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PU-203 / L05 · lecture

Heat Engines and the Second Law

~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the First Law of Thermodynamics, specifically the sign convention for work done by the system, and basic calculus involving line integrals.

Why this matters

The Second Law dictates the direction of time and the limits of technology.

While the First Law tells us that energy is conserved, it does not tell us which processes are possible. A ball does not spontaneously un-break, and heat does not flow from cold to hot. The Second Law introduces entropy as a state function that resolves this ambiguity, establishing that while energy quantity is preserved, its quality degrades in every real process. This lecture defines the heat engine as the canonical device for probing these limits, leading directly to the quantitative statement of the Second Law.

The Direction of Spontaneous Change

Imagine a cup of hot coffee sitting on a table. It cools down until it reaches room temperature. We never observe the reverse: a cup of cold coffee spontaneously heating up while the surrounding air cools down. Both scenarios conserve energy (the First Law is satisfied in both), yet only one happens in nature. This asymmetry is the core of the Second Law.

To quantify this, we look at heat engines. A heat engine is a cyclic device that extracts heat from a hot reservoir, converts some of it into work, and rejects the remainder to a cold reservoir. The efficiency \(\eta\) is defined as the work output divided by the heat input from the hot source. The Second Law asserts that no engine can be 100% efficient; some heat must always be rejected. This is not a limitation of engineering, but a fundamental constraint of nature. The "missing" energy is not lost, but it is rendered unavailable for further work, a concept we formalize through entropy.

Defining the Heat Engine Cycle

A heat engine operates in a cycle, meaning its internal energy \(U\) returns to its initial value after one complete operation. By the First Law, \(\Delta U = Q - W\), where \(Q\) is the net heat added and \(W\) is the net work done by the system. Since \(\Delta U = 0\) for a cycle, we have \(W_{\text{net}} = Q_{\text{net}}\).

Let \(Q_H\) be the heat absorbed from the hot reservoir at temperature \(T_H\), and \(Q_C\) be the heat rejected to the cold reservoir at temperature \(T_C\). We adopt the convention that \(Q_H\) and \(Q_C\) are positive magnitudes. The net heat input is \(Q_H - Q_C\). Therefore, the work output is:

\[ W_{\text{net}} = Q_H - Q_C \]

The efficiency \(\eta\) is the ratio of useful work to the total energy input from the hot source:

\[ \eta = \frac{W_{\text{net}}}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H} \]

This equation reveals a crucial insight: to maximise efficiency, we must minimise the ratio \(Q_C/Q_H\). Ideally, we would like \(Q_C = 0\), implying \(\eta = 1\). However, the Second Law forbids this. It states that it is impossible to construct a cyclic engine that produces no effect other than the extraction of heat from a single reservoir and the performance of an equivalent amount of work. This is the Kelvin-Planck statement of the Second Law.

Entropy and the Clausius Inequality

To move from qualitative statements to a quantitative law, we introduce entropy \(S\). For a reversible process, the change in entropy is defined as the heat transferred divided by the temperature at which the transfer occurs:

\[ dS = \frac{\delta Q_{\text{rev}}}{T} \]

For an irreversible process, the entropy change is greater than the heat transfer divided by temperature. This leads to the Clausius Inequality, which applies to any cyclic process:

\[ \oint \frac{\delta Q}{T} \leq 0 \]

For a heat engine operating between two reservoirs, the cycle consists of heat intake at \(T_H\) and heat rejection at \(T_C\). Applying the inequality:

\[ \frac{Q_H}{T_H} - \frac{Q_C}{T_C} \leq 0 \]

Rearranging this gives the fundamental constraint on the heat flows:

\[ \frac{Q_C}{Q_H} \geq \frac{T_C}{T_H} \]

Substituting this back into the efficiency equation, we find the upper bound for any heat engine:

\[ \eta \leq 1 - \frac{T_C}{T_H} \]

This result is independent of the working substance. It depends only on the temperatures of the reservoirs. The equality holds only for a reversible engine, which is an idealization that assumes no friction, no turbulence, and infinitesimally slow processes.

The Carnot Limit and Reversibility

The theoretical maximum efficiency is achieved by a Carnot engine, which operates on a reversible cycle. The Carnot cycle consists of two isothermal processes (at \(T_H\) and \(T_C\)) and two adiabatic processes. Because the adiabatic processes involve no heat transfer, they do not contribute to the entropy balance of the reservoirs, but they are essential for changing the temperature of the working substance.

For a reversible engine, the entropy lost by the hot reservoir is exactly equal to the entropy gained by the cold reservoir. The total entropy change of the universe (engine + reservoirs) is zero. For any real, irreversible engine, the total entropy change is positive. This means that real engines always generate entropy, leading to a lower efficiency than the Carnot limit.

It is critical to note that the Carnot efficiency is a ceiling, not a target. No real engine reaches it; the best large power stations reach roughly 60–70% of their Carnot limit. The gap between real efficiency and Carnot efficiency is a measure of the irreversibilities in the system, such as friction and heat leaks.

The derivation

◆ DERIVATION EMBED · D-015 · not re-derived here

Carnot's Theorem and Maximum Engine Efficiency

The second law forces every reversible engine between two reservoirs to share the same efficiency 1 - T_c/T_h, bounding all real engines.

η ≤ 1 − TC ⁄ TH
Open full derivation →
Tier 3 · watch out

A common misconception is that the Second Law forbids local decreases in entropy. It does not. Living organisms, for example, maintain low internal entropy by exporting higher entropy to their surroundings. The Second Law applies to the total entropy of the isolated system (the universe), not to subsystems. When calculating efficiency, ensure that \(T_H\) and \(T_C\) are absolute temperatures in Kelvin, not Celsius, as the ratio \(T_C/T_H\) is dimensionless but physically meaningful only in an absolute scale. C

Worked example

Problem. A steam turbine operates between a hot reservoir at \(500\ \mathrm{K}\) and a cold reservoir at \(300\ \mathrm{K}\). The engine absorbs \(1000\ \mathrm{J}\) of heat from the hot reservoir. Calculate the maximum possible work output and the efficiency of the engine.

  1. Identify the given values: \(T_H = 500\ \mathrm{K}\), \(T_C = 300\ \mathrm{K}\), \(Q_H = 1000\ \mathrm{J}\).
  2. Calculate the Carnot efficiency using the formula \(\eta_{\text{max}} = 1 - \frac{T_C}{T_H}\).
  3. Substitute the temperatures: \(\eta_{\text{max}} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4\).
  4. Calculate the maximum work output using \(W_{\text{max}} = \eta_{\text{max}} Q_H\).
  5. Substitute the values: \(W_{\text{max}} = 0.4 \times 1000\ \mathrm{J} = 400\ \mathrm{J}\).
\[ W_{\text{max}} = 400\ \mathrm{J}, \quad \eta_{\text{max}} = 40\% \]

Units check. Efficiency is a ratio of energies, so it is dimensionless. Work is in Joules, consistent with the input heat. The temperatures are in Kelvin, ensuring the ratio is valid.

CHECK YOURSELF

Three quick tests

  1. Can a heat engine have an efficiency of 100%?
    AnswerNo, the Second Law (Kelvin-Planck statement) forbids converting all heat from a single reservoir into work. Some heat must be rejected to a colder reservoir.
  2. If \(T_H\) is increased while \(T_C\) remains constant, does the maximum efficiency increase or decrease?
    AnswerIt increases. The term \(T_C/T_H\) becomes smaller, so \(1 - T_C/T_H\) becomes larger.
  3. What is the sign of the total entropy change for a reversible cycle?
    AnswerZero. In a reversible process, the entropy decrease of the hot reservoir is exactly balanced by the entropy increase of the cold reservoir.
PROBLEMS

Problem set for PU-203 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-015 · Threads: energy, matter, chance