Heat Engines & Carnot Cycle Reference Cheat Sheet

A printable reference covering heat engines, thermal efficiency, Carnot efficiency, and refrigerator coefficients of performance for grades 11-12.

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Heat engines turn thermal energy into useful work by moving heat from a hot reservoir to a cold reservoir. This cheat sheet helps students connect energy flow diagrams, efficiency formulas, and the ideal Carnot cycle in one clear reference. It is useful for solving thermodynamics problems where heat, work, temperature, and efficiency must be compared. The focus is on readable formulas and the meaning behind each quantity. The most important idea is energy conservation, written as $W = Q_H - Q_C$ for one complete engine cycle. Thermal efficiency compares useful work output to heat input using $e = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}$ . A Carnot engine gives the maximum possible efficiency between two reservoirs, $e_C = 1 - \frac{T_C}{T_H}$ , when temperatures are measured in kelvins. Refrigerators and heat pumps are analyzed with coefficients of performance instead of efficiency.

Key Facts

For a heat engine operating in a cycle, the net work output is $W = Q_H - Q_C$ .
Thermal efficiency is $e = \frac{W}{Q_H}$ , so an engine is more efficient when it converts more input heat into work.
Using energy conservation, engine efficiency can also be written as $e = 1 - \frac{Q_C}{Q_H}$ .
The maximum possible efficiency for any engine between two reservoirs is the Carnot efficiency $e_C = 1 - \frac{T_C}{T_H}$ .
Temperatures in Carnot formulas must be absolute temperatures measured in kelvins, so $T_K = T_{^{\circ}\mathrm{C}} + 273.15$ .
For a reversible Carnot cycle, the heat ratio equals the temperature ratio: $\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$ .
For a refrigerator, the coefficient of performance is $\mathrm{COP}_{R} = \frac{Q_C}{W}$ .
For a heat pump, the coefficient of performance is $\mathrm{COP}_{HP} = \frac{Q_H}{W}$ .

Vocabulary

Heat engine: A device that absorbs heat from a hot reservoir, does work, and releases waste heat to a cold reservoir.
Hot reservoir: The high-temperature source that supplies heat $Q_H$ to a heat engine.
Cold reservoir: The low-temperature sink that receives rejected heat $Q_C$ from a heat engine.
Thermal efficiency: The fraction of input heat converted into useful work, given by $e = \frac{W}{Q_H}$ .
Carnot cycle: An ideal reversible cycle made of two isothermal processes and two adiabatic processes.
Coefficient of performance: A measure of refrigerator or heat pump performance, comparing useful heat transfer to work input.

Common Mistakes to Avoid

Using Celsius in $e_C = 1 - \frac{T_C}{T_H}$ is wrong because Carnot efficiency requires absolute temperature in kelvins.
Confusing $Q_H$ and $Q_C$ gives the wrong energy flow because $Q_H$ is heat absorbed from the hot reservoir and $Q_C$ is heat rejected to the cold reservoir.
Writing efficiency as $e = \frac{Q_H}{W}$ is wrong because efficiency is useful work output divided by heat input, $e = \frac{W}{Q_H}$ .
Assuming every engine reaches Carnot efficiency is wrong because real engines have friction, turbulence, and other irreversible losses.
Treating refrigerator $\mathrm{COP}_{R}$ as the same as engine efficiency is wrong because $\mathrm{COP}_{R} = \frac{Q_C}{W}$ can be greater than $1$ .

Practice Questions

1 A heat engine absorbs $1200\ \mathrm{J}$ from a hot reservoir and rejects $750\ \mathrm{J}$ to a cold reservoir. Find the work output $W$ and efficiency $e$ .
2 A Carnot engine operates between $T_H = 600\ \mathrm{K}$ and $T_C = 300\ \mathrm{K}$ . Calculate its maximum efficiency $e_C$ .
3 A refrigerator removes $900\ \mathrm{J}$ of heat from a cold space while using $300\ \mathrm{J}$ of work. Find $\mathrm{COP}_{R}$ .
4 Explain why no real heat engine operating between the same two reservoirs can be more efficient than a Carnot engine.

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