Heat Engines & Carnot

Heat Engines & Carnot

Updated Mar 26, 2026

Your car engine burns fuel, and some of that energy becomes motion — but a lot of it goes straight out the exhaust pipe and radiator as waste heat. A typical gasoline engine is only ~25% efficient. The other 75% literally goes up in heat.

It turns out this isn’t just a matter of engineering imperfection. No matter how clever the engineering, no engine can ever convert all of the input heat into useful work. The second law of thermodynamics forbids it. And the Carnot cycle tells you exactly how close to perfect you could ever theoretically get — given the temperatures of the hot and cold reservoirs you’re working with.

How a Heat Engine Works

A heat engine runs between two thermal reservoirs:

  1. Hot reservoir (THT_{H}) - the energy source (burning fuel, boiling water, the sun)
  2. Cold reservoir (TCT_{C}) - the energy sink (exhaust, cooling water, the atmosphere)

Each cycle, the engine:

  • Absorbs heat QHQ_{H} from the hot reservoir
  • Turns part of it into useful work W
  • Dumps the remaining heat QCQ_{C} into the cold reservoir

By conservation of energy: QHQ_{H} = W + QCQ_{C}

Efficiency

Efficiency measures how much of the input heat becomes useful work:

The Carnot Cycle - Maximum Possible Efficiency

The Carnot cycle is a theoretical ideal - a perfectly reversible cycle made of two isothermal and two adiabatic processes. No real engine can match it, but it sets the absolute ceiling on efficiency:

Worked Example

A power plant operates between a steam temperature of 500 °C and a cooling water temperature of 27 °C. What is the maximum possible efficiency?

Convert to Kelvin: THT_{H} = 500 + 273 = 773 K, TCT_{C} = 27 + 273 = 300 K

eCarnote_{\text{Carnot}} = 1 - TCT_{C} / THT_{H} = 1 - 300773\frac{300}{773} = 1 - 0.388 = 0.612 = 61.2%

Even under ideal Carnot conditions, nearly 40% of the input heat must be wasted. The actual efficiency of a real power plant would be much lower.

The Carnot Cycle on a PV Diagram

The Carnot cycle consists of four steps:

  1. Isothermal expansion at THT_{H} - gas absorbs QHQ_{H} from the hot reservoir and expands
  2. Adiabatic expansion - gas continues expanding, cooling from THT_{H} to TCT_{C} with no heat transfer
  3. Isothermal compression at TCT_{C} - gas dumps QCQ_{C} to the cold reservoir as it is compressed
  4. Adiabatic compression - gas is compressed back to the starting state, warming from TCT_{C} to THT_{H}
Carnot cycle on a PV diagram showing two isothermal processes at the hot and cold reservoir temperatures connected by two adiabatic processes, forming a closed clockwise loop with the enclosed area representing net work output
The Carnot cycle on a PV diagram. Two isotherms (constant temperature curves) are connected by two adiabatic paths (no heat transfer). The enclosed area represents the net work output per cycle. Credit: Wikimedia Commons, CC BY-SA

Why No Real Engine Reaches Carnot Efficiency

Real engines fall short because of:

  • Friction between moving parts (wastes energy as heat)
  • Irreversible heat transfer (heat flows across finite temperature differences)
  • Turbulence in gas flow
  • Non-ideal gas behavior

Typical real-world efficiencies: car engines around 20-30%, power plants around 30-40%, diesel engines around 35-45%.

These engine numbers are brake thermal efficiencies: the fraction of the fuel’s chemical energy that reaches the crankshaft as useful work. Whole-vehicle figures, sometimes quoted as “tank-to-wheels” efficiency, are lower, because they also subtract drivetrain, idling, and accessory losses. The two are measured against different denominators, so they are not the same quantity.

Refrigerators and Heat Pumps

A refrigerator is a heat engine running backward - it uses work input to move heat from cold to hot. The “coefficient of performance” (COP) replaces efficiency:

COP_refrigerator = QCQ_{C} / W

The MCAT rarely tests COP calculations but may describe a refrigerator conceptually.

What is the Carnot efficiency of an engine operating between 600 K and 300 K?
Click to reveal answer
50%. eCarnote_{\text{Carnot}} = 1 - TCT_{C}/THT_{H} = 1 - 300600\frac{300}{600} = 1 - 0.5 = 0.5 = 50%. This means even a perfect engine operating between these temperatures would waste half the input heat.
A heat engine absorbs 1000 J from the hot reservoir and exhausts 600 J to the cold reservoir. What is the efficiency and how much work is done?
Click to reveal answer
W = 400 J, e = 40%. W = QHQ_{H} - QCQ_{C} = 1000 - 600 = 400 J. Efficiency = W/QHQ_{H} = 4001000\frac{400}{1000} = 0.40 = 40%. The remaining 60% of the input energy is dumped as waste heat.
Why must you use Kelvin (not Celsius) in the Carnot efficiency formula?
Click to reveal answer
Because the formula involves a ratio of temperatures. The Carnot formula e = 1 - TCT_{C}/THT_{H} requires an absolute scale. Using Celsius could give negative temperatures and nonsensical ratios. For example, TCT_{C} = -20 °C and THT_{H} = 80 °C gives e = 1 - (-2080\frac{20}{80}) = 1.25, which implies over 100% efficiency - physically impossible.