What does Carnot efficiency represent physically? The theoretical maximum efficiency any heat engine could achieve operating between two given reservoir temperatures. 3) Why does a larger temperature difference between hot and cold reservoirs produce higher Carnot efficiency? Because Tc/Th approaches zero as the gap grows, pushing efficiency closer to 1 (100%). 4) Can a real engine ever exceed its Carnot efficiency limit? No โ€” it's a fundamental theoretical maximum derived from the Second Law. 5) Why must temperatures be in Kelvin, not Celsius, for this formula? Because it's a ratio of absolute temperatures โ€” Celsius's arbitrary zero point would make the ratio physically meaningless.
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