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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=QHQCW = Q_H - Q_C for one complete engine cycle. Thermal efficiency compares useful work output to heat input using e=WQH=1QCQHe = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}. A Carnot engine gives the maximum possible efficiency between two reservoirs, eC=1TCTHe_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=QHQCW = Q_H - Q_C.
  • Thermal efficiency is e=WQHe = \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=1QCQHe = 1 - \frac{Q_C}{Q_H}.
  • The maximum possible efficiency for any engine between two reservoirs is the Carnot efficiency eC=1TCTHe_C = 1 - \frac{T_C}{T_H}.
  • Temperatures in Carnot formulas must be absolute temperatures measured in kelvins, so TK=TC+273.15T_K = T_{^{\circ}\mathrm{C}} + 273.15.
  • For a reversible Carnot cycle, the heat ratio equals the temperature ratio: QCQH=TCTH\frac{Q_C}{Q_H} = \frac{T_C}{T_H}.
  • For a refrigerator, the coefficient of performance is COPR=QCW\mathrm{COP}_{R} = \frac{Q_C}{W}.
  • For a heat pump, the coefficient of performance is COPHP=QHW\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 QHQ_H to a heat engine.
Cold reservoir
The low-temperature sink that receives rejected heat QCQ_C from a heat engine.
Thermal efficiency
The fraction of input heat converted into useful work, given by e=WQHe = \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 eC=1TCTHe_C = 1 - \frac{T_C}{T_H} is wrong because Carnot efficiency requires absolute temperature in kelvins.
  • Confusing QHQ_H and QCQ_C gives the wrong energy flow because QHQ_H is heat absorbed from the hot reservoir and QCQ_C is heat rejected to the cold reservoir.
  • Writing efficiency as e=QHWe = \frac{Q_H}{W} is wrong because efficiency is useful work output divided by heat input, e=WQHe = \frac{W}{Q_H}.
  • Assuming every engine reaches Carnot efficiency is wrong because real engines have friction, turbulence, and other irreversible losses.
  • Treating refrigerator COPR\mathrm{COP}_{R} as the same as engine efficiency is wrong because COPR=QCW\mathrm{COP}_{R} = \frac{Q_C}{W} can be greater than 11.

Practice Questions

  1. 1 A heat engine absorbs 1200 J1200\ \mathrm{J} from a hot reservoir and rejects 750 J750\ \mathrm{J} to a cold reservoir. Find the work output WW and efficiency ee.
  2. 2 A Carnot engine operates between TH=600 KT_H = 600\ \mathrm{K} and TC=300 KT_C = 300\ \mathrm{K}. Calculate its maximum efficiency eCe_C.
  3. 3 A refrigerator removes 900 J900\ \mathrm{J} of heat from a cold space while using 300 J300\ \mathrm{J} of work. Find COPR\mathrm{COP}_{R}.
  4. 4 Explain why no real heat engine operating between the same two reservoirs can be more efficient than a Carnot engine.

Understanding Heat Engines & Carnot Cycle Reference

A working engine needs a temperature difference, not simply a large amount of heat. The hot source may be burning fuel, steam in a boiler, or hot gases in a car cylinder. The cold source may be outside air, cooling water, or a radiator.

Heat naturally spreads from hotter places to cooler places. An engine uses part of this natural flow to push a piston, turn a turbine, or produce electricity. Once both reservoirs reach the same temperature, there is no preferred direction for heat flow and no continuing source of work.

The Carnot cycle is an ideal model that shows the best limit set by nature. It has four stages. During two stages, the working substance changes temperature without receiving or releasing heat.

These are called adiabatic processes. During the other two stages, it absorbs or releases heat while staying at a constant temperature. These are called isothermal processes.

The ideal cycle must happen extremely slowly, with no friction, no turbulence, and no heat transfer across a noticeable temperature difference. Real engines cannot meet these conditions, so their performance stays below the Carnot limit.

A pressure volume graph gives a useful physical picture of work in a cycle. When a gas expands, it can push outward on a piston. When it is compressed, work must be done on the gas.

The area enclosed by the complete loop on a pressure volume graph represents the net work for one cycle. A clockwise loop represents an engine that delivers net work. A counterclockwise loop represents a refrigerator or heat pump that needs work supplied to it.

Students should track the direction of every energy transfer carefully. A common error is to treat all heat entering a system as work output, even though some energy must leave as waste heat.

Refrigerators make cold spaces colder by moving heat out of them. Their useful result is the heat removed from the cold compartment, not the electrical work used by the compressor. Heat pumps use the same basic process, but their useful result is heat delivered indoors.

This explains why a coefficient of performance can be greater than one. It does not mean energy has been created. The device combines electrical work with heat taken from its surroundings.

In calculations, use kelvins whenever temperature appears in a ratio. A change of one degree Celsius equals a change of one kelvin, but zero degrees Celsius is not zero thermal energy. Converting temperatures before substituting values prevents impossible efficiency results.