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

Before any symbols: the internal energy $U$ is just the total thermal energy stored in the working gas — how "hot and full" it is. Over one complete cycle the gas returns to its exact starting state, so its stored energy comes back to where it began; we write that "no net change" as $Δ U = 0$ , where $Δ$ (delta) means "final minus initial".

Throughout, all heats are written as magnitudes (positive numbers): $Q_{H}$ is heat taken in, $Q_{C}$ is heat dumped out, $W$ is work out. The two facts we lean on: $W = Q_{H} - Q_{C} η = \frac{W}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}}$ and the ceiling for any engine between hot temperature $T_{H}$ and cold temperature $T_{C}$ (in kelvin): $η_{Carnot} = 1 - \frac{T _{C}}{T _{H}}$

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

The picture above is the accountant's view of every engine: heat in the red top pipe ( $Q_{H}$ ), a smaller amount out the green side arrow as useful work ( $W$ ), and the leftover leaking out the blue bottom pipe ( $Q_{C}$ ). Whatever goes in must come out — that is the whole First Law here. Keep this three-arrow picture in mind: every exercise below is just asking for one of those three arrows given the other two.

Level 1 — Recognition

Goal: read the definition, put numbers in the right slots.

Recall Solution L1-a

Step 1 — Work. In the figure's language, we know the red pipe ( $Q_{H}$ ) and the blue pipe ( $Q_{C}$ ) and want the green arrow ( $W$ ). Whatever heat is not dumped becomes work: $W = Q_{H} - Q_{C} = 800 - 600 = 200 J .$ Why this step? Over one cycle the gas returns to its start, so its internal energy is unchanged, $Δ U = 0$ , and the First Law forces every joule of net heat to leave as work. Step 2 — Efficiency. "What you get over what you pay": $η = \frac{W}{Q _{H}} = \frac{200}{800} = 0.25 = 25%.$ Cross-check. $1 - Q_{C} / Q_{H} = 1 - 600/800 = 0.25$ ✓ — both forms of $η$ agree.

Recall Solution L1-b

Step 1. Use the temperature ceiling directly (temperatures already in kelvin): $η_{Carnot} = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{300}{400} = 0.25 = 25%.$ Why kelvin? The formula measures how far apart the two reservoirs are on the absolute scale, where $0 K$ means "no thermal energy at all". Only on that scale does a ratio of temperatures carry physical meaning.

Level 2 — Application

Goal: rearrange the same two equations to solve for the unknown that isn't handed to you.

Recall Solution L2-a

Step 1 — Heat in. Turn $η = W / Q_{H}$ inside-out: $Q_{H} = \frac{W}{η} = \frac{500}{0.40} = 1250 MJ .$ Why? $η$ tells us $W$ is $40%$ of $Q_{H}$ ; to recover the whole from a known part, divide by the fraction. Step 2 — Heat dumped. Everything not converted is rejected: $Q_{C} = Q_{H} - W = 1250 - 500 = 750 MJ .$ Insight. The plant throws away $750 MJ$ — more than the $500 MJ$ it delivers. Good engines are still mostly heat-dumpers.

Recall Solution L2-b

Step 1 — Relate $Q_{C}$ to $Q_{H}$ . Since $η = 1 - Q_{C} / Q_{H}$ , the dumped fraction is $Q_{C} / Q_{H} = 1 - η = 0.70$ . So $Q_{H} = \frac{Q _{C}}{1 - η} = \frac{420}{0.70} = 600 J .$ Why this step? We were given the waste number $Q_{C}$ , so we anchor on the waste fraction $1 - η$ , not on $η$ itself. Step 2 — Work. $W = Q_{H} - Q_{C} = 600 - 420 = 180 J .$ Cross-check. $η = W / Q_{H} = 180/600 = 0.30$ ✓.

Level 3 — Analysis

Goal: compare against a limit, or reason about feasibility — no single plug-in.

Recall Solution L3-a

Step 1 — Find the ceiling. $η_{Carnot} = 1 - \frac{300}{500} = 1 - 0.60 = 0.40 = 40%.$ Why? No engine — real or ideal — can exceed the reversible Carnot value between the same two temperatures. Step 2 — Compare. Claimed $0.45 > 0.40$ ceiling → impossible. It exceeds the maximum by $5$ percentage points, which would violate the Second Law. Reading it physically: to beat Carnot you'd have to dump less heat than even a perfectly reversible engine, decreasing the entropy of the universe — nature never allows that.

Recall Solution L3-b

Engine A. $W_{A} = Q_{H} - Q_{C} = 1000 - 700 = 300 J$ (so $η_{A} = 0.30$ ). Engine B. First its efficiency: $η_{B} = 1 - 250/500 = 0.50$ . Then $W_{B} = η_{B} Q_{H} = 0.50 \times 1000 = 500 J .$ Compare. $W_{B} = 500 J > W_{A} = 300 J$ : Engine B does more work. Why this matters: with the same fuel ( $Q_{H}$ ), a higher efficiency directly means more work and less waste — B wastes only $500 J$ versus A's $700 J$ .

Level 4 — Synthesis

Goal: stitch several relations together, sometimes bringing in rate/power or a reversed cycle.

Recall Solution L4-a

Idea. Efficiency is a ratio, so it applies to rates just as to per-cycle energies: divide both sides of $η = W / Q_{H}$ by the time for one cycle to get $η = P / \dot{Q}_{H}$ , where $\dot{Q}$ means "heat per second". Step 1 — Heat-in rate. $\dot{Q}_{H} = \frac{P}{η} = \frac{600}{0.40} = 1500 MW .$ Step 2 — Dump rate. $\dot{Q}_{C} = \dot{Q}_{H} - P = 1500 - 600 = 900 MW .$ Why this step? Same energy balance (the green arrow plus the blue pipe equal the red pipe), now read per second. Insight. $900 MW$ of waste heat pours into the environment continuously — the reason big plants sit beside rivers or cooling towers.

Recall Solution L4-b

Stage 1. $η_{1} = 1 - 400/600 = 1/3$ . So $W_{1} = \frac{1}{3} \times 1200 = 400 J, Q_{C, 1} = 1200 - 400 = 800 J .$ Why this step? $η_{1}$ is the fraction of Stage 1's red-pipe heat that leaves as green-arrow work, so multiply $Q_{H}$ by $η_{1}$ to get $W_{1}$ ; the rest ( $Q_{H} - W_{1}$ ) is Stage 1's blue-pipe reject. Stage 2. Its heat input is Engine 1's reject: $Q_{H, 2} = 800 J$ . Its efficiency $η_{2} = 1 - 300/400 = 0.25$ . So $W_{2} = 0.25 \times 800 = 200 J .$ Why this step? Stage 2's red pipe is Stage 1's blue pipe — the waste heat is recycled as the second engine's fuel — so we again multiply that input by $η_{2}$ to read off its work. Total. $W_{tot} = 400 + 200 = 600 J$ , hence $η_{overall} = \frac{600}{1200} = 0.50 = 50%.$ Beautiful check. A single Carnot engine straight from $600 K$ to $300 K$ gives $η = 1 - 300/600 = 0.50$ — exactly the same. Splitting a reversible drop into reversible stages loses nothing, which is a deep property of the Carnot bound.

Level 5 — Mastery

Goal: full multi-step problems where one slip cascades. Read carefully.

Recall Solution L5-a

Step 1 — Work per cycle. $W = η Q_{H} = 0.22 \times 2500 = 550 J .$ Step 2 — Cycles per second. $2000 cycles/min \div 60 = 33. \overline{3} cycles/s$ . Step 3 — Power. Power is work per second: $P = W \times (cycles/s) = 550 \times \frac{2000}{60} = 18333. \overline{3} W \approx 18.33 kW .$ Why divide by 60 first? Power must be per second; the cycle count was given per minute, so convert the rate before multiplying. Step 4 — Heat dumped per second. Per cycle $Q_{C} = Q_{H} - W = 2500 - 550 = 1950 J$ . Per second: $\dot{Q}_{C} = 1950 \times \frac{2000}{60} = 65000 W = 65 kW .$ Sanity: $\dot{Q}_{H} = 2500 \times \frac{2000}{60} \approx 83.33 kW$ , and indeed $P + \dot{Q}_{C} = 18.33 + 65 = 83.33 kW$ ✓.

Recall Solution L5-b

Step 1 — Carnot ceiling. $η_{Carnot} = 1 - \frac{350}{700} = 0.50 = 50%.$ Step 2 — Second-law efficiency = actual ÷ ideal: $\frac{η}{η _{Carnot}} = \frac{0.35}{0.50} = 0.70 = 70%.$ Meaning: the engine captures $70%$ of what a perfect reversible engine could between these temperatures. Step 3 — Highest allowed $T_{C}$ . A real efficiency of $0.45$ is legal only when the ceiling is at least $0.45$ . Setting the ceiling equal to the target gives the borderline (highest allowed) $T_{C}$ : $1 - \frac{T _{C}}{700} = 0.45 ⟹ \frac{T _{C}}{700} = 0.55 ⟹ T_{C} = 700 \times 0.55 = 385 K .$ Why this step? Lowering $T_{C}$ raises the ceiling (bigger temperature gap = more room for efficiency). So any $T_{C} \leq 385 K$ keeps the ceiling $\geq 0.45$ and makes $η = 0.45$ permissible; above $385 K$ it becomes impossible. Final answer. Highest allowed cold temperature is $T_{C} = 385 K$ . The present $350 K$ (giving a $0.50$ ceiling) already leaves room for the $0.45$ target.

Answer key (numbers only)

Recall Reveal all final answers

L1-a: $W = 200 J$ , $η = 0.25$
L1-b: $η = 0.25$
L2-a: $Q_{H} = 1250 MJ$ , $Q_{C} = 750 MJ$
L2-b: $Q_{H} = 600 J$ , $W = 180 J$
L3-a: ceiling $0.40$ ; claim $0.45$ is impossible (exceeds by $0.05$ )
L3-b: $W_{A} = 300 J$ , $W_{B} = 500 J$ → B wins
L4-a: $\dot{Q}_{H} = 1500 MW$ , $\dot{Q}_{C} = 900 MW$
L4-b: $W_{tot} = 600 J$ , $η_{overall} = 0.50$
L5-a: $W = 550 J$ , $P \approx 18.33 kW$ , $\dot{Q}_{C} = 65 kW$
L5-b: ceiling $0.50$ ; second-law efficiency $0.70$ ; highest allowed $T_{C} = 385 K$

Connections

First Law of Thermodynamics — every $W = Q_{H} - Q_{C}$ balance above.
Second Law of Thermodynamics — the ceiling that kills the L3-a claim.
Carnot Cycle — source of $η = 1 - T_{C} / T_{H}$ used in L1, L3, L4, L5.
Entropy — why staged reversible engines (L4-b) lose nothing.
Refrigerators and Heat Pumps — the same balance run backwards.
PV diagrams and Work — the loop area whose value is the $W$ we keep computing.