Heat engines — efficiency η = 1 − Q_C - Q_H: Visual walkthrough

Step 1 — Draw the machine as a box with three pipes

WHAT. Forget equations. Picture a sealed box. Three things cross its boundary: heat flows in from something hot, heat flows out into something cold, and a shaft pokes out to spin a wheel (that is the useful work).

WHY. Before we can conserve energy, we must name every channel through which energy enters or leaves. Miss one, and the accounting breaks. A picture guarantees we caught them all — you can literally count the arrows crossing the box.

PICTURE. In the figure below:

The amber block on top is the hot reservoir, a huge tank of heat at temperature $T_{H}$ .
The cyan block on the bottom is the cold reservoir at temperature $T_{C}$ .
The big amber arrow pouring down into the box is $Q_{H}$ : the heat we pay for.
The thin cyan arrow leaving the bottom is $Q_{C}$ : the heat we are forced to dump.
The white arrow out the side is $W$ : the work we harvest.

We treat $Q_{H}$ and $Q_{C}$ as magnitudes (always positive) and track their direction with the arrows, not with signs. That choice keeps the pictures clean.

Step 2 — Insist the machine returns to its exact starting state

WHAT. Inside the box is a trapped bit of gas — the working substance. We demand the engine runs in a cycle: after one full loop, that gas is back to the same condition it started in.

WHY. A car engine must fire again and again. If the gas ended each loop hotter or more spread out, it would eventually be "used up." So we require it to reset. This single requirement is the secret lever that forces waste heat to exist — watch it work in Step 5.

Before the picture, two words we will use, so we define them now:

PICTURE. The figure traces the gas's state as a loop on this $P$ – $V$ graph. The path leaves the start dot, wanders as the gas is heated, expands, cools and is compressed, and comes back to the same dot. That closed loop is the visual meaning of "cycle."

Because the loop returns to the same dot, the internal energy ends exactly where it began: $Δ U_{cycle} = 0$

Step 3 — Apply energy conservation to the box (the First Law)

WHAT. We now write down that energy is conserved for the box. First name the one new symbol we are about to use: let $Q_{net}$ stand for the ==net heat that flows into the gas over the cycle== — that is, all the heat added minus all the heat removed, added up as a single number. The energy stored inside then changes by this net heat minus the work that flowed out.

WHY. This is the First Law of Thermodynamics — the only physics we truly need. Everything else is bookkeeping. We use it because it is the exact statement of "energy cannot be created or destroyed," specialised to a system that trades heat and work.

PICTURE. The figure shows a balance scale: on one pan, the energy the gas gains; on the other, net heat in minus work out. They must match.

Now, with $Q_{net}$ defined, the First Law reads:

$change in stored energy of the gas Δ U = net heat that flowed IN Q_{net} - work done BY the gas W$

Read it as a sentence: the energy inside went up by whatever net heat came in, minus whatever work leaked out as a spinning shaft.

Step 4 — Add up the net heat from the two pipes

WHAT. The box has heat flowing in ( $Q_{H}$ ) and heat flowing out ( $Q_{C}$ ). The net heat $Q_{net}$ (just named in Step 3) is their difference.

WHY. In Step 3 the term $Q_{net}$ was a single lump. But our box (Step 1) has two heat pipes pointing in opposite directions. To fill in that lump we must combine them, respecting direction: in counts positive, out counts negative.

PICTURE. The figure lines up the two heat arrows tip-to-tail on a number line: a long amber arrow to the right ( $+ Q_{H}$ ), then a shorter cyan arrow back to the left ( $- Q_{C}$ ). What's left over is $Q_{net}$ .

$Q_{net} = in from hot (+) Q_{H} - out to cold (-) Q_{C}$

The minus sign on $Q_{C}$ is doing real work: it encodes that this heat is leaving, so it subtracts from what the box keeps.

Step 5 — Collide Step 2 and Step 4 to get the work

WHAT. Substitute $Δ U = 0$ (Step 2) and $Q_{net} = Q_{H} - Q_{C}$ (Step 4) into the First Law (Step 3), then solve for $W$ .

WHY. This is the payoff. The cycle condition ( $Δ U = 0$ ) turns the First Law from a statement about stored energy into a statement purely about heat and work — exactly what we can measure at the pipes.

PICTURE. The figure shows the balance scale from Step 3, but now the " $Δ U$ " pan is pinned to zero. That forces the other pan to balance itself: heat-in must equal heat-out plus work-out.

$0 = (Q_{H} - Q_{C}) - W ⟹ W = Q_{H} - Q_{C}$

In words: the work you harvest is exactly the heat you didn't dump. Every joule that entered either left as work or left as waste heat — there is no third exit and nothing stays behind (because $Δ U = 0$ ).

Step 6 — Define efficiency as "got out ÷ paid in"

WHAT. Efficiency $η$ is the fraction of the heat you paid for that came back as work.

WHY. We need a single fair number to compare engines. "What you get divided by what you pay" is the honest measure — it's a pure ratio, unit-free, so a tiny toy engine and a giant power plant can be compared directly.

PICTURE. The figure draws $Q_{H}$ as a full wide pipe entering, and $W$ as the narrower stream that survives to the output. $η$ is the ratio of pipe widths.

$η \equiv \frac{W}{Q _{H}} = \frac{Q _{H} - Q _{C} work (Step 5)}{heat paid in Q _{H}}$

Now split the fraction into two pieces — this is just arithmetic, $\frac{a - b}{a} = 1 - \frac{b}{a}$ :

$η = \frac{Q _{H}}{Q _{H}} - \frac{Q _{C}}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}}$

Term by term: the $1$ is the dream (all heat becomes work); the $\frac{Q _{C}}{Q _{H}}$ is the fraction thrown away, subtracted off. Efficiency is one minus the waste.

Step 7 — Walk every case, including the degenerate ones

WHAT. Check the formula at its extremes so no scenario surprises you.

WHY. A formula you trust is one you've tested at its edges. We push $Q_{C}$ to its two limits and check the answer stays sane.

PICTURE. Three side-by-side "pipe" diagrams: the realistic middle case, and the two impossible extremes at the ends.

Case A — $Q_{C} = 0$ (dump nothing): then $η = 1 - 0 = 1$ , a perfect engine. The picture shows the output stream as wide as the input. Forbidden by the Second Law — this is the upper edge $η$ can approach but never reach.
Case B — realistic, $0 < Q_{C} < Q_{H}$ : then $0 < η < 1$ . The output stream is genuinely narrower than the input. This is every real engine.
Case C — $Q_{C} = Q_{H}$ (dump everything): then $η = 1 - 1 = 0$ . All the heat flows straight through, no work out. The picture shows zero output stream. Useless, but not impossible.

So $η$ is always pinned in the band: $0 \leq η < 1$ It never goes negative (that would need $Q_{C} > Q_{H}$ — more heat out than in, impossible for a heat engine), and it never reaches $1$ .

Step 8 — The ceiling: how close to 1 can you get?

WHAT. For the best-possible (reversible) engine sliding heat between temperatures $T_{H}$ and $T_{C}$ , the waste ratio equals the temperature ratio, $\frac{Q _{C}}{Q _{H}} = \frac{T _{C}}{T _{H}}$ , which drops the efficiency straight to $η = 1 - T_{C} / T_{H}$ .

WHY. Step 7 says $η < 1$ , but how much less? The gap depends on the temperatures — and here is the intuition for why heat carries a temperature "tag":

Intuition Why

$Q_{C} / Q_{H} = T_{C} / T_{H}$ for the best engine (no equations) Think of heat as a crowd of people and temperature as how spread out (disordered) that crowd is. A quantity called Entropy measures that spread; for heat $Q$ swapped at temperature $T$ the entropy handed over is the ratio $Q / T$ — more heat, or a colder handover, means more disorder moved.

The best engine is reversible: it creates no extra disorder. So the disorder it drops off at the cold side must exactly equal the disorder it picked up at the hot side.
Equal disorder in and out means $\frac{Q _{H}}{T _{H}} = \frac{Q _{C}}{T _{C}}$ (entropy taken = entropy given).
Rearrange that single balance and you get $\frac{Q _{C}}{Q _{H}} = \frac{T _{C}}{T _{H}}$ . A real engine makes extra disorder, so it dumps more than this — its $Q_{C}$ is larger and its efficiency lower.

So substituting this best-case ratio into $η = 1 - Q_{C} / Q_{H}$ :

PICTURE. The figure plots the Carnot ceiling $η_{Carnot} = 1 - T_{C} / T_{H}$ against $T_{H}$ for a fixed $T_{C}$ : it rises toward $1$ but only reaches it if $T_{H} \to \infty$ or $T_{C} \to 0$ .

$η_{Carnot} = 1 - \frac{T _{C}}{T _{H}} (T in kelvin)$

The bigger the temperature gap, the closer to the perfect $1$ you creep — but the flat asymptote in the figure is the wall you can never climb over. (The full machine is the Carnot Cycle.)

The one-picture summary

The earlier figures each froze one step. This final figure does something new: it overlays the whole chain of logic onto a single energy river, colour-coding which step owns which piece — so you can read the derivation top to bottom in one glance instead of nine. The heat $Q_{H}$ enters (Step 1), the cycle condition splits it (Steps 2–5), work $W$ branches off, waste $Q_{C}$ drains away, and the width ratio of the work branch to the inflow is the efficiency (Step 6), with the temperature wall of Step 8 marked at the top.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a box with a hot pipe coming in from the top, a cold pipe leaving the bottom, and a shaft sticking out the side that spins a wheel. Heat rushes in the top; some of it grabs the shaft and spins the wheel (that's your work); the rest tumbles out the bottom into the cold sink.

Now the trick: we insist the guts of the box come back to exactly how they started after each loop, so no energy hides inside. That means every joule that came in the top has to leave — either as spin, or as dumped heat. So work = heat in minus heat dumped. Simple counting.

Efficiency is just asking: of all the heat I bought, what fraction came out as spin? That's work over heat-in, which is the same as one minus (dumped over bought). The dumped part can never be zero — nature won't let you turn all the heat into spin — so efficiency always falls short of a perfect $1$ . And the absolute best you could ever do is set by the two temperatures: the best engine spreads out exactly as much disorder as it soaks up, which pins its waste ratio to $T_{C} / T_{H}$ , giving $η = 1 - T_{C} / T_{H}$ . Hotter hot and colder cold buy you more spin, but the wall at $1$ never moves.

Connections

First Law of Thermodynamics — the energy-balance scale of Steps 3–5.
Second Law of Thermodynamics — why the $Q_{C} = 0$ case (Step 7A) is forbidden.
Carnot Cycle — the reversible engine behind Step 8's ceiling.
Entropy — the disorder ratio $Q / T$ that pins $Q_{C} / Q_{H} = T_{C} / T_{H}$ .
PV diagrams and Work — the closed loop of Step 2, and $W$ as its enclosed area.
Refrigerators and Heat Pumps — flip every arrow in the summary figure.