Carnot Efficiency

Equivalent forms

\eta_{\text{Carnot}} = \frac{T_H - T_C}{T_H}

\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}

An absolute ceiling on efficiency — set not by engineering, but by the laws of physics. No amount of technology can surpass it.

Symbol	Name	SI unit	Dimension	Range
$eta$	Carnot efficiencyoutput Maximum theoretical efficiency of a heat engine	dimensionless	1	0 – 1
$T_H$	Hot reservoir temperature Absolute temperature of the heat source	K	Θ	100 – 2000
$T_C$	Cold reservoir temperature Absolute temperature of the heat sink	K	Θ	50 – 500

Unit systems

Symbol	SI	CGS	Imperial
$eta$	dimensionless	dimensionless	dimensionless
$T_H$	K	K	°R
$T_C$	K	K	°R

Where it holds

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).

Dimensional analysis

$[\eta ] = 1 - [T_C]/[T_H]$ → dimensionless $= 1 - \Theta /\Theta$ = dimensionless $\checkmark$

Discovery

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.

Try this

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.

Research status: stable

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

Common misconceptions

Carnot efficiency is not achievable in practice — it requires infinitely slow (quasi-static) processes, meaning zero power output
100% efficiency requires $T_C = 0\,\mathrm{K}$ , 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

Experimental verification

No engine has ever exceeded the Carnot limit. Modern combined-cycle gas turbines

(T_H \approx 1500\,\mathrm{K}

,

T_C \approx 300\,\mathrm{K})

achieve

\sim 63

% of their Carnot limit

(\sim 80

%). The gap is due to irreversibilities (friction, finite-rate heat transfer). Thermoelectric devices and fuel cells also obey this bound.

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

\eta = 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.

Limiting cases

T_C \to 0

⟶

\eta \to 1 (100

%)A cold sink at absolute zero would allow perfect efficiency (unattainable by Third Law).

T_C \to T_H

⟶

\eta \to 0

No temperature difference means no ability to do work.

T_H \to \infty

⟶

\eta \to 1

Infinitely hot source approaches perfect efficiency asymptotically.

What if…

What if you could cool the sink to near absolute zero?

Efficiency would approach 100%, but the Third Law forbids reaching $T_C = 0\,\mathrm{K}$ . Even at $T_C = 3\,\mathrm{K}$ (liquid helium), the energy cost of maintaining such a cold sink far outweighs any efficiency gain.

What if

T_H = T_C

?

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.

What if you run the Carnot cycle in reverse?

You get a Carnot refrigerator with $COP = T_C/(T_H - T_C)$ . At $T_C = 260\,\mathrm{K}$ , $T_H = 300\,\mathrm{K}$ : $COP = 260/40 = 6.5$ , meaning each joule of work moves 6.5 J of heat.

1

Power plant maximum efficiency

Given ·

T H:: 600
T C:: 300

Find · eta

Steps

Identify: $T_H = 600\,\mathrm{K}$ (boiler), $T_C = 300\,\mathrm{K}$ (river cooling water)
Apply Carnot formula: $\eta = 1 - T_C/T_H$
$\eta = 1 - 300/600 = 0.50$ → maximum 50% efficiency

Result ·

\eta = 1 - T_C/T_H = 1 - 300/600 = 0.50 = 50

%

2

Geothermal plant feasibility

Given ·

T H:: 423
T C:: 300

Find · eta

Steps

Geothermal source at $150^{\circ}C \to T_H = 150 + 273 = 423\,\mathrm{K}$
Ambient cooling: $T_C = 300\,\mathrm{K}$
$\eta = 1 - 300/423 = 0.291$ → maximum 29.1%, real efficiency $\sim 10-15$ %

Result ·

\eta = 1 - 300/423 = 0.291 = 29.1

%

First Law of Thermodynamics→Entropy Change→Ideal Gas Law→