Section 02
The Idea, Step by Step
A car engine burns fuel to push you forward — but the hood gets hot and the exhaust pours out heat. Not all the burning becomes motion; much of it escapes as waste heat. Sadi Carnot asked a daring question in 1824: of the heat you pour in, what is the most you could ever turn into useful work? His answer doesn't depend on how clever the engineer is — only on two temperatures.
Name the players. Heat $Q_H$ flows in from a hot source at temperature $T_H$ (the flame). Some of it becomes useful work $W$. The leftover, $Q_C$, is dumped into a cold sink at $T_C$ (the outside air). Efficiency is just what you get out divided by what you paid for, $\eta = W/Q_H$. Carnot's stunning result is that the best efficiency any engine can reach is fixed by the two temperatures alone:
The ceiling on every engine
$$\eta_{\max} = 1 - \frac{T_C}{T_H}$$
Put in numbers: a flame at $T_H = 800$ K rejecting to air at $T_C = 300$ K can convert at most $1 - 300/800 = 0.625$ — that is 62.5% of its heat into work, and not one percent more, no matter the design.
Why that exact formula? Carnot's ideal engine runs four reversible strokes: it absorbs $Q_H$ isothermally at $T_H$, expands adiabatically while cooling to $T_C$, rejects $Q_C$ isothermally at $T_C$, then compresses adiabatically back to the start. "Reversible" means no entropy is created, which forces $Q_H/T_H = Q_C/T_C$ — and that single relation rearranges straight into $\eta = 1 - T_C/T_H$. Real engines have friction and rushed, finite-rate heat flow, so they always generate extra entropy and fall short of this ceiling. In the sim, the sliders set $T_H$, $T_C$, and $Q_H$; the readouts show $\eta$, the work $W = Q_H(1 - T_C/T_H)$, and the rejected heat $Q_C = Q_H\,T_C/T_H$.
Try this in the sim above. First, drag $T_C$ down toward 100 K and watch $\eta$ climb — but notice it only reaches a true 100% if $T_C$ hits absolute zero, which nature forbids. Second, set $T_H$ and $T_C$ nearly equal and see $\eta$ collapse toward zero: no temperature gap, no work. Third, open the T-S Diagram tab — the Carnot cycle draws a perfect rectangle whose enclosed area is exactly $W_{\text{net}} = Q_H - Q_C$.
Section 03
Equations & Derivation
Carnot Efficiency — Theoretical Maximum
$$\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} = \frac{W_{\text{net}}}{Q_H}$$
Carnot Cycle — Four Processes
$$\text{1→2: Isothermal expansion at }T_H;\;\;2→3:\text{Adiabatic expansion}$$
Heat Relationships
$$Q_C = Q_H\frac{T_C}{T_H},\quad W_{\text{net}} = Q_H - Q_C = Q_H\left(1-\frac{T_C}{T_H}\right)$$
Air-Standard Engine Efficiency ($r$ = compression ratio, $r_c$ = cutoff ratio)
$$\eta_{\text{petrol (Otto)}} = 1 - \frac{1}{r^{\gamma-1}},\quad \eta_{\text{diesel}} = 1 - \frac{1}{r^{\gamma-1}}\,\frac{r_c^{\gamma}-1}{\gamma\,(r_c-1)},\quad \gamma = \frac{C_p}{C_v}$$
The Four Carnot Processes
| Symbol | Quantity | SI Unit |
|---|
| $T_H$ | Hot reservoir temperature | K |
| $T_C$ | Cold reservoir temperature | K |
| $\eta_C$ | Carnot efficiency | dimensionless |
| $Q_H$ | Heat absorbed from hot reservoir | J |
| $Q_C$ | Heat rejected to cold reservoir | J |
| $W_{\text{net}}$ | Net work output per cycle | J |
1
Process 1→2: Isothermal expansion at $T_H$. Gas absorbs $Q_H$ from hot reservoir and expands, doing work $W_{12}=Q_H=nRT_H\ln(V_2/V_1)$. Temperature constant.
2
Process 2→3: Adiabatic expansion. $Q=0$. Gas cools from $T_H$ to $T_C$ while expanding and doing work. $W_{23}=nC_v(T_H-T_C)$.
3
Process 3→4: Isothermal compression at $T_C$. Gas rejects $Q_C$ to cold reservoir. Work done on gas: $W_{34}=Q_C=nRT_C\ln(V_3/V_4)$.
4
Process 4→1: Adiabatic compression. $Q=0$. Gas returns to initial state. Net work $W_{\text{net}}=Q_H-Q_C$.
5
Carnot's theorem. All reversible engines operating between the same two temperatures have the same efficiency $\eta_C=1-T_C/T_H$. Any irreversible engine has lower efficiency. This is a consequence of the 2nd Law.
Ref: Halliday, Resnick & Walker 10th Ed., §20-1 to §20-4; Callen — Thermodynamics (2nd Ed.); Fermi — Thermodynamics, Ch. 5.
Section 05
Common Misconceptions
❌ A more powerful engine is more efficient.
✅ Power (energy per unit time) and efficiency (fraction of heat converted to work) are independent. A small efficient diesel generator can be more efficient than a powerful petrol engine. Efficiency depends on $T_H$ and $T_C$; power depends on how quickly cycles are completed.
📖 HRW 10th Ed., §20-1.
❌ Carnot efficiency can be exceeded by clever engineering.
✅ Carnot efficiency is a fundamental thermodynamic limit from the 2nd Law — not an engineering challenge. No engine operating between two fixed temperatures can exceed $\eta_C=1-T_C/T_H$, regardless of the working fluid, cycle design, or engineering ingenuity.
📖 HRW 10th Ed., §20-3; Callen — Thermodynamics.
❌ Heat pumps are less efficient than electric heaters.
✅ An electric heater converts 1 J of electricity to 1 J of heat (COP=1). A heat pump moves heat from outdoors to indoors: Carnot COP=$T_H/(T_H-T_C)$. At $T_C=-10°C$ and $T_H=20°C$: Carnot COP=293/30≈9.8. Real heat pumps achieve COP≈3-5 — delivering 3-5 J of heat per 1 J of electricity.
📖 Serway & Jewett 8th Ed., §22.5.
Misconception research: Arons — A Guide to Introductory Physics Teaching; Linn & Songer (1991), J. Research in Science Teaching.