Second law, heat engines and entropy

Physics · Heat and thermodynamics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

The first law allows any change that conserves energy. The second law tells us which changes actually happen. A heat engine takes heat Q₁ from a hot source, turns part of it into work and must reject the rest, Q₂, to a cold sink. No engine can turn all the heat into work.

The Carnot engine is an ideal engine working in a reversible cycle of two isothermal and two adiabatic processes. Its efficiency depends only on the two temperatures, and no real engine working between the same temperatures can do better.

Entropy measures the disorder of a system. In every natural process the total entropy of the universe increases, and energy becomes less available for doing work.

Rules to remember

Kelvin statement: no engine working in a cycle can convert all the heat taken from a source into work; a sink is necessary
Clausius statement: heat cannot flow by itself from a colder body to a hotter body; work must be done to make it happen
Carnot cycle order: isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression
efficiency = work output ÷ heat input = 1 − Q₂/Q₁; for a Carnot engine η = 1 − T₂/T₁, with temperatures in kelvin
Petrol engine: four strokes (intake, compression, power, exhaust), petrol-air mixture lit by a spark plug, efficiency about 25 to 30%
Diesel engine: no spark plug; air alone is compressed until very hot, then diesel is sprayed in and ignites; efficiency about 35 to 40%
change in entropy = heat transferred ÷ absolute temperature (ΔS = ΔQ/T), unit J K⁻¹; positive when heat is added

Common mistake

Putting Celsius temperatures into η = 1 − T₂/T₁ or ΔS = ΔQ/T. Add 273 first. Also, T₂ is the sink (cold) temperature and goes on top of the fraction.

Worked example

A Carnot engine works between a source at 227 °C and a sink at 27 °C. It takes in 2000 J of heat in each cycle. Find its efficiency and the work done per cycle.

T₁ = 227 + 273 = 500 K and T₂ = 27 + 273 = 300 K.
η = 1 − T₂/T₁ = 1 − 300/500 = 0.40, which is 40%.
Work = η × Q₁ = 0.40 × 2000 = 800 J, so 1200 J is rejected to the sink.

Answer: Efficiency 40%; work done 800 J per cycle.

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