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

Level 1 — Recognition

Goal: plug numbers into the two master formulas. No rearranging, no traps beyond units.

Recall Solution 1.1

WHAT / WHY: Both temperatures are already in kelvin (they carry the K unit), so we drop them straight into $η = 1 - T_{C} / T_{H}$ . No conversion, no rearranging. $η = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 = 40%$ Reading it: $40%$ of the heat pulled from the hot reservoir becomes useful work; the other $60%$ is dumped into the cold one.

Recall Solution 1.2

WHAT / WHY: We know $η$ and $T_{C}$ , we want $T_{H}$ . Start from $η = 1 - T_{C} / T_{H}$ and isolate $T_{H}$ — this is pure algebra, one variable. $0.25 = 1 - \frac{300}{T _{H}} \Rightarrow \frac{300}{T _{H}} = 0.75 \Rightarrow T_{H} = \frac{300}{0.75} = 400 K$

Recall Solution 1.3

WHAT / WHY: Efficiency is defined as $η = W / Q_{H}$ , so multiply to get work. Then use the first law over one cycle ( $Δ U = 0 \Rightarrow W = Q_{H} - Q_{C}$ ) to get the rejected heat. $W = η Q_{H} = 0.4 \times 800 = 320 J$ $Q_{C} = Q_{H} - W = 800 - 320 = 480 J$ Cross-check with the reservoir identity: $Q_{C} = Q_{H} (T_{C} / T_{H})$ . Since $η = 0.4$ , $T_{C} / T_{H} = 0.6$ , so $Q_{C} = 800 \times 0.6 = 480 J$ ✓.

Level 2 — Application

Goal: convert units, combine both master formulas, run a refrigerator backwards.

Recall Solution 2.1

WHAT / WHY: The temperatures are in Celsius, but $η$ uses a ratio, which only means something on the kelvin scale (zero = no thermal energy). So convert first: $T (K) = T (^{\circ} C) + 273$ . $T_{H} = 327 + 273 = 600 K, T_{C} = 27 + 273 = 300 K$ $η = 1 - \frac{300}{600} = 0.5 = 50%$

Recall Solution 2.2

WHAT / WHY: First law gives the absorbed heat: $Q_{H} = W + Q_{C}$ . Then efficiency from its definition. Then use $T_{C} / T_{H} = Q_{C} / Q_{H}$ (reversible identity) to back out $T_{H}$ . $Q_{H} = W + Q_{C} = 600 + 900 = 1500 J$ $η = \frac{W}{Q _{H}} = \frac{600}{1500} = 0.4 = 40%$ From $\frac{Q _{C}}{Q _{H}} = \frac{T _{C}}{T _{H}}$ : $\frac{900}{1500} = \frac{270}{T _{H}} \Rightarrow 0.6 = \frac{270}{T _{H}} \Rightarrow T_{H} = \frac{270}{0.6} = 450 K$ .

Recall Solution 2.3

WHAT / WHY: A refrigerator is a Carnot cycle in reverse: we pay work $W$ to pump heat $Q_{C}$ out of the cold side. The COP measures "heat moved per unit work". We are told COP's formula, so compute it, then solve for $W$ . $COP = \frac{T _{C}}{T _{H} - T _{C}} = \frac{250}{300 - 250} = \frac{250}{50} = 5$ $W = \frac{Q _{C}}{COP} = \frac{500}{5} = 100 J$ Reading it: each joule of electricity moves 5 J of heat out — much better than "1 J of work makes 1 J of heat", because we are moving heat, not creating it.

Level 3 — Analysis

Goal: reason about how $η$ responds to changes, and reconstruct hidden cycle data.

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

Recall Solution 3.1

WHAT / WHY: Compute $η$ for both options and compare. This exposes an asymmetry: because $T_{C}$ sits in the numerator and $T_{H}$ in the denominator of $T_{C} / T_{H}$ , they don't affect $η$ equally. Baseline: $η_{0} = 1 - 300/500 = 0.400$ .

Raise $T_{H}$ to $550$ : $η = 1 - 300/550 = 1 - 0.5455 = 0.4545$ . Gain $\approx 0.0545$ .
Lower $T_{C}$ to $250$ : $η = 1 - 250/500 = 1 - 0.5 = 0.500$ . Gain $= 0.100$ . Conclusion: lowering the cold side wins here (gain $0.100$ vs $0.055$ ). Look at the red curve in the figure: near this operating point, the efficiency is steeper against $T_{C}$ than against $T_{H}$ .

Recall Solution 3.2

WHAT / WHY: During isothermal expansion at $T_{H}$ , $Δ U = 0$ so all absorbed heat becomes work: $Q_{H} = R T_{H} ln (V_{2} / V_{1})$ — the log comes from integrating $\int P d V = \int R T_{H} d V / V$ . Then the adiabats force $V_{3} / V_{4} = V_{2} / V_{1}$ , so $Q_{C} = R T_{C} ln (V_{2} / V_{1})$ with the same log. $Q_{H} = R T_{H} ln 2 = 8.314 \times 400 \times 0.6931 = 2304.9 J$ $Q_{C} = R T_{C} ln 2 = 8.314 \times 300 \times 0.6931 = 1728.7 J$ $W = Q_{H} - Q_{C} = 2304.9 - 1728.7 = 576.2 J$ Cross-check: $η = W / Q_{H} = 576.2/2304.9 = 0.250 = 1 - 300/400$ ✓.

Recall Solution 3.3

WHAT / WHY: "Same efficiency" means $1 - T_{m} /600 = 1 - 300/ T_{m}$ , i.e. the two temperature ratios are equal: $T_{m} /600 = 300/ T_{m}$ . That's a single equation for $T_{m}$ ; cross-multiplying gives a geometric mean. $T_{m}^{2} = 600 \times 300 = 180000 \Rightarrow T_{m} = 180000 \approx 424.3 K$ Reading it: equal-efficiency staging places the middle temperature at the geometric mean $T_{H} T_{C}$ , not the arithmetic mean $450 K$ .

Level 4 — Synthesis

Goal: chain first-law bookkeeping, entropy, and impossible-engine reasoning.

Recall Solution 4.1

WHAT / WHY: For a reservoir that gives/receives heat $Q$ at fixed temperature $T$ , its entropy change is $Δ S = Q / T$ (a reservoir stays at one temperature). The hot reservoir loses $Q_{H}$ (so $Δ S < 0$ ), the cold reservoir gains $Q_{C}$ . First find $Q_{C}$ : reversible identity $Q_{C} = Q_{H} (T_{C} / T_{H}) = 1000 \times 250/500 = 500 J$ . $Δ S_{hot} = - \frac{Q _{H}}{T _{H}} = - \frac{1000}{500} = - 2.00 J/K$ $Δ S_{cold} = + \frac{Q _{C}}{T _{C}} = + \frac{500}{250} = + 2.00 J/K$ $Δ S_{total} = - 2.00 + 2.00 = 0 J/K$ Conclusion: total entropy change is exactly zero, the signature of a reversible cycle — consistent with $\oint d Q_{rev} / T = 0$ . The gas itself returns to its start, so $Δ S_{gas} = 0$ too.

Recall Solution 4.2

WHAT / WHY: The Carnot value is the ceiling any engine between these reservoirs can reach. Compute the claimed efficiency and the maximum, then confirm with entropy (the ultimate arbiter via the second law). Efficiency check: claimed $η = W / Q_{H} = 700/1000 = 0.70$ . Carnot max $= 1 - 300/600 = 0.50$ . Since $0.70 > 0.50$ , it exceeds the ceiling — impossible. Entropy check: $Q_{C} = Q_{H} - W = 1000 - 700 = 300 J$ . $Δ S_{total} = - \frac{Q _{H}}{T _{H}} + \frac{Q _{C}}{T _{C}} = - \frac{1000}{600} + \frac{300}{300} = - 1.667 + 1.000 = - 0.667 J/K$ A negative total entropy change violates the second law (the universe's entropy can never fall). Verdict: impossible.

Recall Solution 4.3

WHAT / WHY: First law gives $Q_{C}$ . Real $η$ from its definition; Carnot $η$ as the benchmark. Because this engine is irreversible we expect $Δ S_{total} > 0$ (strictly positive). $Q_{C} = Q_{H} - W = 1000 - 400 = 600 J$ $η_{real} = \frac{400}{1000} = 0.40, η_{Carnot} = 1 - \frac{300}{600} = 0.50$ Real is below Carnot ( $0.40 < 0.50$ ) — allowed. Entropy generated: $Δ S_{total} = - \frac{1000}{600} + \frac{600}{300} = - 1.667 + 2.000 = + 0.333 J/K$ Conclusion: positive entropy generation confirms an irreversible but legal engine.

Level 5 — Mastery

Goal: prove the theorem, and derive the numbers behind an abstract statement.

Recall Solution 5.1

WHAT / WHY: Assume the opposite and show it lets us break the second law. This is proof by contradiction — the standard weapon when a quantity claims to exceed a ceiling. Setup (see figure): Suppose a "super-engine" X has $η_{X} > η_{C}$ . Run X forward to produce work $W$ , absorbing $Q_{H}^{X}$ from the hot reservoir. Use all of $W$ to drive a Carnot engine backwards as a refrigerator, pumping heat from cold to hot. Match the work: the reversed Carnot engine consumes the same $W$ . Choose it to reject exactly $Q_{H}^{C}$ to the hot reservoir and absorb $Q_{C}^{C}$ from the cold one, with $W = Q_{H}^{C} - Q_{C}^{C}$ . The heat drawn from the hot reservoir by X is $Q_{H}^{X} = W / η_{X}$ . The heat returned to the hot reservoir by the fridge is $Q_{H}^{C} = W / η_{C}$ . Since $η_{X} > η_{C}$ , we get $Q_{H}^{X} = W / η_{X} < W / η_{C} = Q_{H}^{C}$ . The contradiction: the hot reservoir receives $Q_{H}^{C}$ and loses $Q_{H}^{X}$ , a net gain of $Q_{H}^{C} - Q_{H}^{X} > 0$ . By the first law the same net heat is drawn from the cold reservoir — with no external work (the two engines' work cancelled). We have moved heat from cold to hot spontaneously, a perpetual-motion machine of the second kind. This violates the second law. Therefore $η_{X} \leq η_{C}$ . $■$

Recall Solution 5.2

WHAT / WHY: Turn the abstract proof into numbers. Compute how much heat X pulls from the hot side and how much the reversed Carnot fridge returns; their difference is the illegal free heat. $Q_{H}^{X} = \frac{W}{η _{X}} = \frac{300}{0.6} = 500 J (drawn from hot by X)$ $Q_{H}^{C} = \frac{W}{η _{C}} = \frac{300}{0.5} = 600 J (returned to hot by fridge)$ $Net into hot reservoir = Q_{H}^{C} - Q_{H}^{X} = 600 - 500 = 100 J > 0$ Conclusion: $100 J$ per cycle flows cold $\to$ hot with zero net work — impossible. The positive number is the contradiction, confirming $η_{X} \leq 0.5$ .

Recall Solution 5.3

WHAT / WHY: In the parent derivation the working-substance-dependent logs cancelled, leaving $Q_{C} / Q_{H} = T_{C} / T_{H}$ for any reversible engine. So heat ratios of reversible engines give a substance-free thermometer — this is the definition of the Kelvin scale. Fixing $T_{H} = 273.16 K$ as the reference, any other reservoir's temperature is: $T_{C} = T_{H} \cdot \frac{Q _{C}}{Q _{H}} = 273.16 \times 0.500 = 136.58 K$ Reading it: we never measured "molecular speed" or used any gas — just a heat ratio. That is why the Kelvin scale is called absolute: it needs no material.

Recall One-line self-test before you leave

Master fact ::: $η = 1 - T_{C} / T_{H}$ and $Q_{C} / Q_{H} = T_{C} / T_{H}$ , temperatures in kelvin. Reversible signature ::: $Δ S_{total} = 0$ (irreversible $> 0$ , impossible $< 0$ ). Theorem ::: no engine beats Carnot; proof couples a super-engine to a reversed Carnot to break the second law.