Carnot cycle — full derivation, efficiency = 1 − T_C - T_H: Worked examples

Every symbol used here is built in the parent note, but so you never have to flip back, here is the full cast in plain words — including the two ( $Δ U$ and $R$ , $e$ ) the examples lean on.

Recall The symbols, in plain words

$T_{H}$ = temperature of the hot reservoir (the stove), in kelvin.
$T_{C}$ = temperature of the cold reservoir (the ice), in kelvin.
$Q_{H}$ = heat the gas absorbs from the hot reservoir (a positive number of joules).
$Q_{C}$ = heat the gas rejects into the cold reservoir (a positive number of joules).
$W$ = net useful work done in one full cycle, $W = Q_{H} - Q_{C}$ .
$η$ (eta) = efficiency = "fraction of absorbed heat that became work" $= W / Q_{H}$ .
$Δ U$ = change in internal energy of the gas — the energy stored in the random jiggling of its molecules. The symbol $Δ$ (delta) means "final minus initial". For an ideal gas $U$ depends only on temperature, so if temperature returns to its start value, $Δ U = 0$ .
$R$ = the universal gas constant, $R = 8.314 J K^{- 1} mol^{- 1}$ — the fixed number linking pressure, volume and temperature in $P V = R T$ for one mole of gas.
$e$ = Euler's number $\approx 2.718$ , the base of the natural logarithm $ln$ . It is the special number for which $ln e = 1$ ; $e^{x}$ and $ln x$ undo each other.
Kelvin = the absolute temperature scale where $0$ is the coldest possible. Convert: $T (K) = T (^{\circ} C) + 273.15$ (we use $273$ for arithmetic).

Because pictures carry the reasoning here, three figures anchor the examples: an energy-flow diagram for engines, the mirrored flow for a refrigerator, and a curve showing efficiency climbing toward its ceiling.

The scenario matrix

Here is the full list of case-classes this topic can produce. Each worked example below is tagged with the cell it fills.

#	Case class	What makes it tricky	Example
A	Clean kelvin plug-in	nothing — the baseline	Ex 1
B	Input in Celsius	must convert before forming the ratio	Ex 2
C	Find $Q_{C}$ , $W$ from $Q_{H}$	use $Q_{C} / Q_{H} = T_{C} / T_{H}$	Ex 3
D	Reverse direction (refrigerator)	efficiency becomes coefficient of performance	Ex 4
E	Degenerate: $T_{H} = T_{C}$	$η = 0$ , engine does nothing	Ex 5
F	Limiting: $T_{C} \to 0$	$η \to 1$ , why it's unreachable	Ex 6
G	Real-world word problem	strip words down to $T_{H}, T_{C}, Q_{H}$	Ex 7
H	Exam twist: "beat Carnot?" claim	Carnot's theorem catches the trap	Ex 8
I	Work back to a volume ratio	invert $Q_{H} = R T_{H} ln (V_{2} / V_{1})$	Ex 9
J	Sign edge: $T_{C} > T_{H}$	$η < 0$ signals a heat-pump regime	Ex 10

The energy-flow picture below is the backbone of Cells A–H — keep it in view.

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

We now fill every cell.

The worked examples

Worked example Example 2 — Cell B: input in Celsius (the ratio trap)

A power plant works between a boiler at $327^{\circ} C$ and a cooling tower at $27^{\circ} C$ . Find $η$ .

Forecast: A student who forgets to convert would compute $1 - 27/327 \approx 0.92$ . Do you trust that? Guess the real answer first.

Convert both to kelvin: $T_{H} = 327 + 273 = 600 K$ , $T_{C} = 27 + 273 = 300 K$ . Why this step? Efficiency uses the ratio $T_{C} / T_{H}$ , and a ratio depends on where zero sits. Celsius puts zero at freezing water; kelvin puts it at absolute cold. Only the kelvin ratio is physical.
Now plug in: $η = 1 - \frac{300}{600} = 1 - 0.5 = 0.5$ . Why this step? Same formula as Ex 1, but only trustworthy after the conversion.

$η = 0.5 = 50%$

Verify: The Celsius shortcut gave $0.92$ — wildly different from $0.50$ . This gap is exactly the kelvin mistake warned about in the parent note. The real answer $50%$ is much lower and correct.

Worked example Example 3 — Cell C: find

$Q_{C}$ and $W$ from $Q_{H}$ The engine of Ex 2 ( $η = 0.5$ ) absorbs $Q_{H} = 4000 J$ per cycle. Find the work $W$ and the rejected heat $Q_{C}$ .

Forecast: Half of $4000$ becomes work — but where does the other half go? Guess $Q_{C}$ .

Work: $W = η Q_{H} = 0.5 \times 4000 = 2000 J$ . Why this step? By definition $η = W / Q_{H}$ , so $W = η Q_{H}$ . This is the fraction of absorbed heat that became useful.
Rejected heat via the first law: $Q_{C} = Q_{H} - W = 4000 - 2000 = 2000 J$ . Why this step? Over one full cycle the gas returns to its starting temperature, so its internal energy comes back too: $Δ U = 0$ (defined above — final minus initial internal energy). The first law $Q_{net} = Δ U + W$ then becomes $Q_{H} - Q_{C} = W$ , so whatever heat didn't become work must leave as $Q_{C}$ .

$W = 2000 J, Q_{C} = 2000 J$

Verify: Cross-check with the reversible identity $Q_{C} = Q_{H} \frac{T _{C}}{T _{H}} = 4000 \times \frac{300}{600} = 2000 J$ ✓. In the flow figure (s01) the red $Q_{H}$ arrow splits exactly in half: half exits as green $W$ , half continues down as blue $Q_{C}$ . Both routes agree, confirming the first law and the reversible relation $Q_{H} / T_{H} = Q_{C} / T_{C}$ are consistent.

Worked example Example 4 — Cell D: run it backwards (refrigerator)

Now drive the same machine in reverse as a refrigerator between $T_{C} = 300 K$ (inside the fridge) and $T_{H} = 600 K$ (kitchen air). If it dumps $Q_{H} = 4000 J$ into the kitchen, find the coefficient of performance (COP) and the work input.

Forecast: A fridge does the opposite of an engine — you pay work to pump heat out of the cold side. Do you expect the COP (benefit per work spent) to be more or less than 1?

The reversed flow (arrows flipped from s01) is shown here — now work goes in and heat is pumped uphill:

For a reversed Carnot cycle the heats still obey $\frac{Q _{H}}{T _{H}} = \frac{Q _{C}}{T _{C}}$ , so $Q_{C} = Q_{H} \frac{T _{C}}{T _{H}} = 4000 \times \frac{300}{600} = 2000 J$ . Why this step? Reversibility means the same temperature–heat relation holds no matter which way we turn the cycle; only the arrows flip (compare s01 with s02).
Work you must supply: $W = Q_{H} - Q_{C} = 4000 - 2000 = 2000 J$ . Why this step? First law again: the kitchen heat $Q_{H}$ equals the heat pulled from inside plus the work you paid.
Fridge COP $= \frac{Q _{C}}{W} = \frac{2000}{2000} = 1.0$ . Why this step? For a fridge the "benefit" is heat removed from the cold space $Q_{C}$ , and the "cost" is the work $W$ .
Turn this into a temperature-only formula. Start from $COP = \frac{Q _{C}}{W} = \frac{Q _{C}}{Q _{H} - Q _{C}}$ . Divide top and bottom by $Q_{H}$ : $COP = \frac{Q _{C} / Q _{H}}{1 - Q _{C} / Q _{H}}$ . Now use $Q_{C} / Q_{H} = T_{C} / T_{H}$ : $COP = \frac{T _{C} / T _{H}}{1 - T _{C} / T _{H}}$ . Multiply top and bottom by $T_{H}$ : $COP = \frac{T _{C}}{T _{H} - T _{C}} = \frac{300}{600 - 300} = 1.0$ . Why this step? We derived the temperature formula rather than quoting it — dividing by $Q_{H}$ let us swap in the reversible ratio $T_{C} / T_{H}$ , and multiplying by $T_{H}$ cleared the fractions.

$Q_{C} = 2000 J, W = 2000 J, COP = 1.0$

Verify: Both routes ( $Q_{C} / W$ and $T_{C} / (T_{H} - T_{C})$ ) give $1.0$ ✓. Notice COP is not $1 - T_{C} / T_{H}$ — a refrigerator is a different question ("heat moved per work") than an engine ("work per heat absorbed").

Worked example Example 5 — Cell E: degenerate case

$T_{H} = T_{C}$ What efficiency does a Carnot engine reach if the hot and cold reservoirs are at the same temperature, say both $400 K$ ?

Forecast: No temperature gap at all. Can any work come out?

Plug in: $η = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{400}{400} = 1 - 1 = 0$ . Why this step? When $T_{C} = T_{H}$ the ratio is exactly $1$ , killing the whole expression.
Interpret: $η = 0$ means $W = η Q_{H} = 0$ — the engine produces no net work. Why this step? An engine's whole job is to exploit a temperature difference. With none, heat has no "downhill" direction to flow while doing work.

$η = 0$

Verify: By the second law, you cannot extract net work from a single reservoir in a cycle (that would be a perpetual motion machine of the second kind). $η = 0$ is exactly this law at the boundary. In the flow figure (s01) the green $W$ arrow would shrink to zero width. ✓

Worked example Example 6 — Cell F: limiting case

$T_{C} \to 0$ Suppose the cold reservoir could be cooled toward absolute zero, $T_{C} \to 0 K$ , with $T_{H} = 500 K$ fixed. What does $η$ approach, and why can we never reach it?

Forecast: As the cold side gets colder, the ratio $T_{C} / T_{H}$ shrinks. What is the ceiling on $η$ ?

Take the limit: as $T_{C} \to 0$ , $\frac{T _{C}}{T _{H}} \to 0$ , so $η = 1 - \frac{T _{C}}{T _{H}} \to 1 - 0 = 1$ . Why this step? $T_{H}$ is fixed and finite, so only the numerator of the ratio shrinks toward zero.
Build the sanity table below (fixing $T_{H} = 500$ ) and read off how $η$ creeps toward $1$ but never touches it while $T_{C} > 0$ . Why this step? A limit is easier to trust when you watch discrete values approach it.

$T_{C}$ (K)	$T_{C} / T_{H}$	$η = 1 - T_{C} / T_{H}$
250	0.50	0.50
50	0.10	0.90
5	0.01	0.99
$\to 0$	$\to 0$	$\to 1$

The same climb is drawn as a curve here — notice it presses against the dashed $η = 1$ ceiling but never reaches it:

$η \to 1 as T_{C} \to 0, but η < 1 always$

Verify: $η = 1$ would require $T_{C} = 0 K$ exactly, consistent with the parent's " $η < 1$ always." ✓

Recall Why is absolute zero unreachable? (the third-law footnote)

Cooling works by removing disorder (entropy) from a substance — but each step of cooling can only extract a fraction of what remains. As you approach $0 K$ , the leftover entropy you must still remove shrinks, yet so does the amount any single step can pull out, so you need ever more steps and never arrive. Intuition: it is like trying to reach a wall by always stepping halfway there. This is the third law of thermodynamics, and it is why $η = 1$ is a limit you approach but never achieve.

Worked example Example 7 — Cell G: real-world word problem

A geothermal plant taps steam at $200^{\circ} C$ and rejects waste heat into a river at $20^{\circ} C$ . It absorbs $6 \times 1 0^{5} J$ of heat every second from the steam. What is the maximum possible power output (work per second)?

Forecast: The temperatures are modest — do you expect a high or low efficiency for a real plant like this?

Strip the words to numbers, convert to kelvin: $T_{H} = 200 + 273 = 473 K$ , $T_{C} = 20 + 273 = 293 K$ . Why this step? "Steam" and "river" are just the hot and cold reservoirs. Everything else is decoration; the physics needs only $T_{H}$ , $T_{C}$ , and $Q_{H}$ .
Maximum efficiency (Carnot, the upper bound for any real engine): $η = 1 - \frac{293}{473}$ . Compute: $\frac{293}{473} = 0.6194$ , so $η = 1 - 0.6194 = 0.3806 \approx 0.381$ . Why this step? "Maximum possible" is the signal to use Carnot — no real engine between these reservoirs can beat it.
Power = work per second $= η \times (heat absorbed per second) = 0.3806 \times 6 \times 1 0^{5} \approx 2.28 \times 1 0^{5} W$ . Why this step? Since $η = W / Q_{H}$ holds per cycle, it also holds per second when we use rates: $\dot{W} = η \dot{Q}_{H}$ .

$η \approx 38.1%, P_{m a x} \approx 2.28 \times 1 0^{5} W$

Verify: Units: $[dimensionless] \times [J/s] = [W]$ ✓. And $\approx 38%$ is a believable low-temperature-plant ceiling — real geothermal plants run below it, around $10$ – $20%$ .

Worked example Example 8 — Cell H: exam twist ("I beat Carnot!")

An inventor claims an engine that takes $Q_{H} = 1000 J$ from a $600 K$ source, rejects only $Q_{C} = 400 J$ to a $400 K$ sink, and pockets $W = 600 J$ . Is this possible?

Forecast: Their claimed efficiency is $600/1000 = 60%$ . Compare that to what Carnot allows between these reservoirs before deciding.

Carnot ceiling for these reservoirs: $η_{Carnot} = 1 - \frac{400}{600} = 1 - 0.6667 = 0.3333 = 33.3%$ . Why this step? Carnot's theorem says no engine between two fixed reservoirs can beat the reversible efficiency $1 - T_{C} / T_{H}$ .
Inventor's claimed efficiency: $η_{claim} = \frac{W}{Q _{H}} = \frac{600}{1000} = 0.6 = 60%$ . Why this step? Compute their claim on its own terms so we can compare like with like.
Compare: $0.60 > 0.333$ — the claim exceeds the Carnot limit, so it is impossible. Why this step? Beating Carnot would let you build a perpetual motion machine of the second kind, violating the second law.

$Impossible: 60% > η_{Carnot} = 33.3%$

Verify (entropy route): Here is why the entropy criterion is the right test. Each reservoir's entropy changes by (heat it receives) $/$ (its temperature). The hot source loses $Q_{H}$ at $T_{H}$ , so its entropy changes by $- Q_{H} / T_{H}$ . The cold sink gains $Q_{C}$ at $T_{C}$ , so its entropy changes by $+ Q_{C} / T_{C}$ . The gas returns to its start, so its own entropy change over a cycle is zero. The second law demands the total entropy of the universe never decreases: $Δ S_{univ} = \frac{Q _{C}}{T _{C}} - \frac{Q _{H}}{T _{H}} \geq 0.$ Test the claim: $\frac{400}{400} - \frac{1000}{600} = 1 - 1.667 = - 0.667 < 0$ ✗. A negative total entropy change is forbidden, confirming the engine cannot exist. ✓

Worked example Example 9 — Cell I: work back to a volume ratio

One mole of ideal gas undergoes the isothermal expansion (step 1) at $T_{H} = 500 K$ and absorbs $Q_{H} = 2882 J$ . Take $R = 8.314 J K^{- 1} mol^{- 1}$ (the gas constant, defined above). Find the volume ratio $V_{2} / V_{1}$ .

Forecast: The heat absorbed grows with $ln (V_{2} / V_{1})$ . If $Q_{H}$ is a bit less than $R T_{H}$ , do you expect the ratio to be near $e$ (about $2.7$ ), or much larger?

During an isothermal step the temperature is constant, so $Δ U = 0$ (internal energy depends only on temperature) and all absorbed heat becomes expansion work. Integrating $P = R T_{H} / V$ over the volume gives the log formula $Q_{H} = R T_{H} ln \frac{V _{2}}{V _{1}} .$ Why this step? $\int d V / V = ln V$ , so the logarithm is exactly what pops out when pressure falls as $1/ V$ — this is why $ln$ , and not any other function, appears here.
Solve for the ratio: $ln \frac{V _{2}}{V _{1}} = \frac{Q _{H}}{R T _{H}} = \frac{2882}{8.314 \times 500} = \frac{2882}{4157} = 0.6933$ . Why this step? Dividing isolates the log; we "undo" the multiplication by $R T_{H}$ .
Exponentiate to remove the log: $\frac{V _{2}}{V _{1}} = e^{0.6933} = 2.000$ . Why this step? $ln$ and $e$ are inverse operations — $e^{l n x} = x$ (that is the whole point of $e$ , defined above) — which releases the ratio trapped inside the logarithm.

$\frac{V _{2}}{V _{1}} = 2.0$

Verify: Plug back: $R T_{H} ln 2 = 8.314 \times 500 \times 0.6931 = 2882 J$ ✓ recovers the given $Q_{H}$ .

Recall Why the compression step must share this ratio

The two adiabatic legs of the cycle obey $T V^{γ - 1} = const$ . Writing that for the hot→cold leg ( $T_{H} V_{2}^{γ - 1} = T_{C} V_{3}^{γ - 1}$ ) and the cold→hot leg ( $T_{H} V_{1}^{γ - 1} = T_{C} V_{4}^{γ - 1}$ ) and dividing one by the other cancels the temperatures and the exponent, forcing $V_{2} / V_{1} = V_{3} / V_{4}$ . So here $V_{3} / V_{4} = 2.0$ as well — and that equality is why the two logs cancel and only $T_{C} / T_{H}$ survives in the efficiency.

Worked example Example 10 — Cell J: sign edge

$T_{C} > T_{H}$ (negative $η$ ) By mistake someone labels the reservoirs backwards and plugs $T_{H} = 300 K$ , $T_{C} = 400 K$ into the engine formula. What comes out, and what does the sign mean?

Forecast: The "cold" reservoir is now hotter than the "hot" one. Do you expect $η$ above 1, exactly 0, or below 0?

Blindly apply the formula: $η = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{400}{300} = 1 - 1.3333 = - 0.3333$ . Why this step? When the reservoir labelled "cold" is actually the hotter one, the ratio $T_{C} / T_{H}$ exceeds $1$ , so $1$ minus it goes negative.
Read the sign physically. A negative $η$ cannot be a real engine efficiency (you cannot get negative work out for positive heat in). It signals that heat naturally flows from the hotter body ( $400 K$ ) to the cooler one ( $300 K$ ) on its own — no engine here. To move heat the other way you must supply work: this is the heat-pump / refrigerator regime (Cell D), not an engine. Why this step? The formula has no built-in referee; the sign is the referee. A negative result is the maths telling you the two reservoirs are swapped and you are in the reverse (heat-pump) situation.

$η = - 0.333 \Rightarrow not an engine; this is a heat-pump regime$

Verify: Swap the labels back to the physical roles ( $T_{H} = 400$ , $T_{C} = 300$ ) and the honest engine efficiency is $1 - 300/400 = 0.25 > 0$ ✓ — a sensible positive number. The negative value was purely a symptom of mislabelling, and its sign correctly flagged the reversed regime.

Recall Which cell did each example fill?

Cell A (clean) ::: Ex 1 Cell B (Celsius conversion) ::: Ex 2 Cell C (find $Q_{C}$ , $W$ ) ::: Ex 3 Cell D (refrigerator / COP) ::: Ex 4 Cell E (degenerate $T_{H} = T_{C}$ ) ::: Ex 5 Cell F (limit $T_{C} \to 0$ ) ::: Ex 6 Cell G (word problem) ::: Ex 7 Cell H (beat-Carnot twist) ::: Ex 8 Cell I (invert to volume ratio) ::: Ex 9 Cell J (sign edge $T_{C} > T_{H}$ ) ::: Ex 10