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Heat Engines & Carnot Cycle

Q: A Carnot engine operates with T_H = 800 K and T_C = 400 K. What is its efficiency?

E_Carnot = 1 − T_C/T_H. Substitute values.

Q: Why is the actual efficiency of a gasoline engine always less than the Carnot efficiency for the same temperature range?

You can work it out this way: consider friction, heat loss through cylinder walls, and non-quasi-static compression. What do these have in common thermodynamically?

Q: If the cold reservoir temperature is lowered from 300 K to 200 K while T_H stays at 800 K, how does efficiency change?

You can work it out this way: calculate e_Carnot for both T_C values and compare. What is the trend?

Discover the upper efficiency limit set by the second law of thermodynamics

Fill your car with 100 joules' worth of gasoline and only about 25 of them ever push the pistons. The other 75 leave through the radiator and the exhaust as heat you paid for and cannot use — and the hard truth is that no engineering, however brilliant, can fix this. In 1824 a 28-year-old French engineer named Sadi Carnot proved why: any engine that turns heat into work must run between a hot place and a cold place, and the second law of thermodynamics taxes every single cycle. The tax rate depends on nothing but the two temperatures. In this lab you run Carnot's ideal engine yourself. A piston pumps through four strokes — two isothermal, two adiabatic — while a P-V diagram traces the loop whose enclosed area is the work you keep, and an energy-flow panel splits every 100 J absorbed from the hot reservoir into work delivered and heat rejected. At the default 600 K and 300 K the split is exactly fifty-fifty: the ceiling reads 50.0%. You will find the only two levers that move it — a hotter source, a colder sink — and discover they are not equal: the same hundred kelvins buys more efficiency on the cold side than on the hot. By the end, a warm radiator will read to you as a law of nature, not a design flaw.

What you'll be able to do

Describe a heat engine as a cyclic device that absorbs Q_H from a hot reservoir, converts part of it to net work W, and rejects Q_C to a cold reservoir, and apply the first law over one full cycle (ΔU = 0, so W = Q_H − Q_C) to close the energy books on any reservoir pair
Calculate thermal efficiency e = W/Q_H and Carnot efficiency e_C = 1 − T_C/T_H using absolute temperatures, and identify the four steps of the Carnot cycle — isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression — on a P-V diagram, reading net work as the enclosed area
Explain why Carnot efficiency is an absolute ceiling that depends only on the two reservoir temperatures (Carnot's principle), and why real engines fall below it through irreversibilities such as friction, turbulence, and finite-temperature-difference heat transfer

Formulas

e = \frac{W_{net}}{Q_H} = 1 - \frac{Q_C}{Q_H}

Thermal efficiency: fraction of heat input converted to work

e_{Carnot} = 1 - \frac{T_C}{T_H}

Carnot efficiency: maximum possible for given reservoir temperatures

W_{net} = Q_H - Q_C

Net work equals heat absorbed minus heat rejected

\text{COP} = \frac{Q_C}{W}

Coefficient of performance for a refrigerator

Make a prediction

Default setup: the piston runs a Carnot cycle between a 600 K hot reservoir and a 300 K cold one, absorbing a normalized 100 J of heat per cycle (energy-flow panel, top row). Before the first cycle completes — how much of that 100 J becomes work, and how much is rejected to the cold reservoir?

No grading here — commit to a guess, then scroll down and test it yourself.

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Drag T_H from 600 K up to 800 K with T_C fixed at 300 K: watch dp-eff climb 50.0% → 62.5%, W/cycle rise 1.00 → 1.25 kJ, and the energy panel's blue W bar grow from 50 J to about 63 J while the red Q_H row stays pinned at 100 J. Every extra joule of work is a joule the cold reservoir no longer receives.
Reset to defaults, then drag T_C from 300 K down to 200 K: dp-eff climbs 50.0% → 66.7%. Now compare levers — adding 200 K on the hot side bought 12.5 points, while removing just 100 K on the cold side buys 16.7. Which reservoir is the cheaper lever per kelvin, and can you see why straight from η = 1 − T_C/T_H?
Keep your eyes on the Energy Flow panel through a full cycle at any settings: Q_H (absorbed) never leaves 100 J, and the blue W and green Q_C bars always sum to exactly 100 J. Cross-check on the data panel's ×2 kJ scale — W/cycle + Q_C/cycle = 2.00 kJ = Q_H/cycle at every setting you try.
Slow the cycle down (Cycles/sec slider to 0.5, or the 0.5x button) and match the four color-coded P-V segments to the piston: red isothermal expansion with the piston rising, orange adiabatic expansion as it keeps rising but the gas cools, blue isothermal compression, teal adiabatic compression back to point A. Catch the white dot crossing each segment boundary exactly as the Phase readout flips.

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Your prediction

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Answer: 50 J of work and 50 J rejected — the Carnot ceiling η = 1 − 300/600 = 0.50 splits the 100 J exactly in half, and even this ideal engine cannot do better

The engine's entire efficiency is set by the two reservoir temperatures: η = 1 − T_C/T_H = 1 − 300/600 = 0.5000. The energy-flow panel renders exactly that: Q_H (absorbed) pinned at 100 J, W (work out) = η × 100 = 50 J, Q_C (rejected) = (1 − η) × 100 = 50 J — and the data panel says the same on its ×2 kJ scale: Q_H/cycle = 2.00 kJ, W/cycle = 1.00 kJ, Q_C/cycle = 1.00 kJ, with dp-eff at 50.0%. Option B is the targeted misconception: reversible means lossless, not tax-free. Even a perfectly reversible Carnot engine must reject Q_C = Q_H · T_C/T_H = 50 J per cycle; the second law forbids complete conversion of heat to work in any cyclic process, and 100% efficiency would require a cold reservoir at absolute zero (College Physics 2e §15.3–15.4). Option C invokes the working-substance misconception: Carnot's principle gives every reversible engine between the same two reservoirs the same maximum efficiency, 'valid no matter what the working substance is' (Univ. Physics Vol 2 §4.5) — the gas cannot rescue you.

Quiz (0/3)

A Carnot engine operates with T_H = 800 K and T_C = 400 K. What is its efficiency?

Why is the actual efficiency of a gasoline engine always less than the Carnot efficiency for the same temperature range?

If the cold reservoir temperature is lowered from 300 K to 200 K while T_H stays at 800 K, how does efficiency change?

You can now

Describe a heat engine as a cyclic device that absorbs Q_H from a hot reservoir, converts part of it to net work W, and rejects Q_C to a cold reservoir, and apply the first law over one full cycle (ΔU = 0, so W = Q_H − Q_C) to close the energy books on any reservoir pair
Calculate thermal efficiency e = W/Q_H and Carnot efficiency e_C = 1 − T_C/T_H using absolute temperatures, and identify the four steps of the Carnot cycle — isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression — on a P-V diagram, reading net work as the enclosed area
Explain why Carnot efficiency is an absolute ceiling that depends only on the two reservoir temperatures (Carnot's principle), and why real engines fall below it through irreversibilities such as friction, turbulence, and finite-temperature-difference heat transfer