Second Law and engines (A-level only)

AQA·Physics·3.11.2.5

Heat engines and the Second Law

A heat engine operates in a cycle: after each cycle, it returns to its initial state and can repeat the same process. During one cycle, it receives energy $$Q_H$$ by heating from a hot source at temperature $$T_H$$, delivers work $$W$$, and transfers energy $$Q_C$$ by heating to a colder sink at temperature $$T_C$$. The source is hotter than the sink, so $$T_H > T_C$$.

Because the engine returns to its initial state, its internal energy has no net change over a complete cycle. The First Law of thermodynamics therefore gives:

QH=W+QCQ_H = W + Q_C

The energy received from the hot source is accounted for by the work delivered and the energy transferred to the cold sink. Here, $$Q_H$$ and $$Q_C$$ are the amounts of energy transferred by heating.

Why a heat engine needs a cold sink

Energy conservation alone does not require any energy to be transferred to the sink. If $$Q_C = 0$$, the First Law would still be satisfied by $$W = Q_H$$: all the energy received from the hot source would be delivered as work.

The Second Law of thermodynamics rules out such a cyclic engine. A device cannot operate in a cycle with the sole effect of taking energy by heating from a single source and converting it entirely into work. A heat engine must therefore transfer some energy by heating to a colder sink as well as deliver work.

Heat engine efficiency

A heat engine receives energy by heating from a hot source. It delivers some of that energy as useful work and transfers the rest as heat to a cold sink. Efficiency measures the fraction of the energy received that becomes useful work.

For energy transfers measured over the same cycle, let $$Q_H$$ be the heat received from the hot source, $$Q_C$$ the heat rejected, and $$W$$ the work delivered. Since the engine returns to its starting state after a complete cycle, $$W = Q_H - Q_C$$. Its efficiency is therefore:

η=WQH=QH−QCQH\eta = \frac{W}{Q_H} = \frac{Q_H-Q_C}{Q_H}

Efficiency $$\eta$$ is a fraction; multiplying it by 100 gives the percentage of the supplied energy delivered as work. The less heat rejected for a given heat input, the greater the efficiency.

Maximum heat engine efficiency

The theoretical maximum efficiency depends on the absolute temperatures of the hot source and cold sink:

ηmax⁡=TH−TCTH\eta_{\max} = \frac{T_H-T_C}{T_H}

Here, $$T_H$$ is the hot-source temperature and $$T_C$$ is the cold-sink temperature. Both temperatures must be in kelvin.

An ideal reversible engine operating between these temperatures has the maximum theoretical efficiency. The limit increases if the hot source is made hotter while the sink temperature stays fixed, or if the sink is made colder while the source temperature stays fixed.

Practical engines have lower efficiencies because real processes are irreversible. Friction dissipates energy, and heat transfer across a finite temperature difference is irreversible. Unwanted heat losses also reduce the useful work obtained from the supplied energy.

Combined heat and power

An engine produces useful work, but it also releases heat to its surroundings. In combined heat and power (CHP), some of that heat is recovered and supplied for heating instead of being wasted. The engine’s work is commonly used to generate electricity.

For example, heat recovered from the engine’s exhaust or cooling system can be used to heat a building. The system therefore provides two useful outputs from the energy supplied: work, commonly as electricity, and heat.

Overall energy utilisation

The overall utilisation measures how much of the supplied energy is delivered as useful work and useful heat:

Overall utilisation=useful work output+useful heat outputenergy supplied\text{Overall utilisation} = \frac{\text{useful work output} + \text{useful heat output}}{\text{energy supplied}}

All three quantities must be measured over the same period and in the same energy units. The result can be multiplied by 100 to express it as a percentage. Recovering more heat for a useful purpose increases the overall utilisation, even if the work output stays the same.

Overall utilisation and work efficiency

The engine’s work efficiency counts only the useful work output, not the heat supplied for heating:

Work efficiency=useful work outputenergy supplied\text{Work efficiency} = \frac{\text{useful work output}}{\text{energy supplied}}

Recovering rejected heat does not turn that heat into additional work. CHP can therefore have a higher overall utilisation than work efficiency. It does not remove the Second Law of thermodynamics limit on how much of the supplied energy a heat engine can convert into work.