Carnot cycle — full derivation, efficiency = 1 − T_C - T_H: Question bank

Reminders of the symbols used below (all built in the parent note, restated so this page stands alone):

$T_{H}, T_{C}$ — hot and cold reservoir temperatures, in kelvin (never Celsius).
$Q_{H}$ — heat absorbed from the hot side (a positive number); $Q_{C}$ — heat rejected to the cold side (also stored as a positive number).
$W$ — net work out per cycle; $η = W / Q_{H} = 1 - T_{C} / T_{H}$ — the efficiency.
$V_{1} \to V_{2}$ — the volumes during the hot isothermal; $V_{3} \to V_{4}$ — the volumes during the cold isothermal (see the figures below).

The cycle you are reasoning about

Before the traps, look at the actual loop. The first figure is the $P$ – $V$ diagram of the four strokes; every question below refers to a leg of this picture.

Figure — Carnot cycle — full derivation, efficiency = 1 − T_C - T_H

The net work per cycle is the area enclosed by the loop. The direction you travel matters, so the next figure shows the sign convention: a clockwise loop is an engine (work out), a counterclockwise loop is a refrigerator/heat pump (work in).

True or false — justify

A Carnot engine's efficiency depends on which gas you use

False — the adiabatic links force

V_{2} / V_{1} = V_{3} / V_{4}

, so the volume logs cancel and only the ratio

T_{C} / T_{H}

survives, independent of the working substance.

Doubling both

T_{H}

and

T_{C}

leaves the efficiency unchanged

True —

η = 1 - T_{C} / T_{H}

depends only on the ratio, and doubling both leaves

2 T_{C} /2 T_{H} = T_{C} / T_{H}

the same.

A Carnot engine can reach

η = 1

T_{H}

is made large enough

False —

η = 1 - T_{C} / T_{H}

approaches

1

only as

T_{H} \to \infty

, never equals it, and

T_{C} > 0

always leaves a nonzero fraction to be rejected.

Over one full cycle the internal energy change

Δ U

is zero

True — internal energy is a state function and the gas returns to its exact starting state, so

Δ U = 0

regardless of the path taken.

Heat is absorbed only during the isothermal expansion, never elsewhere

True — the two adiabatics exchange no heat by definition, and the cold isothermal rejects heat, so absorption happens only in step 1 at

T_{H}

Running a Carnot engine backwards turns it into an ideal refrigerator

True — because every step is reversible, reversing the cycle (counterclockwise loop) pumps heat

Q_{C}

from cold to hot using work

W

, the definition of a Carnot refrigerator.

The total entropy change of the universe over one Carnot cycle is positive

False — a Carnot cycle is fully reversible, so

\oint d Q_{r e v} / T = 0

and the total entropy change of gas + reservoirs + surroundings is exactly zero.

Spot the error

"Since

η = 1 - T_{C} / T_{H}

with

T_{C} = 2 7^{\circ}

C and

T_{H} = 22 7^{\circ}

η = 1 - 27/227 = 0.88

The error is using Celsius in a ratio; convert first —

T_{C} = 300

T_{H} = 500

K, giving

η = 1 - 300/500 = 0.40

, far from

0.88

"The gas absorbs

Q_{H}

during the adiabatic expansion because it is expanding and doing work."

An adiabatic exchanges no heat by definition (

d Q = 0

); the work done in expansion comes from the gas's own internal energy, which is why it cools.

Q_{C} / Q_{H} = (T_{C} / T_{H}) \cdot ln (V_{3} / V_{4}) / ln (V_{2} / V_{1})

, so efficiency depends on the volumes."

An adiabatic of an ideal gas obeys

T V^{γ - 1} = const

(from

C_{V} d T = - \frac{R T}{V} d V

integrated), so step 2 gives

T_{H} V_{2}^{γ - 1} = T_{C} V_{3}^{γ - 1}

and step 4 gives

T_{H} V_{1}^{γ - 1} = T_{C} V_{4}^{γ - 1}

; dividing these yields

V_{2} / V_{1} = V_{3} / V_{4}

, making the log ratio exactly

1

so the volumes cancel and

Q_{C} / Q_{H} = T_{C} / T_{H}

"A cleverly engineered irreversible engine between the same two reservoirs beat Carnot, reaching

η = 0.6

where Carnot gives

0.5

This violates Carnot's theorem — no engine between two reservoirs can exceed the reversible value, since chaining it with a reversed Carnot engine would build a perpetual-motion machine of the second kind.

"Heat can be transferred isothermally and reversibly across a

10 K

gap between gas and reservoir."

Reversible heat transfer requires the gas and reservoir to be at the same temperature (an infinitesimal gap); a finite

10 K

gap makes the step irreversible and wastes available work.

"Because

Δ U = 0

over the cycle, the engine does no net work."

Δ U = 0

means net heat equals net work (

W = Q_{H} - Q_{C}

), not zero work; the work equals the enclosed area of the clockwise loop on the

P

–

V

diagram.

Why questions

Why must the heat-exchange steps be isothermal rather than, say, isobaric?

To stay reversible, the gas must sit at the reservoir's temperature the whole time it takes heat, which forces

T

constant during exchange; an isobaric step would let

T

drift and open a temperature gap.

Why do the intermediate temperature changes use adiabatics instead of just touching a warm reservoir?

Touching an intermediate-temperature reservoir would transfer heat across a finite gap (irreversible); insulating the gas lets it change its own temperature via work, keeping every step reversible.

Why does lowering

T_{C}

by one degree help more than raising

T_{H}

by one degree (near typical values)?

Compare the partial derivatives of

η = 1 - T_{C} / T_{H}

\partial η / \partial T_{C} = - 1/ T_{H}

and

\partial η / \partial T_{H} = + T_{C} / T_{H}^{2}

; their magnitudes' ratio is

(1/ T_{H}) / (T_{C} / T_{H}^{2}) = T_{H} / T_{C} > 1

whenever

T_{H} > T_{C}

, so a one-degree drop in

T_{C}

always changes

η

more — e.g. at

T_{H} = 600, T_{C} = 300

the cold side is twice as effective per degree.

Why is the identity

Q_{H} / T_{H} = Q_{C} / T_{C}

called "the birth of entropy"?

Rearranged as

Q_{H} / T_{H} - Q_{C} / T_{C} = 0

, it says

\oint d Q_{r e v} / T = 0

for the reversible loop, meaning

d Q_{r e v} / T

integrates to a state function — that function is entropy.

Why can no real engine reach the Carnot efficiency in practice?

Real processes involve friction, finite temperature gaps, and finite speeds — all irreversible — so they generate entropy and fall short of the ideal reversible bound.

Why does the efficiency formula quietly assume the absolute temperature scale?

Because

T_{C} / T_{H}

is a ratio, only a scale whose zero is the true absence of thermal energy (kelvin) gives a physically meaningful fraction; shifting the zero point (Celsius) corrupts the ratio.

Why does traversing the loop counterclockwise flip the sign of the net work?

On the clockwise path you expand at higher pressure than you compress at, so

\oint P d V > 0

(work out); reversing the direction swaps which legs are expansion vs compression, so

\oint P d V < 0

and you must supply work — the loop becomes a refrigerator.

Edge cases

What happens to

η

in the limit

T_{C} \to 0

η \to 1 - 0 = 1

, a perfect engine — but

T_{C} = 0

K is unreachable by the third law, so this is a limit, never an achievable value.

What happens to

η

T_{H} \to T_{C}

(reservoirs equal)?

η \to 1 - T_{C} / T_{C} = 0

; with no temperature difference there is no driving gap, the loop collapses to zero enclosed area, and no net work can be extracted.

If you tried to run the "engine" with

T_{C} > T_{H}

(cold source hotter than hot), what does the formula say?

η = 1 - T_{C} / T_{H}

becomes negative, signalling that heat would flow the wrong way — you'd need to input work, i.e. it is no longer an engine but a heat pump.

Is the reversible identity

Q_{H} / T_{H} = Q_{C} / T_{C}

valid for an irreversible engine?

No — for irreversible operation

Q_{H} / T_{H} < Q_{C} / T_{C}

(the entropy rises), so equality is the special reversible boundary that only Carnot achieves.

What is the enclosed

P

–

V

area for a cycle where

T_{H} = T_{C}

Zero — with equal temperatures the two isotherms merge and the two adiabatics have no temperature span, so the loop pinches to a line of zero enclosed area and zero work.

Recall One-line self-test

Cover every answer and re-derive the reason, not the verdict. If any "why" took more than one sentence of struggle, reread that section of the parent note.