Heat engines — efficiency η = 1 − Q_C - Q_H: Worked examples

Before touching numbers, four reminders in plain words so no symbol is a mystery:

Definition The letters, in words

$Q_{H}$ = the heat (energy) poured in from the hot side, always a positive count of joules.
$Q_{C}$ = the heat dumped out to the cold side, also a positive count of joules.
$W$ = the useful work the engine hands you, in joules.
$Δ U$ = the change in internal energy of the gas — how much energy is stored inside the working substance now versus at the start. (The symbol $Δ$ means "change in"; $U$ is the stored internal energy.) Over one complete cycle the gas comes back to exactly its starting state, so this stored energy is unchanged: $Δ U = 0$ .
$T_{H}$ , $T_{C}$ = the absolute (kelvin) temperatures of the hot and cold reservoirs. "Absolute" means measured from true zero, $0 K$ ; never Celsius, because we will divide temperatures and only kelvin ratios are physically meaningful. All heats and work are magnitudes (sizes) here — we never write a negative $Q_{H}$ ; direction is baked into which letter it is.

Intuition The one picture behind every example (see below)

Before any arithmetic, burn this into your eye: heat "falls" from hot to cold like water down a wheel, and the engine skims off work $W$ on the way down. The figure shows the flow — a fat yellow pipe $Q_{H}$ coming down from the hot block, a thinner blue pipe $Q_{C}$ leaving to the cold block (the thinner pipe is labelled "smaller" right on the figure), and the pink arrow $W$ tapped off the side. Every example is just this picture with different numbers on the pipes.

Figure description (alt-text): A vertical chalkboard sketch. A pale-yellow HOT reservoir block sits at the top, a blue COLD reservoir block at the bottom, and a chalk-outlined ENGINE box in the middle. A thick yellow downward arrow labelled "Q_H (heat in, big)" enters the engine from the hot block; a thinner blue downward arrow labelled "Q_C (dumped, smaller)" leaves the engine into the cold block; a pink rightward arrow labelled "W = Q_H − Q_C (work out)" leaves the engine's side. The relation "eta = W / Q_H" is written beside the engine.

The scenario matrix

Every heat-engine question is one (or a blend) of these cells. Each row is a "type of shot"; the last column names the example that hits it. (Reminder throughout: $η$ means $W / Q_{H}$ — useful work out over heat paid in.)

Cell	What's given / what's weird	What you must find	Example
A. Forward basics	$Q_{H}, Q_{C}$ given	$W$ , $η = W / Q_{H}$	Ex 1
B. Reverse solve	$η$ and $W$ given	$Q_{H}$ , $Q_{C}$	Ex 2
C. Carnot ceiling check	$T_{H}, T_{C}$ , claimed $η = W / Q_{H}$	Is it legal?	Ex 3
D. Degenerate: $Q_{C} = 0$	zero waste heat	Is it possible?	Ex 4
E. Degenerate: $Q_{C} = Q_{H}$	all heat dumped	value of $η$	Ex 4
F. Limit: $T_{C} \to 0$ or $T_{H} \to \infty$	extreme temperatures	limiting $η$	Ex 5
G. Unit trap: Celsius given	°C temperatures	correct Carnot $η$	Ex 6
H. Real-world word problem	power (watts), time, fuel	efficiency, dumped heat rate	Ex 7
I. Exam twist: two engines in series	waste of one feeds next	combined $η$	Ex 8
J. PV-loop area	net work as loop area	$η$ from a cycle sketch	Ex 9

We work them in order; each is labelled with its cell.

Worked examples

Cell A — Forward basics

Worked example Example 1 (Cell A)

An engine takes in $Q_{H} = 800 J$ and rejects $Q_{C} = 600 J$ each cycle. Find $W$ and $η$ .

Forecast: Guess — is this a good engine or a poor one? (It dumps most of what it eats, so expect low $η$ .)

Step 1. $W = Q_{H} - Q_{C} = 800 - 600 = 200 J$ . Why this step? Over a full cycle the gas returns to its start, so its stored internal energy is unchanged, $Δ U = 0$ (the "change in internal energy" defined above). By the First Law of Thermodynamics, with no net change in stored energy, every joule of net heat becomes work.

Step 2. $η = \frac{W}{Q _{H}} = \frac{200}{800} = 0.25 = 25%$ . Why this step? Efficiency is what you get ( $W$ ) over what you pay ( $Q_{H}$ ).

Verify: Second form $1 - Q_{C} / Q_{H} = 1 - 600/800 = 1 - 0.75 = 0.25$ ✓. Units: J/J is dimensionless, correct for a fraction. As forecast, it's a poor engine.

Cell B — Reverse solve

Cell C — Carnot ceiling check

Worked example Example 3 (Cell C)

A salesman's engine runs between $T_{H} = 500 K$ and $T_{C} = 350 K$ (both kelvin, as flagged in the definitions above) and claims $η = 0.32$ (recall $η = W / Q_{H}$ ). Legal or fraud?

Forecast: Guess whether $0.32$ can squeeze under the reversible ceiling.

Step 1. Best possible: $η_{Carnot} = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{350}{500} = 1 - 0.70 = 0.30$ . Why this step? The Carnot Cycle gives the absolute ceiling for any engine between these two temperatures. We use kelvin so the ratio $T_{C} / T_{H}$ is meaningful.

Step 2. Compare: claimed $0.32 > 0.30 = η_{Carnot}$ — impossible. Why this step? Beating the reversible bound would violate the Second Law of Thermodynamics.

Verify: $η_{Carnot} = 0.30$ ; the claim exceeds it by $0.02$ , so it is fraudulent. (If they had claimed $0.29$ , it would be legal though impressively close to reversible.)

Cells D & E — Degenerate inputs

Worked example Example 4 (Cells D and E)

Examine two extreme "engines": (a) one that dumps nothing, $Q_{C} = 0$ ; (b) one that dumps everything, $Q_{C} = Q_{H}$ . Find $η$ and say whether each can exist.

Forecast: One is a dream machine, one is useless. Guess which $η$ is 1 and which is 0.

Step 1 — case (a), $Q_{C} = 0$ . $η = 1 - \frac{0}{Q _{H}} = 1 = 100%$ . Why this step? Plug the degenerate value straight into $η = 1 - Q_{C} / Q_{H}$ .

Step 2 — is (a) real? No. $Q_{C} = 0$ means heat fully converted to work with no other effect — forbidden by the Kelvin–Planck form of the Second Law of Thermodynamics. It's the unreachable ceiling $η = 1$ .

Step 3 — case (b), $Q_{C} = Q_{H}$ . $η = 1 - \frac{Q _{H}}{Q _{H}} = 1 - 1 = 0 = 0%$ . Why this step? Same formula, other extreme.

Step 4 — meaning of (b). With $η = 0$ , $W = Q_{H} - Q_{C} = 0$ : heat flows in and straight back out doing no work. Physically allowed (it's just heat leaking through), but it isn't an engine — it's a conductor.

Verify: $W_{(a)} = Q_{H} - 0 = Q_{H}$ (all work, forbidden); $W_{(b)} = Q_{H} - Q_{H} = 0$ (no work). The whole allowed range is $0 \leq η < 1$ , and these two cases sit exactly on its endpoints. ✓

Cell F — Limiting behaviour

Worked example Example 5 (Cell F)

A Carnot engine keeps $T_{H} = 400 K$ fixed. (a) Push the cold reservoir toward absolute zero, $T_{C} \to 0 K$ . (b) Instead fix $T_{C} = 300 K$ and push $T_{H} \to \infty$ . What does $η$ approach in each? (All temperatures kelvin.)

Forecast: Both should push efficiency up — but to what number?

Step 1 — case (a). $η = 1 - \frac{T _{C}}{T _{H}} \to 1 - \frac{0}{400} = 1$ . Why this step? As $T_{C} \to 0$ the wasted-fraction term $T_{C} / T_{H} \to 0$ , so $η$ climbs to its ceiling.

Step 2 — sample point for (a). At $T_{C} = 40 K$ : $η = 1 - 40/400 = 0.90 = 90%$ . Colder cold ⇒ higher $η$ . Why this step? A finite numeric point makes the trend visible: we shrink $T_{C}$ and watch $η$ rise toward the ceiling, so we can trust the limit isn't just algebra.

Step 3 — is the case-(a) ceiling reachable? No. Reaching $T_{C} = 0 K$ exactly is forbidden by the third law of thermodynamics, so $η = 1$ is approached but never met. Why this step? A limit tells us where we're heading, not whether we can arrive; we must separately state physics that blocks the endpoint, exactly as the Second Law blocked $Q_{C} = 0$ in Example 4.

Step 4 — case (b). $η = 1 - \frac{300}{T _{H}}$ . As $T_{H} \to \infty$ , the fraction $300/ T_{H} \to 0$ , so $η \to 1$ . Sample: at $T_{H} = 3000 K$ , $η = 1 - 300/3000 = 0.90 = 90%$ . Why this step? This is the mirror-image limit — instead of chilling the cold side we heat the hot side, and the same wasted-fraction term vanishes, confirming "bigger temperature gap ⇒ higher $η$ ."

Verify: Both limits give $η \to 1$ but never $= 1$ ; the sample points $0.90$ and $0.90$ confirm the trend "bigger temperature gap ⇒ higher efficiency" from the parent's water-wheel picture. ✓

Cell G — The Celsius trap

Worked example Example 6 (Cell G)

An engine works between $T_{H} = 227^{\circ} C$ and $T_{C} = 27^{\circ} C$ . Find the maximum efficiency. (Watch the units!)

Forecast: A student who forgets to convert will compute $1 - 27/227 \approx 0.88$ . Guess whether the true answer is higher or lower.

Step 1. Convert to kelvin: $T (K) = T (^{\circ} C) + 273.15$ . $T_{H} = 227 + 273.15 = 500.15 K$ , $T_{C} = 27 + 273.15 = 300.15 K$ . Why this step? The Carnot formula divides temperatures, and only absolute temperature makes the ratio physically meaningful. Celsius has an arbitrary zero, so ratios of °C are nonsense.

Step 2. $η_{Carnot} = 1 - \frac{300.15}{500.15} = 1 - 0.60012 = 0.39988 \approx 0.40 = 40%$ . Why this step? Apply the ceiling formula with correct units.

Verify: The wrong Celsius answer would be $1 - 27/227 = 0.881$ — far too optimistic. The true $\approx 40%$ is much lower, confirming the trap. (Rounding to whole degrees, $T = 500, 300$ gives exactly $0.40$ .) ✓

Cell H — Real-world word problem

Worked example Example 7 (Cell H)

A power plant outputs $P_{W} = 300 MW$ of electrical power at efficiency $η = 0.40$ . (a) At what rate does it absorb heat from its fuel? (b) At what rate does it dump waste heat into the river? (c) How much fuel-heat does it use in $1$ hour?

Forecast: With 40% efficiency, guess whether the river-dumping rate is bigger or smaller than the 300 MW of useful output.

Step 1 — heat-in rate. Power is just energy per second, so the same algebra as Example 2 works on rates: $\dot{Q}_{H} = \frac{P _{W}}{η} = \frac{300}{0.40} = 750 MW$ . Why this step? Dividing every quantity in $η = W / Q_{H}$ by the cycle time turns joules into watts; the ratio $η$ is unchanged. (The dot in $\dot{Q}$ just means "per second".)

Step 2 — waste-heat rate. $\dot{Q}_{C} = \dot{Q}_{H} - P_{W} = 750 - 300 = 450 MW$ . Why this step? Energy conservation applied to rates: what isn't turned into work is dumped, per second.

Step 3 — fuel-heat in one hour. First recall the prefix and time facts: $1 MW = 1 MJ per second$ (a megawatt is a megajoule each second), and $1 h = 3600 s$ . Energy $=$ power $\times$ time, so $Q_{H} = 750 MW \times 3600 s = 750 \frac{MJ}{s} \times 3600 s = 2700000 MJ = 2.7 \times 1 0^{6} MJ .$ Since $1 MJ = 1 0^{6} J$ , converting once more: $2.7 \times 1 0^{6} MJ = 2.7 \times 1 0^{6} \times 1 0^{6} J = 2.7 \times 1 0^{12} J$ . Why this step? Writing $MW = MJ / s$ makes the seconds cancel cleanly, and stating $1 MJ = 1 0^{6} J$ shows exactly how the two forms of the same answer relate.

Verify: $\dot{Q}_{C} = 450 MW > P_{W} = 300 MW$ — the plant dumps more into the river than it delivers as electricity, matching the parent note's warning. Rate check: $300/750 = 0.40 = η$ ✓. Energy check: $750 \times 3600 = 2700000 MJ$ ✓.

Cell I — Exam twist: two engines in series

Worked example Example 8 (Cell I)

Engine 1 runs between $600 K$ and $400 K$ ; its waste heat feeds engine 2, which runs between $400 K$ and $300 K$ . Both are ideal (Carnot). Engine 1 absorbs $Q_{H} = 1200 J$ . Find the total work and the combined efficiency (work out ÷ original heat in).

Forecast: Guess whether stacking two engines beats a single engine running directly from $600 K$ to $300 K$ .

Step 1 — engine 1. $η_{1} = 1 - 400/600 = 1/3$ . So $W_{1} = η_{1} Q_{H} = \frac{1}{3} (1200) = 400 J$ . Why this step? Each engine individually obeys the Carnot ceiling.

Step 2 — heat passed on. Engine 1 dumps $Q_{C, 1} = Q_{H} - W_{1} = 1200 - 400 = 800 J$ into the $400 K$ reservoir, which is engine 2's input. Why this step? "Waste feeds the next engine" means $Q_{H, 2} = Q_{C, 1}$ .

Step 3 — engine 2. $η_{2} = 1 - 300/400 = 1/4$ . So $W_{2} = η_{2} \cdot 800 = \frac{1}{4} (800) = 200 J$ . Why this step? Same Carnot rule with engine 2's temperatures.

Step 4 — totals. $W_{total} = W_{1} + W_{2} = 400 + 200 = 600 J$ . Combined efficiency $η_{tot} = \frac{W _{total}}{Q _{H}} = \frac{600}{1200} = 0.50$ . Why this step? Combined efficiency uses the original heat input $Q_{H} = 1200 J$ .

Verify: A single Carnot engine straight from $600 K$ to $300 K$ gives $1 - 300/600 = 0.50$ — identical. Two ideal engines in series equal one big ideal engine across the full temperature drop. ✓ Algebraic cross-check, spelling out every factor: $η_{tot} = η_{1} + η_{2} \cdot \frac{Q _{C, 1}}{Q _{H}}$ . Here $\frac{Q _{C, 1}}{Q _{H}} = \frac{800}{1200} = \frac{2}{3}$ (this $2/3$ is exactly the fraction of the original heat that survives engine 1 and enters engine 2). So $η_{tot} = \frac{1}{3} + \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ ✓.

Cell J — Efficiency from a PV loop area

Worked example Example 9 (Cell J)

A gas runs the rectangular cycle sketched below on a PV diagram, traversed clockwise. The enclosed rectangle has width $Δ V = 2 \times 1 0^{- 3} m^{3}$ and height $Δ P = 1 \times 1 0^{5} Pa$ . Heat is absorbed only during the top leg and the left-hand rising leg, totalling $Q_{H} = 500 J$ . Find $W$ and $η$ .

Figure description (alt-text): A chalkboard PV diagram (pressure on the vertical axis, volume on the horizontal). A rectangle is drawn and its interior shaded pink. Small arrows show the cycle running clockwise: the top edge is traversed left-to-right (yellow, labelled "heat absorbed Q_H"), the right edge downward (blue), the bottom edge right-to-left (chalk, labelled "heat rejected Q_C"), and the left edge upward (blue). The shaded area is labelled "W = area = dP x dV = 200 J", with the width labelled "dV = 2e-3 m^3" and the height "dP = 1e5 Pa".

Forecast: The net work is the area of the loop. Guess roughly what fraction of the 500 J it will be.

Step 1 — why those legs carry $Q_{H}$ . On the top and left-rising legs the gas is expanding and/or being pushed to higher pressure, so its temperature (hence internal energy) rises and it does outward work — both require energy in, so heat flows into the gas there ( $Q_{H}$ ). On the right-falling and bottom legs the gas contracts and cools, so heat flows out ( $Q_{C}$ ). That is why $Q_{H} = 500 J$ is assigned to the outward legs. Why this step? The First Law $Q = Δ U + W_{by gas}$ tells us heat enters exactly where internal energy and outward work both demand energy; naming which legs are $Q_{H}$ is not a guess, it follows from the direction of travel.

Step 2 — net work = loop area. For a rectangle, area $= Δ P \times Δ V = (1 \times 1 0^{5}) (2 \times 1 0^{- 3}) = 200 J$ . Why this step? On a PV diagram, work done by the gas is $\int P d V$ ; over a clockwise closed loop this integral equals the enclosed area (a clockwise loop gives positive net work — look at the shaded pink region in the figure). Units: $Pa \cdot m^{3} = J$ .

Step 3 — efficiency. $η = \frac{W}{Q _{H}} = \frac{200}{500} = 0.40 = 40%$ . Why this step? Same master definition; the loop area is the useful work.

Step 4 — dumped heat. $Q_{C} = Q_{H} - W = 500 - 200 = 300 J$ . Why this step? Energy conservation closes the books.

Verify: $1 - Q_{C} / Q_{H} = 1 - 300/500 = 0.40$ ✓. Area came out to $200 J$ , exactly $40%$ of the $500 J$ input — matching the forecast that only part of the heat becomes the loop's area. ✓

Active recall

Recall Which cell is which? (cover the answers)

"Given both heats, find $η$ " ::: Cell A (Ex 1).
"Given $η$ and $W$ , find the heats" ::: Cell B (Ex 2).
"Is a claimed efficiency legal?" ::: Cell C — compare to Carnot (Ex 3).
" $Q_{C} = 0$ gives $η = ?$ and is it real?" ::: $η = 1$ , forbidden by Second Law (Ex 4).
"Two engines in series across the full temperature drop" ::: equals one Carnot engine across that drop (Ex 8).
"Net work from a PV loop" ::: the enclosed area of a clockwise loop (Ex 9).

Connections

Parent topic (Hinglish) — the master formulas worked here.
First Law of Thermodynamics — the source of $W = Q_{H} - Q_{C}$ .
Second Law of Thermodynamics — kills Cells C and D (no $η \geq η_{Carnot}$ , no $η = 1$ ).
Carnot Cycle — the ceiling used in Ex 3, 5, 6, 8.
Entropy — the deeper reason the ceiling exists.
Refrigerators and Heat Pumps — run any of these engines backwards.
PV diagrams and Work — the loop-area method of Ex 9.