The Second Law and Engine Efficiency Revision Notes | A Level AQA Physics

1. Why a Perfect Engine is Impossible

Picture an engine that takes in heat and converts every last joule into useful work, with nothing wasted. The First Law of Thermodynamics — conservation of energy — would not object. Energy in equals energy out, so the books balance. And yet, no such engine has ever been built, and the laws of physics guarantee it never will be.

The First Law tells you how much energy is involved, but it says nothing about the direction in which processes naturally occur. It cannot distinguish between a process that happens spontaneously and one that never does. That gap is precisely what the Second Law fills.

The First Law permits an engine that takes in $Q_H$ and outputs $W = Q_H$ with $Q_C = 0$. Energy is conserved, so there is no violation. The reason this is still impossible is that it directly violates the Kelvin-Planck statement of the Second Law, which forbids converting all absorbed heat into work in a complete cycle — this would correspond to a net decrease in the entropy of the universe, which the Second Law forbids.

2. The Second Law and the Source-Sink Requirement

The Second Law of Thermodynamics, applied to heat engines, states that no heat engine can convert all of the heat it absorbs into work in a complete cycle. There must always be a cold sink to which some heat $Q_C$ is rejected. In other words, a heat engine must operate between a hot source at temperature $T_H$ and a cold sink at temperature $T_C$, with $T_H > T_C$.

Think of it like water flowing through a turbine: the water only flows and does work because there is a height difference. Remove the lower reservoir and the flow stops. In the same way, a heat engine needs a temperature difference — a thermal gradient — to drive the flow of energy that produces work.

This is not a practical limitation that better engineering might overcome. It is a fundamental constraint rooted in the statistical behaviour of large numbers of particles — entropy always increases in an isolated system, and rejecting heat to the sink is the mechanism by which the engine satisfies that requirement.

3. Engine Efficiency: The Key Equations

Now that the source-sink requirement is established, the question becomes: how do we quantify how well an engine converts heat into work? The answer is thermal efficiency, $\eta$.

The second form, $(Q_H - Q_C)/Q_H$, is particularly useful because exam questions often give you the heat flows rather than the work output directly. Since $W = Q_H - Q_C$ by the First Law, both forms are equivalent — choose whichever the data supports.

Notice that $\eta_{max}$ approaches 1 (100%) only when $T_C \to 0$ K or $T_H \to \infty$ — neither of which is physically achievable. Increasing $T_H$ or decreasing $T_C$ both raise the maximum efficiency, which is why engineers push for higher operating temperatures in turbines and lower condenser temperatures in power stations.

4. Worked Examples: Applying the Efficiency Equations

Theory is straightforward, but exam questions test whether you can select the right form of the equation and handle unit conversions without error. Work through both examples below before moving on.

Worked example: Calculating thermal efficiency from heat flows

A heat engine absorbs $Q_H = 5000$ J from a hot source each cycle and rejects $Q_C = 3500$ J to a cold sink. Calculate the thermal efficiency of the engine. [3 marks]

1. $\eta = \frac{Q_H - Q_C}{Q_H}$ — [M1] Selecting the correct form of the efficiency equation. Both forms are equivalent; this one uses the given heat flows directly.
2. $\eta = \frac{5000 - 3500}{5000} = \frac{1500}{5000}$ — [M1] Correct substitution with consistent units (J). The numerator gives the net work output: $W = 5000 - 3500 = 1500$ J.
3. $\eta = 0.30 = 30\%$ — [A1] Efficiency is dimensionless; expressing as a percentage is acceptable and often expected. The engine converts 30% of the heat input into useful work.

Answer: $\eta = 0.30$ (30%)

Worked example: Calculating maximum theoretical efficiency from temperatures

A steam turbine operates with steam entering at $600\,^\circ\text{C}$ and exhausting to a condenser at $40\,^\circ\text{C}$. Calculate the maximum theoretical efficiency. [3 marks]

1. $T_H = 600 + 273 = 873$ K $T_C = 40 + 273 = 313$ K — Convert by adding 273 to each Celsius value. This step is non-negotiable — the Carnot formula requires absolute temperatures.
2. $\eta_{max} = \frac{T_H - T_C}{T_H}$ — [M1] Selecting the correct formula for maximum theoretical efficiency.
3. $\eta_{max} = \frac{873 - 313}{873} = \frac{560}{873}$ — [M1] Correct substitution. The numerator is the temperature difference: $873 - 313 = 560$ K.
4. $\eta_{max} = 0.642 = 64.2\%$ — [A1] To 3 significant figures. This is the theoretical ceiling — no engine operating between these temperatures can exceed 64.2% efficiency.

Answer: $\eta_{max} = 0.642$ (64.2%)

5. Why Practical Engines Fall Short

The Carnot efficiency is a theoretical ceiling, not a target that engineers expect to reach. Every real engine operates below it, often substantially so. Understanding why is both a spec requirement and a rich source of exam questions asking you to evaluate or suggest improvements.

The Carnot result assumes a perfectly reversible cycle — one where every process occurs infinitely slowly, with no friction, no turbulence, and no heat transfer across a finite temperature difference. Real engines violate all of these conditions simultaneously.

• Friction between moving parts (pistons, bearings, turbine blades) converts kinetic energy into heat, which is dissipated rather than doing useful work.
• Heat losses through cylinder walls, exhaust pipes, and engine casings transfer energy to the environment before it can be converted to work.
• Combustion is not perfectly efficient: incomplete combustion wastes fuel energy, and real gas behaviour deviates from the ideal assumptions of the Carnot cycle.
• Turbulence and irreversible mixing in the working fluid increase entropy beyond the theoretical minimum, reducing the work extractable per cycle.
• Finite temperature differences are needed to drive heat transfer at a practical rate — a real engine cannot operate quasi-statically.

A typical petrol engine achieves around 25–35% thermal efficiency; a modern combined-cycle gas turbine power station reaches around 55–60%. Both are well below the Carnot limit for their operating temperatures, illustrating how far irreversibilities push real performance below the theoretical ceiling.

6. Maximising the Use of W and Q_H: Combined Heat and Power

Even a well-engineered engine rejects a substantial fraction of $Q_H$ as waste heat $Q_C$. In a conventional power station, this heat is dumped into a river or cooling tower and lost entirely. Combined heat and power (CHP) schemes take a different approach: rather than discarding $Q_C$, they capture it and put it to use.

In a CHP system, the engine still produces electrical work $W$ as its primary output. The heat that would otherwise be wasted — $Q_C$ — is instead piped to nearby buildings for space heating, or supplied to industrial processes that require thermal energy. The result is that the total useful energy delivered ($W + Q_C$) is a much larger fraction of the original fuel input $Q_H$ than the electrical efficiency alone would suggest.

To see the numbers: an engine with 35% electrical efficiency wastes 65% of $Q_H$ as heat. If a CHP scheme captures 50 percentage points of that waste heat for useful heating, the overall system efficiency rises to 85%. The thermodynamic constraint — that $Q_C$ must be rejected — is not circumvented; instead, that rejected heat is redirected to a purpose rather than wasted.

CHP is particularly effective in hospitals, universities, and industrial sites where there is a simultaneous and sustained demand for both electricity and heat. The closer the heat consumer is to the engine, the lower the distribution losses and the higher the overall system efficiency.

7. Exam Technique and Mark-Winning Strategies

The efficiency equations are straightforward to apply, but the marks in longer questions go to students who demonstrate conceptual understanding alongside correct arithmetic. Here is where the marks are won and lost.

Extended-response questions on this topic typically ask you to evaluate the efficiency of a real engine compared to the Carnot limit, or to assess the advantages of CHP. Structure your answer as: opening claim, supporting physics (with equations where relevant), counter-consideration, and a final judgement. Avoid vague phrases like 'energy is wasted' — name the mechanism and link it to the efficiency formula.