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Physics · Thermodynamics

Heat Engines

A heat engine is the device that turns the disordered energy of heat into orderly mechanical work — the steam engine that moved trains, the petrol engine that moves cars. NCERT introduces it alongside the second law because the engine is exactly where that law bites: heat cannot be converted into work without limit. This deep-dive builds the source–engine–sink model, derives the efficiency $\eta = 1 - Q_2/Q_1$ from the first law, explains why $\eta$ is forever below unity, and drills the NEET traps around it.

What a heat engine is

A heat engine is a device that converts heat into mechanical work, and does so repeatedly by taking a working substance through a closed cycle. NCERT puts it plainly in §11.9: in a steam engine, the heat of the steam is used to do useful work in moving the pistons, which in turn rotate the wheels of the train. The word "cycle" is load-bearing — the engine is useful precisely because the working substance returns to its starting state and the process can be repeated without limit.

The defining feature is that an engine does not convert all the heat it draws into work. It absorbs a quantity of heat $Q_1$ from a hot body, converts part of it into work $W$, and rejects the leftover heat $Q_2$ to a cold body. The very existence of that rejected heat $Q_2$ is the thread that connects heat engines to the second law of thermodynamics.

Source, sink and working substance

Every heat engine, however complicated in hardware, reduces to three idealised elements. NIOS lists them as the three essential requirements of any heat engine. Memorise the trio — NEET stems are written around it.

ComponentRoleKey quantity
Source (hot reservoir)Supplies heat to the engine; held at the higher temperature $T_1$. Large enough that its temperature is unaffected by the heat drawn.Heat absorbed $Q_1$ at $T_1$
Working substanceThe matter (steam, fuel–air mixture, ideal gas) that absorbs heat, expands to do work, and is restored to its initial state each cycle.Net work output $W$
Sink (cold reservoir)Receives the waste heat the engine cannot convert; held at the lower temperature $T_2$. In practice it is the atmosphere or cooling water.Heat rejected $Q_2$ at $T_2$

The reservoirs are idealised as so large that absorbing or releasing heat does not change their temperature — the source stays at $T_1$ and the sink at $T_2$ throughout. This is what lets us treat the engine as operating "between two temperatures", a phrasing NEET uses constantly.

The energy-flow diagram

The standard schematic of a heat engine is a block diagram: heat flows down from the source, the engine taps off work sideways, and the remainder drops into the sink. Every quantity on it is positive by convention, and the arrows encode energy conservation.

Figure 1 — Energy flow in a heat engine SOURCE (hot reservoir) Temperature T₁ ENGINE working substance SINK (cold reservoir) Temperature T₂ Q₁ Q₂ W (work) = Q₁ − Q₂
Heat $Q_1$ is drawn from the source at $T_1$; the engine converts part of it into work $W$ and rejects the rest as $Q_2$ to the sink at $T_2$. Over one cycle, energy conservation forces $Q_1 = W + Q_2$.

Because the working substance returns to its initial state each cycle, its internal energy is unchanged: $\Delta U = 0$. The first law $\Delta Q = \Delta U + \Delta W$ then collapses to net heat = net work:

$$ W = Q_1 - Q_2 \qquad \text{(energy conservation over one cycle)} $$

Thermal efficiency

The figure of merit of a heat engine is its thermal efficiency $\eta$ — the fraction of the absorbed heat that is converted into useful work. NIOS defines it as the ratio of heat converted into work in a cycle to the heat taken from the source:

$$ \eta = \frac{W}{Q_1} $$

Substituting $W = Q_1 - Q_2$ from energy conservation gives the form NEET tests most often:

$$ \eta = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} $$

This expression is completely general — it holds for any heat engine, real or ideal, regardless of the working substance or the cycle. The two forms below are the same equation; choose whichever the data favours.

Form of efficiencyUse it when you knowValidity
$\eta = \dfrac{W}{Q_1}$Work output and heat absorbedAny heat engine
$\eta = 1 - \dfrac{Q_2}{Q_1}$Heat absorbed and heat rejectedAny heat engine
$\eta = 1 - \dfrac{T_2}{T_1}$Source and sink absolute temperaturesIdeal (Carnot) engine only

The third row, $\eta = 1 - T_2/T_1$, is the ceiling — the maximum efficiency any engine working between $T_1$ and $T_2$ can reach. It belongs to the reversible Carnot engine and follows from the universal Carnot relation $Q_1/Q_2 = T_1/T_2$. We develop it fully in the Carnot engine efficiency note; for now, keep it separate from the always-true $\eta = 1 - Q_2/Q_1$.

Why efficiency is always below 1

From $\eta = 1 - Q_2/Q_1$, efficiency would equal $1$ — full, 100% conversion — only if $Q_2 = 0$, i.e. if the engine rejected no heat at all and turned every joule absorbed into work. The second law of thermodynamics forbids exactly this. NCERT states the Kelvin–Planck statement: "No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work."

So $Q_2 > 0$ is mandatory in any real cyclic engine: some heat must be dumped into the sink. That makes $Q_2/Q_1 > 0$ and therefore $\eta < 1$ always. NCERT summarises it bluntly — the efficiency of a heat engine can never be unity. The waste heat $Q_2$ is not a fixable engineering defect; it is a structural demand of the second law.

The first law permits an engine of 100% efficiency — it would merely conserve energy. The second law forbids it. The gap between "permitted by energy conservation" and "permitted by nature" is the whole content of the second law, and the heat engine is where that gap is measured.
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Why the law exists

The full Kelvin–Planck and Clausius statements, and why they are equivalent, are unpacked in the second law of thermodynamics.

The cyclic P–V loop

On an indicator (pressure–volume) diagram, the working substance of an engine traces a closed loop once per cycle, returning to its start. NIOS notes the consequence directly: the work done in one cycle is represented on a P–V diagram by the area enclosed by the cycle. A clockwise loop encloses positive net work — this is an engine. (A counter-clockwise loop encloses negative work and represents a refrigerator.)

Figure 2 — Net work as the area of a closed cycle V P Net work W = enclosed area clockwise = engine
The working substance returns to its initial state each cycle, so $\Delta U = 0$. The net work per cycle equals the area enclosed by the loop; a clockwise traversal gives $W > 0$ (an engine), since heat is absorbed on the high-pressure expansion and less is rejected on compression.

The enclosed area makes the engine's bookkeeping visual: heat $Q_1$ enters along the expansion strokes, heat $Q_2$ leaves along the compression strokes, and the area between them is the net work $W = Q_1 - Q_2$. A path that does not close (a non-cyclic process) cannot drive an engine, because the substance would drift away from its starting state and could not repeat.

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Prerequisite

Each stroke of an engine cycle is one of the standard processes — isothermal, adiabatic, isobaric, isochoric. Review them in thermodynamic processes.

Steam and internal-combustion engines

Two real engines anchor the abstract model. They differ in hardware but obey the same $Q_1 = W + Q_2$ accounting.

EngineSource of $Q_1$Working substanceSink for $Q_2$
Steam engine (external combustion)Furnace burning coal/wood heats a separate boilerSteam (water vapour) driving a pistonCondenser / surrounding air
Petrol & diesel engine (internal combustion)Fuel burnt inside the cylinder itselfHot fuel–air combustion gasesExhaust gases vented to atmosphere

In the steam engine the heat is generated outside the working substance and transferred to it — external combustion. In the petrol or diesel engine the fuel is burnt inside the cylinder, so the combustion gases are the working substance — internal combustion. In both, the exhaust carries away the unavoidable rejected heat $Q_2$; you feel it as the warmth of a car's exhaust. That heat is the second law made tangible. NEET treats these qualitatively — you are expected to identify source, sink and working substance, not to compute cylinder thermodynamics.

Two threads lead out of this page. The first asks: of all engines working between $T_1$ and $T_2$, which is the most efficient, and what is its efficiency? The answer is the reversible Carnot engine, with $\eta = 1 - T_2/T_1$ — the subject of the Carnot engine efficiency note. No engine can beat it; that is Carnot's theorem.

The second thread runs the engine backwards. Supply work $W$ and the device pumps heat from cold to hot — a refrigerator or heat pump. Its performance is rated not by efficiency but by a coefficient of performance, and the Clausius statement of the second law sets the limit there. The two devices are the same physics read in opposite directions.

Quick recap

Heat engines in one breath

  • A heat engine converts heat into work over a repeating cycle, using three elements: source ($T_1$), working substance, sink ($T_2$).
  • It absorbs $Q_1$, outputs work $W$, rejects $Q_2$. Energy conservation over a cycle ($\Delta U = 0$) gives $Q_1 = W + Q_2$.
  • Thermal efficiency $\eta = \dfrac{W}{Q_1} = 1 - \dfrac{Q_2}{Q_1}$ — valid for every heat engine.
  • $\eta = 1 - \dfrac{T_2}{T_1}$ is the Carnot maximum only, and needs $T$ in kelvin.
  • $\eta < 1$ always: the Kelvin–Planck statement forbids $Q_2 = 0$, so some heat must be rejected.
  • On a P–V diagram the net work per cycle is the enclosed area; clockwise loop = engine.
  • Steam engine = external combustion; petrol/diesel = internal combustion. Same $Q_1 = W + Q_2$ accounting.

NEET PYQ Snapshot — Heat Engines

Three genuine NEET PYQs on engine efficiency and the $Q_1 = W + Q_2$ balance. Same toolkit each time: write energy conservation, then apply the right form of $\eta$.

NEET 2018

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

  1. 26.8%
  2. 20%
  3. 6.25%
  4. 12.5%
Answer: (1) 26.8%

Ideal engine ⇒ use the temperature form. Convert to kelvin: $T_2 = 0^\circ\text{C} = 273\ \text{K}$ (freezing), $T_1 = 100^\circ\text{C} = 373\ \text{K}$ (boiling). Then $\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{273}{373} = 1 - 0.732 = 0.268 = 26.8\%$. The trap is forgetting to convert Celsius to kelvin.

NEET 2023

A Carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is:

  1. 200°C
  2. 27°C
  3. 15°C
  4. 100°C
Answer: (2) 27°C

Temperature form, run in reverse. $T_1 = 327^\circ\text{C} = 600\ \text{K}$. From $\eta = 1 - \dfrac{T_2}{T_1}$ with $\eta = 0.5$: $\dfrac{T_2}{600} = 1 - 0.5 = 0.5 \Rightarrow T_2 = 300\ \text{K} = 27^\circ\text{C}$. Note the source temperature was given in Celsius — convert before substituting.

NEET 2017

A Carnot engine having an efficiency of $\tfrac{1}{10}$ as a heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:

  1. 100 J
  2. 1 J
  3. 90 J
  4. 99 J
Answer: (3) 90 J

Efficiency fixes $Q_1$, energy conservation fixes $Q_2$. Running as an engine, $\eta = \dfrac{W}{Q_1} \Rightarrow Q_1 = \dfrac{W}{\eta} = \dfrac{10}{1/10} = 100\ \text{J}$. Then $Q_2 = Q_1 - W = 100 - 10 = 90\ \text{J}$. Reversed as a refrigerator, the heat absorbed from the cold reservoir is this same $Q_2 = 90\ \text{J}$. This is $\eta = W/Q_1$ and $Q_1 = W + Q_2$ working together.

FAQs — Heat Engines

Short answers to the heat-engine questions NEET aspirants get wrong most often.

What is a heat engine in thermodynamics?
A heat engine is a device that converts heat into mechanical work by carrying a working substance through a cyclic process. In each cycle it absorbs heat Q₁ from a hot reservoir (source) at temperature T₁, converts a part of it into useful work W, and rejects the remaining heat Q₂ to a cold reservoir (sink) at temperature T₂. Because the process is cyclic, the working substance returns to its initial state and the engine can repeat the cycle indefinitely.
Why can a heat engine never be 100% efficient?
Because some heat must always be rejected to the sink. The Kelvin–Planck statement of the second law says no process can have, as its sole result, the absorption of heat from a reservoir and its complete conversion into work. So Q₂ can never be zero in a real cyclic engine, which makes η = 1 − Q₂/Q₁ strictly less than 1. Efficiency η = 1 (or 100%) would require Q₂ = 0, contradicting the second law.
What is the formula for the efficiency of a heat engine?
Thermal efficiency is η = W/Q₁, the ratio of net work output to the heat absorbed from the source. Using energy conservation over one cycle, W = Q₁ − Q₂, so η = 1 − Q₂/Q₁. This form is valid for any heat engine. Only for the ideal Carnot engine does it further reduce to η = 1 − T₂/T₁, where T₁ and T₂ are the source and sink absolute temperatures.
Why does ΔU = 0 for a heat engine over one cycle?
Internal energy U is a state variable, so its value depends only on the state of the working substance, not on the path. In a cyclic process the substance returns to its initial state, so the change in internal energy over the complete cycle is zero. The first law ΔQ = ΔU + ΔW then gives net heat = net work, i.e. Q₁ − Q₂ = W.
What are the three essential parts of a heat engine?
Three components: (1) a source or hot reservoir at temperature T₁, from which heat Q₁ is drawn; (2) a sink or cold reservoir at temperature T₂, into which waste heat Q₂ is rejected; and (3) a working substance (such as steam or a fuel–air mixture) that absorbs heat, does work in expanding, and is restored to its initial state to complete the cycle.
Is a refrigerator a heat engine?
A refrigerator is a heat engine run in reverse. External work W is supplied to the device, which extracts heat Q₂ from the cold reservoir and delivers Q₁ = Q₂ + W to the hot reservoir. Instead of efficiency, its performance is measured by the coefficient of performance. The Clausius statement of the second law forbids a perfect refrigerator that moves heat from cold to hot with no work input.
Go deeper
Thermodynamics — Parent chapter Full chapter map: laws, processes, engines, entropy. Carnot Engine Efficiency The ideal reversible engine; η = 1 − T₂/T₁; Carnot's theorem. Refrigerators & Heat Pumps The engine in reverse; coefficient of performance; Clausius limit. Second Law of Thermodynamics Kelvin–Planck and Clausius statements; why η < 1. First Law of Thermodynamics ΔQ = ΔU + ΔW; the energy bookkeeping behind every cycle. Thermodynamic Processes Isothermal, adiabatic, isobaric, isochoric, cyclic strokes.