Thermal Efficiency — η = W/Qh

Thermal efficiency η = W/Qh explained: what heat-engine efficiency means, the second-law limit, and a worked example.

\eta = \frac{W}{Q_h}

Thermal efficiency equals useful work output divided by heat input.

What it means

Thermal efficiency answers the only question that matters about a heat engine: of the heat you paid for, what fraction became useful work? The second law of thermodynamics guarantees the answer is always below 100% — some heat must always be rejected to a cold reservoir, and the Carnot limit sets the absolute ceiling for given reservoir temperatures. Real engines fall well short of Carnot: friction, finite-time heat transfer and combustion irreversibilities each take their cut. Efficiency is dimensionless, usually quoted as a percentage.

Variables

Symbol	Name	Unit	Meaning
$η$	Thermal efficiency	—	Fraction of heat converted to useful work (dimensionless)
$W$	Work output	J	Net useful mechanical work delivered per cycle
$Qh$	Heat input	J	Heat absorbed from the hot reservoir per cycle

Assumptions

Engine operates in a steady cycle
W is the net useful work per cycle
Qh is the heat absorbed from the hot reservoir per cycle

Derivation

Energy conservation for one cycle gives $Q_h = W + Q_c$ : heat in equals work out plus waste heat rejected to the cold reservoir. Efficiency is defined as benefit over cost, $\eta = W/Q_h$ , which rearranges to $\eta = 1 - Q_c/Q_h$ — showing waste heat is unavoidable.

Worked example

Problem

A heat engine absorbs 1000 J of heat each cycle and delivers 400 J of useful work. What is its thermal efficiency?

Given:

W = 400 J of useful work per cycle
Qh = 1000 J of heat input per cycle

Write the definition: $\eta = W/Q_h$
Substitute: $\eta = 400 / 1000$
Divide: $\eta = 0.40 = 40\%$

η = 0.40, or 40%

Only 400 of the 1000 joules become work; the other 600 J must be dumped as waste heat. That 40% is actually respectable — most car engines manage 25–30%. No heat engine can reach 100%, no matter how clever the design.