Thermodynamics and Heat Engines

Concise answer

The first law, $\Delta E_{\text{int}} = Q - W$ , is energy bookkeeping; the second law says entropy of a closed system never decreases, which caps every heat engine at the Carnot efficiency $1 - T_c/T_h$ and makes the named ideal-gas processes the standard test cases.

Definitions

Internal energy: The total microscopic energy of a system; for an ideal gas it depends only on temperature.
First law of thermodynamics: Energy conservation for thermal systems: $\Delta E_{\text{int}} = Q - W$, with $Q$ the heat added to the system and $W$ the work done by the system.
Heat engine: A cyclic device that absorbs heat $Q_h$ from a hot reservoir, does work $W$, and rejects heat $Q_c$ to a cold reservoir.
Carnot engine: The reversible engine composed of two isotherms and two adiabats; it sets the maximum efficiency between two reservoir temperatures.
Entropy: A state variable whose change along a reversible path is $\int dQ/T$; the second law says the total entropy of a closed system never decreases.

Intuition

The first law is a checkbook: heat in is a deposit, work done by the gas is a withdrawal, and internal energy is the balance. For an ideal gas the balance is read directly off the temperature.

The Carnot limit exists because an engine must dump entropy: the heat $Q_h$ arrives carrying entropy $Q_h/T_h$ , and the only way to get rid of it is to reject heat at $T_c$ . Some heat must always be thrown away, so efficiency can never reach 100%.

Concept walkthrough

The first law, $\Delta E_{\text{int}} = Q - W$ , tracks energy through any process. The named ideal-gas processes are the standard special cases: isochoric (no work, so $\Delta E_{\text{int}} = Q$ ), isobaric ( $W = p\Delta V$ ), isothermal ( $\Delta E_{\text{int}} = 0$ , so $Q = W$ ), and adiabatic ( $Q = 0$ , so $\Delta E_{\text{int}} = -W$ , with $pV^{\gamma}$ constant). On a $pV$ diagram, work is the area under the curve and depends on the path; internal energy change does not.

A heat engine runs in a cycle, so its internal energy change per cycle is zero and $W = Q_h - Q_c$ . Its efficiency is $e = W/Q_h = 1 - Q_c/Q_h$ . The Carnot cycle — two isotherms and two adiabats, all reversible — achieves the maximum possible efficiency $e_C = 1 - T_c/T_h$ , which depends only on the reservoir temperatures (in kelvins).

Entropy makes the second law quantitative. For reversible heat transfer, $\Delta S = \int dQ/T$ ; for any closed system the total entropy change is $\ge 0$ , with equality only for reversible processes. An engine claim can be tested by bookkeeping entropy: if the numbers require total entropy to fall, the engine is impossible.

After this page, you should be able to

Apply the first law with a consistent sign convention for heat and work.
Classify isothermal, isobaric, isochoric, and adiabatic ideal-gas processes and identify which term of the first law vanishes in each.
Compute heat engine efficiency from heats or from work, and the Carnot bound from reservoir temperatures.
Calculate entropy changes for reversible heat transfer and use the second law to rule out impossible engines.

Formulas and assumptions

First law of thermodynamics

\Delta E_{\text{int}} = Q - W

Variables

Delta E_int: internal energy change in joules
Q: heat added to the system in joules
W: work done by the system in joules

Assumptions

W is the work done by the system; texts using work done on the system write Delta E_int = Q + W instead.
Q and W are path-dependent; Delta E_int is not.

Ideal gas law

pV = nRT

Variables

p: absolute pressure in pascals
V: volume in cubic meters
n: amount of gas in moles
R: gas constant
T: absolute temperature in kelvins

Assumptions

The gas is ideal: point particles with no intermolecular forces.
T must be in kelvins.

Heat engine efficiency

e = \dfrac{W}{Q_h} = 1 - \dfrac{Q_c}{Q_h}

Variables

e: dimensionless efficiency
W: net work per cycle in joules
Q_h: heat absorbed from the hot reservoir in joules
Q_c: heat rejected to the cold reservoir in joules

Assumptions

The engine operates in a cycle, so W = Q_h - Q_c by the first law.

Carnot efficiency

e_C = 1 - \dfrac{T_c}{T_h}

Variables

e_C: maximum efficiency between the two reservoirs
T_c: cold reservoir temperature in kelvins
T_h: hot reservoir temperature in kelvins

Assumptions

Temperatures are absolute (kelvins).
No engine between the same reservoirs can exceed this efficiency; reversible engines reach it.

Entropy change (reversible heat transfer)

\Delta S = \int \dfrac{dQ}{T} \quad\Rightarrow\quad \Delta S = \dfrac{Q}{T} \ \text{(constant } T\text{)}

Variables

Delta S: entropy change in joules per kelvin
dQ: reversible heat added in joules
T: absolute temperature in kelvins

Assumptions

The integral is taken along a reversible path; entropy is a state variable, so the result applies to any process between the same states.
For a closed system the total entropy change is greater than or equal to zero.

Adiabatic ideal-gas relation

pV^{\gamma} = \text{constant}

Variables

gamma: heat capacity ratio C_p / C_v
p: pressure in pascals
V: volume in cubic meters

Assumptions

The process is quasi-static and adiabatic (Q = 0) for an ideal gas.
Equivalently T V^(gamma - 1) is constant.

Worked example

Carnot engine between 500 K and 300 K

A Carnot engine operates between reservoirs at $T_h = 500\,\text{K}$ and $T_c = 300\,\text{K}$ and absorbs $Q_h = 1000\,\text{J}$ per cycle. Find the efficiency, the work per cycle, and the heat rejected, and verify the entropy bookkeeping.

1Efficiency: $e_C = 1 - T_c/T_h = 1 - 300/500 = 0.40$ , i.e. 40%.
2Work per cycle: $W = e_C Q_h = (0.40)(1000) = 400\,\text{J}$ .
3Heat rejected: $Q_c = Q_h - W = 1000 - 400 = 600\,\text{J}$ .
4Entropy check: the hot reservoir loses $Q_h/T_h = 1000/500 = 2.0\,\text{J/K}$ and the cold reservoir gains $Q_c/T_c = 600/300 = 2.0\,\text{J/K}$ , so the total entropy change is zero — consistent with a reversible cycle.

$e_C = 40\%$ , $W = 400\,\text{J}$ , and $Q_c = 600\,\text{J}$ per cycle, with zero net entropy change as required for a reversible engine.

Common traps

Using Celsius temperatures in $e_C = 1 - T_c/T_h$ or $\Delta S = Q/T$ ; both require kelvins.
Mixing sign conventions: with $\Delta E_{\text{int}} = Q - W$ , the $W$ is work done by the gas, so compression makes $W$ negative.
Treating heat or work as state variables; both depend on the path on the $pV$ diagram, while internal energy does not.
Applying $\Delta E_{\text{int}} = 0$ to any slow process; it holds for isothermal ideal-gas processes, not adiabatic ones (where $Q = 0$ instead).
Concluding entropy can never decrease anywhere; a subsystem's entropy can fall as long as the total for the closed system does not.

GRE Physics overview Energy and Momentum Conservation Quantum Mechanics Foundations Sign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
University Physics Volume 2, Section 3.3: First Law of Thermodynamics — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 3.4: Thermodynamic Processes — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 4.2: Heat Engines — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 4.5: The Carnot Cycle — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 4.6: Entropy — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

Review and maintenance

Source-checked by publisher

Publisher record: Keiko Study editorial owner; Publisher placeholder; no individual credential claim is made yet.
Reviewer record: Technical reviewer pending; Placeholder only; this content is not labeled as reviewed by a named specialist.
Last source check: 2026-08-02
Next scheduled review: 2026-11-02

Recheck ETS content areas and the OpenStax thermodynamics references before each major GRE Physics preparation cycle.

Report an issue with this page

Thermodynamics and Heat Engines

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

First law of thermodynamics

Ideal gas law

Heat engine efficiency

Carnot efficiency

Entropy change (reversible heat transfer)

Adiabatic ideal-gas relation

Worked example

Carnot engine between 500 K and 300 K

Common traps

Related pages and practice

Sources

Review and maintenance