Playground
Interactive Carnot engine schematic: adjust hot and cold reservoir temperatures to see efficiency, heat flow, and work output with animated energy arrows and efficiency gauge.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Carnot efficiencyoutput Maximum theoretical efficiency of a heat engine | dimensionless | 1 | 0 – 1 | |
| Hot reservoir temperature Absolute temperature of the heat source | K | Θ | 100 – 2000 | |
| Cold reservoir temperature Absolute temperature of the heat sink | K | Θ | 50 – 500 |
Deep dive
Derivation
Consider a reversible cycle between reservoirs at T_H and T_C. By the second law, for a reversible engine: Q_H/T_H = Q_C/T_C. Efficiency η = W/Q_H = (Q_H − Q_C)/Q_H = 1 − Q_C/Q_H = 1 − T_C/T_H. No irreversible engine can exceed this because it would violate the Clausius statement of the second law.
Experimental verification
No engine has ever exceeded the Carnot limit. Modern combined-cycle gas turbines (T_H ≈ 1500 K, T_C ≈ 300 K) achieve ~63% of their Carnot limit (~80%). The gap is due to irreversibilities (friction, finite-rate heat transfer). Thermoelectric devices and fuel cells also obey this bound.
Common misconceptions
- Carnot efficiency is not achievable in practice — it requires infinitely slow (quasi-static) processes, meaning zero power output
- 100% efficiency requires T_C = 0 K, which is forbidden by the Third Law of Thermodynamics
- Carnot efficiency applies only to heat engines — fuel cells and solar panels can theoretically exceed it because they are not Carnot-limited
Real-world applications
- Power plant design — determines maximum possible efficiency from steam temperature
- Geothermal energy feasibility analysis (low T_H limits efficiency)
- Automotive engine thermodynamic analysis
- Climate science — Earth's atmospheric heat engine
Worked examples
Power plant maximum efficiency
Given:
- T_H:
- 600
- T_C:
- 300
Find: eta
Solution
η = 1 − T_C/T_H = 1 − 300/600 = 0.50 = 50%
Geothermal plant feasibility
Given:
- T_H:
- 423
- T_C:
- 300
Find: eta
Solution
η = 1 − 300/423 = 0.291 = 29.1%
Scenarios
What if…
- scenario:
- What if you could cool the sink to near absolute zero?
- answer:
- Efficiency would approach 100%, but the Third Law forbids reaching T_C = 0 K. Even at T_C = 3 K (liquid helium), the energy cost of maintaining such a cold sink far outweighs any efficiency gain.
- scenario:
- What if T_H = T_C?
- answer:
- Efficiency drops to zero. No work can be extracted from a system in thermal equilibrium — you need a temperature difference to drive a heat engine. This is equivalent to saying perpetual motion machines of the second kind are impossible.
- scenario:
- What if you run the Carnot cycle in reverse?
- answer:
- You get a Carnot refrigerator with COP = T_C/(T_H − T_C). At T_C = 260 K, T_H = 300 K: COP = 260/40 = 6.5, meaning each joule of work moves 6.5 J of heat.
Limiting cases
- condition:
- T_C → 0
- result:
- η → 1 (100%)
- explanation:
- A cold sink at absolute zero would allow perfect efficiency (unattainable by Third Law).
- condition:
- T_C → T_H
- result:
- η → 0
- explanation:
- No temperature difference means no ability to do work.
- condition:
- T_H → ∞
- result:
- η → 1
- explanation:
- Infinitely hot source approaches perfect efficiency asymptotically.
Context
Sadi Carnot · 1824
Published in 'Réflexions sur la puissance motrice du feu,' Carnot proved this limit at age 28. He died of cholera at 36, his work largely ignored until Clapeyron revived it.
Hook
Why can't any engine convert all heat into useful work?
A power plant operates between a 600 K boiler and a 300 K river. Find the maximum possible efficiency.
Dimensions: [η] = 1 − [T_C]/[T_H] → dimensionless = 1 − Θ/Θ = dimensionless ✓
Validity: Universal upper bound for all heat engines operating between two thermal reservoirs. Real engines always have lower efficiency due to irreversibilities (friction, finite-rate heat transfer, turbulence).