Heat Engines and Thermal Reservoirs

A coal-fired power plant burns fuel at over 600 K and dumps waste heat into a river near 300 K. Only about 40% of that fuel energy ever reaches the electrical grid as usable work. This is not an engineering failure to be fixed with better materials. It is a law of nature. Every device that turns heat into continuous, repeatable work — a car engine, a steam turbine, a refrigerator run in reverse — is a heat engine. Heat engines are the bridge between the First Law, which only tracks energy bookkeeping, and the Second Law, which tells us energy has a direction. Understanding them explains why perpetual motion of the "second kind" is impossible, and why every real engine wastes some heat no matter how cleverly it is built. This article defines heat engines, hot and cold reservoirs, work output, and thermal efficiency. It builds the vocabulary and the first quantitative relation you need before meeting the Carnot cycle and the Kelvin and Clausius statements of the Second Law.

Key takeaways

A heat engine is a cyclic device that absorbs heat from a hot reservoir and converts part of it into work.
A thermal reservoir is idealized as infinite: it exchanges heat without changing temperature.
Thermal efficiency is defined as $\eta = w_{\text{cyc}}/q_h$ , the fraction of absorbed heat converted to work.
The First Law forces every cycle to reject some heat to a cold reservoir: $q_h = w_{\text{cyc}} + q_c$ .
No real or ideal engine reaches $\eta = 1$ ; some heat rejection is unavoidable in a cyclic process.
Key idea: because the working substance must return to its initial state each cycle, some absorbed heat must be dumped to a cold sink — efficiency is always less than 100%.

The core insight

A heat engine converts heat into work only by operating in a cycle between two reservoirs at different temperatures, and because the working substance must return to its starting state, some of the heat it absorbs from the hot reservoir must always be expelled — unconverted — to the cold reservoir, which caps efficiency below unity.

The physical picture

Picture a cylinder of gas fitted with a piston. If you let the gas expand once, pushing the piston outward, you get useful work — but the gas has now cooled and expanded, and the piston has moved to a new position. You cannot repeat the trick indefinitely with that one expansion; a continuous supply of work requires the system to return, again and again, to its starting point. That repeatable loop is a cycle, and any device that performs work by repeatedly cycling a working substance between two temperatures is a heat engine. To make the picture tractable, we idealize the environment the engine talks to. A thermal reservoir is a body so large that it can absorb or release any amount of heat relevant to the process without its temperature changing measurably — think of the ocean absorbing the exhaust heat of a power station, or a furnace supplying heat to a boiler. Every heat engine operates between (at least) two reservoirs: a hot reservoir at temperature $T_h$ , which supplies heat $q_h$ , and a cold reservoir at temperature $T_c < T_h$ , which absorbs the rejected heat $q_c$ . The working substance — steam, air, a refrigerant — cycles between them, and in each complete cycle it returns exactly to its initial thermodynamic state: same internal energy, same entropy, same everything. Because the working substance returns to its initial state after one cycle, $\Delta U_{\text{cyc}} = 0$ . The First Law then demands that the net work delivered by the engine equal the net heat absorbed:

w_{\text{cyc}} = q_h - q_c

This single bookkeeping equation is the seed of the entire theory of engines. It says nothing yet about how much of $q_h$ can become work — that question is answered by the Second Law, previewed here and made rigorous in the Carnot cycle and the Kelvin–Clausius statements. The engine's usefulness is captured by its thermal efficiency, $\eta$ : the fraction of the heat drawn from the hot reservoir that emerges as net work, rather than being dumped as waste heat to the cold reservoir. A real automobile engine might manage $\eta \approx 0.25$ – $0.30$ ; a modern combined-cycle gas power plant can reach $\eta \approx 0.60$ . No device, however cleverly engineered, reaches $\eta = 1$ , and understanding why is the payoff of this whole topic.

Figure

Heat engine schematic: energy flow between two reservoirs

Figure

Efficiency as heat is partitioned between work and waste

The core relations

\Delta U_{\text{cyc}} = 0 \quad \Rightarrow \quad w_{\text{cyc}} = q_h - q_c

where: $w_{\text{cyc}}$ is net work delivered per cycle, $q_h$ heat absorbed from the hot reservoir, $q_c$ heat rejected to the cold reservoir (magnitude, taken positive here for clarity).

\eta = \frac{w_{\text{cyc}}}{q_h}

where: $\eta$ is thermal efficiency, the fraction of input heat converted to net work (dimensionless, $0 \le \eta < 1$ ).

\eta = 1 - \frac{q_c}{q_h}

where: this is the same definition rewritten using the energy balance; it shows directly that $\eta = 1$ only if $q_c = 0$ .

\eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h}

where: the theoretical maximum efficiency for any engine operating between fixed reservoir temperatures $T_h$ and $T_c$ (derived rigorously in the Carnot cycle article; quoted here only to show where the story is heading).

Derivation AWhy q_c cannot be zero: the First Law says less than the Second Law will demand

Goal. Show that the requirement of cyclicity, combined with the First Law alone, already forces us to write $q_h = w_{\text{cyc}} + q_c$ with $q_c$ present as an unavoidable term — and see why the Second Law (previewed) forbids setting $q_c = 0$ even though the First Law alone does not.

Step 1 — Apply the First Law over one complete cycle. For any process, $\Delta U = q - w$ , using the convention that $w$ is work done by the system. Over one full cycle the working substance returns to its initial state, so all state functions return to their initial values:

\Delta U_{\text{cyc}} = 0

This is true regardless of how many reservoirs are involved or how complicated the path is — it only requires that the path be closed.

Step 2 — Separate the total heat exchanged into hot-reservoir input and cold-reservoir output. Suppose the engine contacts exactly two reservoirs during the cycle: it absorbs heat $q_h$ from the hot one during part of the cycle, and it must reject heat $q_c$ to the cold one during another part, to return to its starting entropy and volume. Writing the net heat absorbed as $q_h - q_c$ (defining both $q_h, q_c \ge 0$ as magnitudes) and substituting into the First Law:

0 = (q_h - q_c) - w_{\text{cyc}}

w_{\text{cyc}} = q_h - q_c

Step 3 — Ask whether $q_c = 0$ is allowed by the First Law alone. Inspect the equation: if $q_c = 0$ , then $w_{\text{cyc}} = q_h$ exactly, and energy is perfectly conserved. Nothing in the First Law forbids this — it is just as consistent with $\Delta U_{\text{cyc}} = 0$ as any other split. This is the crucial and often-missed point: the impossibility of $q_c = 0$ is not a First Law statement.

Step 4 — Bring in the entropy argument (foreshadowing the Second Law). The working substance must return to its initial entropy $S_i$ after the cycle, so $\Delta S_{\text{cyc}} = 0$ . During the hot, heat-absorbing leg, the substance's entropy tends to rise (heat added at high $T$ ); during the cold, heat-releasing leg its entropy must fall back down. If the only heat exchange were $q_h$ absorbed with no compensating rejection, the substance's entropy would keep increasing every cycle and could never return to $S_i$ — violating the requirement that entropy is a state function. Rejecting some heat $q_c$ to the cold reservoir is exactly what removes the entropy gained from $q_h$ , restoring the cycle. A quantitative version of this argument (via the Clausius inequality) shows precisely how much $q_c$ is required.

Step 5 — Combine into the efficiency statement. Since $q_c > 0$ is required in any repeating cycle exchanging heat with the surroundings, divide the Step 2 result by $q_h$ :

\eta = \frac{w_{\text{cyc}}}{q_h} = 1 - \frac{q_c}{q_h} < 1

\boxed{\eta < 1 \text{ always, because cyclicity plus the Second Law forces } q_c > 0}

Worked examples

Example 1 — Foundational: computing efficiency from measured heats. A laboratory demonstration heat engine absorbs $q_h = 850\ \text{J}$ from a hot reservoir maintained at $T_h = 500\ \text{K}$ during each cycle, and it is measured to reject $q_c = 550\ \text{J}$ to a cold reservoir at $T_c = 300\ \text{K}$ . Step 1: net work per cycle.

w_{\text{cyc}} = q_h - q_c = 850\ \text{J} - 550\ \text{J} = 300\ \text{J}

Step 2: thermal efficiency.

\eta = \frac{w_{\text{cyc}}}{q_h} = \frac{300\ \text{J}}{850\ \text{J}} = 0.353

So the engine converts 35.3% of the absorbed heat into work; equivalently 64.7% ( $=550/850$ ) is rejected as waste heat. Sanity check against the theoretical ceiling. The Carnot limit for reservoirs at 500 K and 300 K (quoted from the next article, used here only as a check) is:

\eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h} = 1 - \frac{300}{500} = 0.400

Our engine's efficiency, 0.353, is comfortably below 0.400 — consistent with the Second Law, since no real engine can exceed the Carnot efficiency for its operating reservoirs. If we had instead computed $\eta > 0.400$ from the data, we would know an arithmetic or measurement error had occurred.

Example 2 — Applied: sizing a power plant's fuel input and waste-heat rejection. A power station's turbine-generator set is designed to deliver a steady electrical output of $P = 750\ \text{MW}$ . Engineering tests show the plant operates with an overall thermal efficiency of $\eta = 0.38$ (typical for a modern coal-fired steam plant), converting heat from combustion into net shaft work. Step 1: required rate of heat input from the "hot reservoir" (the furnace/boiler). Treat power as work delivered per unit time, so $w_{\text{cyc}}/\Delta t = P$ . Rearranging the efficiency definition for rate of heat input $\dot q_h$ :

\eta = \frac{P}{\dot q_h} \quad\Rightarrow\quad \dot q_h = \frac{P}{\eta} = \frac{750\ \text{MW}}{0.38} = 1974\ \text{MW}

Step 2: rate of heat rejected to the cold reservoir (the cooling water/river). From the energy balance $\dot q_h = P + \dot q_c$ :

\dot q_c = \dot q_h - P = 1974\ \text{MW} - 750\ \text{MW} = 1224\ \text{MW}

Step 3: check with the direct efficiency formula.

1 - \eta = \frac{\dot q_c}{\dot q_h} = \frac{1224}{1974} = 0.620

1 - 0.38 = 0.62 \quad\checkmark

Consistent — 62% of the input thermal power (1224 MW) must be rejected as waste heat, typically warming cooling water discharged to a river or cooling tower, even though the plant is well-engineered and efficient by industrial standards. Interpretation. To generate 750 MW of usable electricity, the plant must actually burn enough fuel to supply nearly 2000 MW of thermal power — almost two-thirds of it is unavoidably discarded as low-grade heat, a direct real-world consequence of $\eta < 1$ .

Limiting cases & common traps

Situation	What happens	Why
$T_h \to T_c$ (reservoirs at same temperature)	$\eta \to 0$	No temperature difference to drive net work; Carnot bound collapses to zero
$T_c \to 0\ \text{K}$ (idealized, unreachable)	$\eta \to 1$	Third Law forbids reaching absolute zero, so this limit is never physically attained
Single-reservoir engine (only one temperature)	Impossible as a continuous cyclic device	Violates the Kelvin statement of the Second Law, previewed here
$q_c$ reported as zero in a real dataset	Signals an error, not a discovery	No cyclic engine has ever achieved $q_c = 0$ ; check units, sign conventions, or the cycle's closure

The chemist behind it

The theory of heat engines was born from a very practical question: why do steam engines, however well built, always waste fuel? Sadi Carnot, a French military engineer trained in the tradition of applied physics, set out in 1824 to answer this with pure reasoning rather than trial-and-error tinkering. His memoir Réflexions sur la puissance motrice du feu ("Reflections on the Motive Power of Fire") introduced the idealized reversible cycle now bearing his name and showed — using only the caloric theory of heat available at the time, later corrected by Clausius and Kelvin once energy conservation was established — that an engine's maximum efficiency depends only on the two operating temperatures, never on the substance used or the mechanism's design. Carnot's insight, that there is a fundamental ceiling on efficiency independent of engineering, is precisely the idea this article has been building toward, and it sets up the rigorous treatment of the Carnot cycle and the Second Law statements that follow.

Origins · 1 minNicolas Léonard Sadi Carnot and how this was discovered

Nicolas Léonard Sadi Carnot

1796–1832 · French engineer whose 1824 analysis of idealized heat engines founded the theory that would become the Second Law of Thermodynamics

Ask the tutor: How does the Carnot cycle actually achieve the maximum possible efficiency, step by step, using an ideal gas?

What exactly counts as a "thermal reservoir" in real life?

Any object large enough that absorbing or releasing the heat involved in the process does not measurably change its temperature. A lake, the atmosphere, or a large furnace all qualify as reservoirs for a power plant, because their heat capacities dwarf the heat exchanged each cycle. The idealization breaks down only if the "reservoir" is small — a coffee cup cooling a hot metal block is not a good reservoir, because its own temperature rises noticeably.

Is thermal efficiency the same as the efficiency you calculate for a car's fuel economy?

Not quite. Thermal efficiency, $\eta = w_{\text{cyc}}/q_h$ , is a thermodynamic ratio comparing net work output to heat input for the ideal cycle. Fuel economy also includes mechanical losses in the drivetrain, accessory loads, and combustion inefficiencies before heat even reaches the working gas. A car's overall "tank-to-wheels" efficiency is always lower than its engine's thermal efficiency alone.

Can a heat engine ever operate using only one reservoir?

No — this is exactly the content of the Kelvin statement of the Second Law, covered next. A device that took heat from a single reservoir and converted all of it to work, with no cold sink, would violate the Second Law even though it would perfectly satisfy the First Law's energy bookkeeping. This impossibility is what distinguishes heat engines from simple energy-conservation devices.

Does thermal efficiency depend on what working substance the engine uses — steam, air, refrigerant?

For the theoretical maximum (the Carnot efficiency), no — it depends only on $T_h$ and $T_c$ , not on the substance. This is a famous and surprising result first shown by Carnot. Real engines, however, achieve different fractions of that theoretical maximum depending on the substance and design, because irreversibilities like friction and rapid heat transfer differ between working fluids.

Why is efficiency always written as a positive fraction less than 1, not a percentage over 100?

Because $\eta = w_{\text{cyc}}/q_h$ is a ratio of the engine's output work to its input heat, and the Second Law guarantees $q_c > 0$ , meaning $w_{\text{cyc}} < q_h$ always. Writing $\eta$ as a percentage (e.g., 35%) is just a convenient rescaling of this same dimensionless fraction between 0 and 1.