Heat engines — efficiency η = 1 − Q_C - Q_H: Question bank

Reminder of the symbols (all defined in the parent): $Q_{H}$ = heat drawn from the hot reservoir (a positive magnitude), $Q_{C}$ = heat dumped into the cold reservoir (positive magnitude), $W$ = useful work done by the engine, $η = W / Q_{H} = 1 - Q_{C} / Q_{H}$ = efficiency of any engine, $T_{H}, T_{C}$ = reservoir temperatures in kelvin.

Figure — Heat engines — efficiency η = 1 − Q_C - Q_H

True or false — justify

A heat engine can convert 100% of the absorbed heat into work in a single non-cyclic process.

True — a one-shot isothermal expansion keeps the gas at constant temperature (so its internal energy is unchanged), and by the First Law all the absorbed heat becomes work; but the gas is now left expanded in a different state, not reset, so this is a single process, not a repeatable cycle. The Kelvin–Planck ban applies only to a repeating cycle.

Over one complete cycle the working gas's internal energy change is zero.

True — internal energy is a state function, and a cycle ends exactly where it began, so

Δ U_{cycle} = 0

regardless of the path taken.

Q_{C} = 0

the engine would have

η = 1

, and this is achievable with clever engineering.

False — algebraically

η = 1

needs

Q_{C} = 0

, but the Second Law (Kelvin–Planck) forbids fully converting cyclic heat to work; no engineering removes this, it is a law of nature.

Doubling both

T_{H}

and

T_{C}

leaves the Carnot efficiency unchanged.

True —

η_{C} = 1 - T_{C} / T_{H}

depends only on the ratio; scaling both temperatures by the same factor keeps

T_{C} / T_{H}

fixed.

A real engine can exceed the Carnot efficiency if it is very well lubricated.

False — Carnot is the ceiling for any engine (reversible ones reach it, irreversible ones fall short); friction only lowers you further below the bound, never above it.

The quantity

Q_{C} / Q_{H}

equals the efficiency of the engine.

False —

Q_{C} / Q_{H}

is the fraction wasted; efficiency is

1 - Q_{C} / Q_{H}

. The two sum to exactly 1.

An engine running between reservoirs of equal temperature (

T_{H} = T_{C}

) can still do net work.

False —

η_{C} = 1 - T_{C} / T_{H} = 0

when

T_{H} = T_{C}

; with no temperature "drop" for heat to fall through, no net work is extractable.

Making the cold reservoir 1 K colder raises efficiency more than making the hot reservoir 1 K hotter.

True whenever $T_{C} < T_{H}$ — treat

η_{C} = 1 - T_{C} / T_{H}

and change one variable at a time. Lowering

T_{C}

d T

raises

η_{C}

d T / T_{H}

. Raising

T_{H}

d T

raises

η_{C}

T_{C} d T / T_{H}^{2} = (T_{C} / T_{H}) (d T / T_{H})

. Since

T_{C} / T_{H} < 1

, the cold-side change wins by exactly the factor

T_{C} / T_{H}

The area enclosed by the cycle on a PV diagram equals the net work

W

done by the engine.

True — see PV diagrams and Work; the loop's enclosed area is

\oint P d V = W_{net}

, positive for a clockwise loop.

Spot the error

η = 1 - Q_{C} / Q_{H} = 1 - 273/373 = 0.27

for an engine with

T_{H} = 100° C, T_{C} = 0° C

The formula mixes up heats with temperatures — the temperature form is

η_{C} = 1 - T_{C} / T_{H}

; also the numbers use kelvin correctly but the symbols

Q

are wrong.

"Since heat is absorbed, the gas's internal energy rises over the cycle, so

Δ U > 0

Heat is absorbed and rejected within one loop; the gas heats up then cools back to its start, giving

Δ U_{cycle} = 0

despite net heat inflow.

"The engine is 40% efficient, so it wastes only 40% of the heat."

Backwards — 40% efficient means 40% becomes work and 60% is dumped as

Q_{C} / Q_{H} = 1 - η = 0.60

"Carnot efficiency for

T_{H} = 500° C, T_{C} = 20° C

1 - 20/500 = 0.96

Celsius is illegal here; convert first:

1 - 293/773 \approx 0.62

. Using °C inflates the answer wildly.

"Because

W = Q_{H} - Q_{C}

, an engine that dumps more heat does more work."

For fixed

Q_{H}

, larger

Q_{C}

means less work; the statement inverts the relationship —

W

shrinks as

Q_{C}

grows.

"An ideal reversible engine is best because it runs the fastest."

Reversibility maximises efficiency, not speed; truly reversible processes are infinitely slow (quasi-static), so the real trade-off is efficiency versus power.

Why questions

Why must a cyclic heat engine reject some heat to a cold reservoir?

A cycle must return the gas to its exact start state, so any energy you poured in as

Q_{H}

that did not leave as work must leave as heat — otherwise

Δ U

over the loop would not be zero. The Kelvin–Planck statement sharpens this: no cycle can exist whose only effect is turning heat fully into work, so

Q_{C} > 0

is forced. (See the energy-flow figure above.)

Why is Kelvin (absolute) temperature required in the Carnot formula?

The ratio

T_{C} / T_{H}

must reflect true thermodynamic temperature; only an absolute scale makes such ratios physically meaningful. On the Celsius scale the ratio near freezing is meaningless (and division by

T_{H}

blows up as

T_{H} \to 0° C

, which is not the physical zero); the true singular point is

0

K, absolute zero.

Why does a bigger temperature difference between reservoirs give higher efficiency?

A larger gap raises

1 - T_{C} / T_{H}

; intuitively heat "falls" through a greater drop, so more of it can be diverted into work — like a taller waterfall spinning a wheel harder.

Why is the Carnot engine the maximum-efficiency benchmark?

It is fully reversible, generating zero entropy; any real irreversibility (friction, finite temperature differences) creates entropy that must be paid for with extra dumped heat, lowering

η

. See Entropy.

Why can't we just insulate the engine perfectly to stop heat loss and reach

η = 1

The rejected heat

Q_{C}

is not an accidental "leak" you can insulate away — it is a thermodynamically mandatory dump required by the Second Law to complete the cycle.

Why do power plants still throw away more energy than they use even when "efficient"?

A typical

η \approx 0.4

means

Q_{C} / Q_{H} = 0.6

; over half the input heat must leave as waste to satisfy the cycle constraint, no matter how modern the plant.

Why is efficiency defined as

W / Q_{H}

and not

W / (Q_{H} + Q_{C})

You only pay for the fuel that supplies

Q_{H}

; the dumped

Q_{C}

is a loss, not a purchased input, so the honest "value for money" ratio is

W / Q_{H}

Edge cases

What is the efficiency when

Q_{C} = Q_{H}

η = 1 - Q_{C} / Q_{H} = 0

; all absorbed heat is dumped, no work is produced — the machine cycles pointlessly.

What is the efficiency when

Q_{C} \to 0

(the forbidden limit)?

η \to 1

, the theoretical maximum, but this violates Kelvin–Planck; it is a limit approached only conceptually, never reached.

What happens to

η_{C}

T_{C} \to 0

K (absolute zero cold reservoir)?

η_{C} \to 1

; a cold reservoir at absolute zero would allow perfect conversion, but absolute zero is unattainable (Third Law), so this too is only a limit.

Can efficiency ever be negative, and what would that mean?

For an engine, no — a negative

η

would mean

W < 0

(work done on the gas), which describes a refrigerator, measured by a COP, not an efficiency.

What is

W

if the engine absorbs heat but the cycle is run in reverse?

Reversing the loop makes

W < 0

: you supply work to pump heat from cold to hot — that device is a heat pump/refrigerator, not an engine.

An engine with

T_{H} = T_{C}

but genuinely different

Q_{H}

and

Q_{C}

— possible?

No — with equal temperatures

η_{C} = 0

, so any real engine there must have

W \leq 0

; you cannot extract net work without a temperature difference.

If an inventor claims

η > η_{C}

for their reservoirs, what is instantly wrong?

They are claiming to beat the reversible bound, which would let you build a machine that decreases the universe's entropy — a direct Second Law violation, so the claim is false without checking any details.

Connections

Parent: Heat engines — efficiency
First Law of Thermodynamics — the $Δ U = 0$ cycle argument behind $W = Q_{H} - Q_{C}$ .
Second Law of Thermodynamics — the source of every "impossible" verdict here.
Carnot Cycle — the efficiency ceiling used in the edge cases.
Entropy — why irreversibility costs efficiency.
Refrigerators and Heat Pumps — where negative $W$ / COP takes over.
PV diagrams and Work — the loop-area interpretation of $W$ .