Thermodynamics

Conversion of organized work into internal modes with macroscopic entropy production. Total energy remains conserved. For a passive linear material, the loss coefficient must give nonnegative supplied work over a completed cycle.

Pressure-temperature path-work identity ( $W_{c} = (W_{a} + W_{b}) /2$ )

Suppose a small volume change obeys

d V = - V_{1} κ_{T} d P + V_{1} α d T

, with its displayed coefficients held constant. Work on the body is

δ W = V_{1} κ_{T} P d P - V_{1} α P d T

. Two rectangular pressure-temperature paths between the same endpoints differ by

W_{a} - W_{b} = V_{1} α Δ P Δ T

, where path a changes temperature first. The straight interpolation samples the mean pressure during its temperature change, giving

W_{c} = (W_{a} + W_{b}) /2

. Variable material coefficients need not preserve that average relation.

Thermal energy

The characteristic thermal energy per microscopic degree-of-freedom scale is

k_{B} T

. Equipartition gives

k_{B} T /2

per independent quadratic term under its classical assumptions.

Maxwell's demon

A thought experiment about measurement, selective control and the Second law of thermodynamics. An active stochastic cooling system uses and dissipates resources while reducing beam deviations.

State of matter

A state of matter is a broad category of material behavior. A solid resists a sustained change of shape, a liquid flows while remaining difficult to compress, and a gas expands to fill its container. These distinctions are useful first descriptions, but they do not enumerate all thermodynamic phases. Plasma adds collective behavior of charged particles, while liquid crystals show that flow and structural order need not disappear together.

Distinguish three questions: what broad state of matter does the material resemble, which thermodynamic phase is present, and what are its temperature, pressure and other state variables? Two samples at different temperatures can be in the same liquid thermodynamic phase. Conversely, crystal polymorphism gives different solid thermodynamic phases of the same substance, distinguished by their crystal structures. A phase transition between them need not change the broad state-of-matter category.

For an imagined substance with crystalline forms A and B, melting A produces a solid–liquid phase transition, while converting A to B produces a solid–solid phase transition. A sample containing A and B has two thermodynamic phases even though all its material is solid. Cutting one uniform crystal into two pieces does not create a second thermodynamic phase: count distinct equilibrium forms, rather than pieces or grains. This distinction matters when applying the phase rule.

A phase diagram locates stable thermodynamic phases using temperature and pressure. Its liquid–gas phase coexistence curve ends at a liquid-gas critical point, around which a path connects liquid and gas without a first-order phase transition. Broad categories can therefore remain useful without a sharp boundary.^[ref]

Solid

 Tagged

Crystal structure

Crystal

A crystal is a solid with spatially ordered constituents.

Magnetic unit cell

A magnetic unit cell repeats the ordered magnetic moments as well as the atomic positions. It can exceed the chemical primitive unit cell when translations interchange oppositely ordered magnetic sublattices. For a simple-cubic Néel pattern

(- 1)^{n_{x} + n_{y} + n_{z}}

, translations preserving the sum's parity form an FCC lattice with conventional parameter twice the chemical spacing and a two-spin primitive basis.

Polycrystalline material

A solid containing many differently oriented crystalline crystal grains. grain boundaries impose compatibility and act as transport paths or sinks for mobile defects.

Grain in a polycrystal

A region with an approximately common crystal orientation, separated from other such regions by grain boundaries.

Grain aspect ratio

A measure of crystal grain elongation. It changes directional boundary spacings and can influence anisotropic defect transport independently of average grain size.

Grain size

A measure of the spatial scale of crystal grains. Its definition must be stated for elongated crystal grains; size, aspect ratio and orientation distribution have distinct physical effects.

Crystal packing

The spatial arrangement and intermolecular contacts of components in a crystal.

Cocrystal

A crystalline phase containing two or more distinct molecular components in one lattice.

Hexagonal close packing

Close packing of triangular layers with ABAB stacking. Its nearest-layer geometry agrees with face-centered cubic stacking, but the further-layer registry differs.

Face-centered cubic

A cubic crystal structure with lattice sites at cube corners and face centres. Its close-packed layers are (111) planes, stacked ABCABC.

Wulff construction

The fixed-volume equilibrium crystal shape obtained by minimizing orientation-dependent surface free energy. Facet support-plane distances are proportional to their interfacial free energies under the construction assumptions.

Winterbottom construction

An equilibrium construction for a supported particle that modifies the Wulff construction using particle-support and support-medium interfacial energies. The support truncates the corresponding free-particle shape.

Single crystal

Liquid

Radial distribution function ( $d N_{r} = 4 πn r^{2} g (r) d r$ )

In a homogeneous isotropic one-component fluid,

g (r)

is the conditional number density at separation

r

from a tagged particle divided by the bulk number density

n

. Thus the displayed shell contains the mean number of neighbours. For an uncorrelated ideal gas

g (r) = 1

away from the tagged particle; repulsive cores suppress small separations, while liquid coordination shells can produce damped oscillations approaching one at large distance. This is a pair correlation, not a normalized one-particle radial probability density. The static structure factor measures its Fourier-space counterpart.

Virial pressure from pair correlations ( $p = n k_{B} T - (2 π n^{2} /3) \int_{0}^{\infty} r^{3} v^{'} (r) g (r) d r$ )

For a differentiable central pair potential in a classical homogeneous three-dimensional fluid, the mechanical virial theorem gives the displayed pressure. The radial distribution function accounts for the probability of each pair separation. In the dilute limit,

g (r) = e^{- v (r) / (k_{B} T)} + O (n)

, and integration by parts yields

B_{2} (T) = - 2 π \int_{0}^{\infty} [e^{- v (r) / (k_{B} T)} - 1] r^{2} d r

. Thus the second virial coefficient controls

p / (n k_{B} T) = 1 + B_{2} n + \dots

. A few virial terms are a low-density approximation, not generally a converged description of a dense liquid. A discontinuous hard core needs its contact form or a limiting argument.

Pair-correlation expression for internal energy ( $U / N = 3 k_{B} T /2 + 2 πn \int_{0}^{\infty} r^{2} v (r) g (r) d r$ )

For a classical monatomic homogeneous fluid with pairwise central pair potential

v (r)

, the mean interaction energy is one half the sum over neighbours of a tagged particle. Using the radial distribution function gives the displayed formula; the factor one half prevents counting each pair twice. The kinetic term follows from the equipartition theorem. Its use requires the relevant integral to exist and a thermodynamic state specified by number density and temperature. Quantum kinetic contributions, many-body forces and internal molecular modes require modifications.

Gas

Elastic fluid

A historical physical name for a gas or vapour, emphasizing its mobility and appreciable compressibility. It does not imply an exactly ideal gas or prevent a vapour from undergoing condensation.

Gas law

A relation among quantities such as pressure, volume, temperature and amount of substance for a gas.

Vapor

Polymorphism (materials science)

Crystal polymorphism is the existence of different crystal structures for the same chemical substance. Each crystal polymorph can have its own mass density, melting behavior and range of stability on a phase diagram. A different external shape of the same crystal structure alone does not establish a different crystal polymorph.^[ref]

At fixed temperature and pressure, compare the chemical potentials of the possible crystal polymorphs. The lower value determines the stable bulk thermodynamic phase; equality permits phase coexistence when neither form is beaten by another thermodynamic phase. A higher-chemical potential form can persist through metastability, so finding it in a sample does not by itself locate an equilibrium boundary.

Crystal polymorph

A solid thermodynamic phase with a particular crystal structure, among distinct crystalline forms of the same chemical substance. Crystal polymorphism names the existence of these alternatives; a crystal polymorph names one of the forms.

Thermodynamic system

A thermodynamic system is the part of the physical world selected for an energy and matter balance. Draw its boundary before classifying an interaction as heat, work or transported matter. The boundary may move: a sealed piston encloses a closed system even while its volume changes. An isolated system additionally excludes all external energy transfers.^[ref]

Boundary choice changes the ledger. If an electrical heater warms water, choosing only the water counts incoming heat; including the heater counts electrical work entering through its leads. The combined internal energy records storage, without separate stored categories for the two transfer mechanisms.^[ref]

Thermodynamic state

An equilibrium thermodynamic state specifies sufficient independent macroscopic properties, together with material identity and constraints, to determine the other equilibrium properties. For a uniform fixed-composition ideal gas, its amount of substance, temperature and volume determine its pressure through the ideal gas law. These properties specify a state, without specifying a thermodynamic process that prepared it.^[ref]

Thermodynamic process

A thermodynamic process changes a thermodynamic system between states. A quasistatic process supplies equilibrium intermediate states and therefore a path on equilibrium coordinates. An irreversible thermodynamic process can have equilibrium endpoints while lacking a uniform intermediate temperature or pressure. Endpoint values still determine changes in a state function; computing heat and work requires the actual interaction history.

Simple compressible system

A thermodynamic system for which the reversible mechanical work is entirely pressure-volume work. At fixed composition its internal energy is specified by entropy and volume; additional work modes require additional state variables.

Thermodynamic equilibrium

A thermodynamic system is in thermodynamic equilibrium when its macroscopic state has no spontaneous tendency to change through any exchange allowed by its constraints. For an isolated system, stable thermodynamic equilibrium maximizes entropy over the accessible states at fixed total energy and conserved material inventories. A wall can forbid an exchange, so always specify what can cross or move. Thermal, mechanical and chemical conditions answer different questions about that same constrained state. ^[ref]

For example, two compartments separated by a rigid, impermeable wall that transmits heat can reach thermal equilibrium while retaining different pressures and compositions. Allowing an unloaded piston to move adds a mechanical equilibrium condition. Allowing one chemical species to cross adds chemical potential equality for particle exchange; allowing a chemical reaction adds its chemical equilibrium condition. A fixed barrier can therefore support a constrained thermodynamic equilibrium that ceases to be an equilibrium when the barrier is removed.

Uniformity is not the definition of thermodynamic equilibrium. An isothermal atmosphere in hydrostatic equilibrium can have a common temperature while its pressure and mass density vary with height: the pressure gradient balances gravity. In an external field, the exchange condition must include the relevant potential energy; for charged matter this leads to the electrochemical potential. Relativistic gravitational corrections to thermal equilibrium are described separately by the Tolman temperature law.

A steady state only says that the macroscopic description is constant in time. A bar maintained between hot and cold thermal reservoirs has steady heat flux and produces entropy, so it is not in thermodynamic equilibrium. Even matching temperatures at one perfect-contact interface can coexist with heat transfer through the surrounding material; matching those two local values does not establish equilibrium of the whole apparatus. See thermal conduction for that spatial transport problem.

The useful simplification is that equilibrium conditions predict an endpoint without computing every collision or flow during relaxation. The first law of thermodynamics supplies the conserved quantities, and the Second law of thermodynamics supplies the direction of change. For a system held at fixed temperature and volume, the corresponding criterion is minimization of Helmholtz free energy; at fixed temperature and pressure, it is minimization of Gibbs free energy, with the material and exchange constraints still imposed.

Phase equilibrium

Coexistence without a thermodynamic driving force for further exchange. Phases able to exchange heat and constituents have equal temperature and chemical potentials of the exchanging species. Mechanical balance is also needed; interface curvature can require a pressure jump rather than equal bulk pressures.

Mechanical equilibrium

Mechanical equilibrium requires force balance for the mechanical displacements that are allowed. Two uniform fluids separated by a freely moving, unloaded, frictionless piston with equal face areas require equal pressures. A supported piston can sustain a pressure difference, and a fluid in gravity instead obeys hydrostatic equilibrium. For an upward height coordinate

z

, that balance is

d p / d z = - ρ g

, where

ρ

is mass density and

g

is the downward gravitational acceleration. Mechanical equilibrium alone does not establish thermal equilibrium or chemical equilibrium.

Pressure equilibrium ( $p_{1} = p_{2}$ )

Two fluids at a stationary, freely movable interface require equal pressures when surface tension and other normal stresses are negligible. This local pressure equilibrium does not require equal temperatures or mass densities. Curvature with nonzero surface tension or an applied load changes the normal-stress balance. Comparing points at different heights also requires the hydrostatic equilibrium condition.

Thermodynamic stability

Mechanical stability ( $F^{''} (x_{*}) > 0$ )

For a mechanical coordinate

x

at fixed temperature, total volume and particle numbers, the relevant thermodynamic potential is the Helmholtz free energy

F (x)

. Its generalized force is

f = - F^{'} (x)

. At mechanical equilibrium,

f (x_{*}) = 0

, and a linearly restoring force has

f^{'} (x_{*}) < 0

, equivalently

F^{''} (x_{*}) > 0

. A Taylor expansion then proves a strict local free-energy minimum. If

F

is a strictly convex function throughout the accessible domain, the stationary state is the unique global minimum.

Closed system

A closed system exchanges no matter with its surroundings, but can exchange energy through heat or work. A sealed cylinder with a moving piston is a typical idealized example. “Closed” fixes the material inventory, rather than the volume or the internal energy. An insulated cylinder can receive compression work; a rigid sealed vessel can receive heat. Neither is necessarily an isolated system.

Isolated system

An isolated system exchanges neither matter nor energy with its surroundings. Its total energy is fixed, while the Second law of thermodynamics requires its total entropy to be nondecreasing. Its internal energy alone is fixed when bulk kinetic energy and external potential energy also remain unchanged. For example, an initially spinning stirrer inside an insulated rigid vessel can lose kinetic energy while the contents gain internal energy, with no external input.

An isolated system need not yet be in thermodynamic equilibrium. Two enclosed bodies at different temperatures can exchange heat internally until they reach thermal equilibrium; their combined energy stays constant. Their internal transfer is absent from the outer boundary ledger. A vacuum-insulated container is only an approximation to isolation: residual thermal conduction and thermal radiation can still carry energy across its boundary.

Quasistatic process

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance

n

of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature

T

. Reduce the external load in infinitesimal steps so that the piston expands from

V_{0}

2 V_{0}

. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law gives

p (V) = \frac{n RT}{V}, n RT = 1 kJ, V_{0} = 1 L = 1 0^{- 3} m^{3}, p_{0} = \frac{n RT}{V _{0}} = 1 MPa .

Here

R

is the molar gas constant; specifying

n RT

fixes an energy scale, without choosing

n

and

T

separately. The initial and final equilibrium states are

(V_{0}, p_{0}, T)

and

(2 V_{0}, p_{0} /2, T)

. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:

W_{out, rev} = \int_{V_{0}}^{2 V_{0}} p (V) d V = n RT ln 2 \approx 0.693147 kJ .

The integral of

1/ V

supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature,

Δ U = 0

; the first law of thermodynamics,

Δ U = Q_{in} - W_{out}

, gives

Q_{in, rev} = 0.693147 kJ

to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Figure 1.
Reversible isothermal expansion: work is the area beneath the equilibrium pressure curve
. Fixed amount of substance of ideal gas, with $n RT = 1 kJ$ and $V_{0} = 1 L$ . The blue curve gives every intermediate equilibrium state; its shaded area is $0.693147 kJ$ . The arrow indicates expansion, without specifying a time dependence. **Details**.

Now start from the same initial state, but abruptly lower the external pressure to the constant value

p_{ext} = p_{0} /2 = 0.5 MPa

. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature

T

, mechanical equilibrium and the ideal gas law give

V_{f} = n RT / p_{ext} = 2 V_{0}

. The work delivered to the constant load is

W_{out, load} = p_{ext} (V_{f} - V_{0}) = (0.5 MPa) (1 L) = 0.5 kJ, Δ U = 0, Q_{in, load} = 0.5 kJ .

This work follows directly from load force times piston displacement, using

d V = A d x

for piston area

A

. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither

p (V) = n RT / V

at the bath temperature nor

p (V) = p_{ext}

describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

Figure 2.
The same equilibrium endpoints give different work for reversible and constant-load expansion
. The blue curve is the reversible ideal gas path. The dashed orange line is constant external pressure, whose rectangle gives $0.5 kJ$ of load work. The additional blue shading is $0.193147 kJ$ available in the reversible case. Black points mark the common gas endpoints; the open orange point marks the initial external load, and the dotted gap indicates the initial imbalance. No irreversible internal gas path is drawn. **Details**.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:

Δ S_{gas} = \int_{V_{0}}^{2 V_{0}} \frac{δ Q _{rev}}{T} = n R ln 2.

The thermal reservoir stays at

T

and loses heat

Q_{in} = W_{out}

, so

Δ S_{bath} = - W_{out} / T

. Assume the ideal load stores work without an entropy change. The combined entropy production is therefore

Δ S_{total} = n R ln 2 - \frac{W _{out}}{T} = {0, n R (ln 2 - 1/2) > 0, reversible expansion, constant external load .

(5)

No numerical entropy unit is implied by

n RT = 1 kJ

alone:

n R

T

must also be specified. The lost work opportunity is

T Δ S_{total} = 0.193147 kJ

for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume

V_{0}

inside a rigid, insulated vessel of total volume

2 V_{0}

. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so

W_{out} = Q_{in} = 0

and

Δ U = 0

. After equilibration the temperature is again

T

, the pressure is

p_{0} /2

, and the gas entropy has increased by

n R ln 2

. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

State function

A state function assigns a value to each equilibrium thermodynamic state of a specified thermodynamic system. Its change depends only on the initial and final states. Internal energy

U

, entropy

S

and enthalpy

H = U + p V

are state functions; here

p

is pressure and

V

is volume. Transferred heat

Q

and work

W

belong to a thermodynamic process, rather than to a state. A system stores internal energy; it does not store separate inventories of “heat received” and “work received”.^[ref]

Use the explicit convention

Q > 0

for heat entering and

W_{by} > 0

for work done by the thermodynamic system on its surroundings. For a closed system with unchanged endpoint bulk kinetic energy and external potential energy, the first law of thermodynamics reads

Δ U = Q - W_{by}, d U = δ Q - δ W_{by} .

The notation

d U

denotes an exact differential of a state function;

δ Q

and

δ W_{by}

denote incremental transfers. Along a reversible thermodynamic process of a simple compressible system,

δ W_{by, rev} = p d V, δ Q_{rev} = T d S,

where

T

is absolute temperature. These are rules for particular thermodynamic processes, not definitions of stored heat or stored work. In particular, a quasistatic process with friction can remain near thermodynamic equilibrium while producing entropy.^[ref]

For any state function

F

and two equilibrium states

A, B

, an integral of its exact differential gives

\int_{A}^{B} d F = F (B) - F (A), \oint d F = 0.

The proof is telescoping: add the changes

F (x_{i + 1}) - F (x_{i})

along a subdivided path, and all intermediate values cancel. The smooth limit gives the integral. Conversely, if an integrable differential form has zero integral around every accessible closed path in a connected region, define

F (B)

by integrating from a reference state. Two routes differ by a closed path, so the definition is unambiguous. This argument needs every accessible loop, rather than a local test that might miss a hole in the domain; see path independence.

The same cancellation does not hold separately for heat and work. A thermodynamic cycle has

Δ U = 0

but can have nonzero net transfers, with

\oint δ Q = \oint δ W_{by}

. Nor does the fact that entropy is a state function make

Q / T

its general change: the formula

d S = δ Q_{rev} / T

uses a reversible thermodynamic process and its varying temperature.

This distinction makes endpoint calculations powerful. We can compute an internal energy, entropy or enthalpy difference along a convenient reversible reference path even when the actual thermodynamic process is irreversible. Its actual heat and work must still be obtained from its own boundary interactions. The appealing cancellation is visible in two reversible ideal-gas paths between the same states: a different mechanical output requires a matching difference in thermal input, while stored internal energy arrives at the same value.

Two reversible ideal-gas paths between the same states

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume

C_{V} = 3 n R /2

. Stipulate amount of substance

n = 1 mol

and molar gas constant

R = 8.314 J mo l^{- 1} K^{- 1}

for this calculation. Set initial temperature

T_{0} = 300 K

and volume

V_{0} = 0.010 m^{3}

, and define

E_{0} = n R T_{0} = 2.4942 kJ

and

p_{0} = E_{0} / V_{0} = 249.42 kPa

. The specified equilibrium endpoints are

A : (V, T, p) = (V_{0}, T_{0}, p_{0}), B : (V, T, p) = (3 V_{0}, 2 T_{0}, 2 p_{0} /3) .

Thus

B

has volume

0.030 m^{3}

, temperature

600 K

and pressure

166.28 kPa

. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure in

W_{by, rev} = \int_{path} p d V .

An endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from

A

C = (V_{0}, 2 T_{0}, 2 p_{0})

, then undergoes isothermal expansion at

2 T_{0}

B

. The lower route undergoes isothermal expansion at

T_{0}

from

A

D = (3 V_{0}, T_{0}, p_{0} /3)

, then heats at fixed volume to

B

. The tuples use the same

(V, T, p)

ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

Figure 1.
Two reversible ideal gas paths connect the same equilibrium endpoints
. Original idealized pressure-volume diagram. The upper route is $A \to C \to B$ and the lower route is $A \to D \to B$ ; the shaded gap is the difference in work, not an experimental uncertainty. Both are reversible thermodynamic processes under the stated piston and thermal reservoir assumptions. **Details**.

The internal energy of an ideal gas gives the same endpoint difference for both routes:

Δ U = \frac{3}{2} n R (2 T_{0} - T_{0}) = \frac{3}{2} E_{0} = 3.7413 kJ .

On either fixed-volume heating leg,

W_{by} = 0

and

Q = 3 E_{0} /2

. On each isothermal expansion,

Δ U = 0

, so the first law of thermodynamics gives

Q = W_{by}

. The ideal gas law and the integral of

1/ V

then give

W_{by, upper} W_{by, lower} = \int_{V_{0}}^{3 V_{0}} \frac{2 E _{0}}{V} d V = 2 E_{0} ln 3, = \int_{V_{0}}^{3 V_{0}} \frac{E _{0}}{V} d V = E_{0} ln 3, Q_{upper} Q_{lower} = \frac{3}{2} E_{0} + 2 E_{0} ln 3, = \frac{3}{2} E_{0} + E_{0} ln 3.

Here

ln

is the natural logarithm. Rounded to

0.001 kJ

, the upper route has

(Q, W_{by}, Δ U) = (9.222, 5.480, 3.741) kJ

and the lower route has

(6.481, 2.740, 3.741) kJ

. Subtracting separately rounded entries can differ by

0.001 kJ

; the unrounded balances are exact in the stipulated model.

Figure 2.
Different heat and work transfers leave the same internal energy increment
. Original calculated energy ledger for the two reversible paths. Each incoming heat bar is divided into outgoing work and the common stored increment $Δ U$ . Labels are rounded; bar lengths use unrounded values. These are process totals, not inventories of stored heat or stored work. **Details**.

The shaded area between the curves explains the entire difference:

W_{by, upper} - W_{by, lower} = Q_{upper} - Q_{lower} = E_{0} ln 3 \approx 2.740159 kJ .

(5)

Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy

H = 5 n RT /2

in this model,

Δ H = \frac{5}{2} E_{0} = 6.2355 kJ, Δ S = \frac{3}{2} n R ln 2 + n R ln 3 \approx 17.7781 J K^{- 1} .

(6)

To obtain the entropy change along either route, integrate

C_{V} d T / T

on its heating leg, giving

C_{V} ln 2

, and

n R d V / V

on its isothermal expansion, giving

n R ln 3

. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total

Q

divided by its final temperature equals

Δ S

, and neither total

Q

equals

Δ H

: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

Mechanical input and thermal storage in a calorimeter

The UCSB demonstration uses an aluminum cylinder turned by a crank, with a wrapped rope supporting a hanging mass. An embedded electrical thermometer connects through sliding contacts to an electrical resistance meter; a supplied electrical resistance–temperature table or fitted equation converts the readout. A counter records turns. Record the initial temperature, crank steadily, then stop and convert the minimum recorded electrical resistance to the final temperature.^[ref]

The proposed control is to start below room temperature and finish comparably above it, approximately balancing incoming and outgoing environmental heat. Neglected accessory heat capacities and rough temperature estimates limit accuracy. The mechanical input and thermal response test an energy balance; warming alone does not establish quantitative equivalence.^[ref]

This is a modern companion to Joule's paddle-wheel experiment, with a different apparatus. Choose the cylinder as the closed system. When endpoint bulk mechanical energy changes are negligible, a model for its thermal storage is

Δ U ≃ C_{eff} (T_{f} - T_{i}) = W_{on} + Q_{room}, W_{by} = - W_{on},

where

C_{eff}

is its participating heat capacity, approximately constant over the run, and

Q_{room}

is signed environmental heat. The cylinder is not an isolated system, since mechanical work enters. The temperature rise estimates a change in internal energy, rather than an amount of externally supplied heat. This boundary distinction connects a measured thermal response to the same stored quantity used in two reversible ideal-gas paths between the same states.

Heat engine

A heat engine uses a repeating thermodynamic cycle to turn part of an incoming heat transfer into net outgoing work. In the two-thermal reservoir model, the hot thermal reservoir supplies heat

Q_{h} > 0

at fixed temperature

T_{h}

, and the cold thermal reservoir receives heat of magnitude

Q_{c} > 0

at fixed temperature

T_{c} < T_{h}

. A Stirling engine makes the repeated motion visible; the abstract energy budget applies independently of its particular mechanism.^[ref]

Choose the entire cyclic device as a closed system, with no net change in bulk kinetic energy or potential energy. Count heat entering as positive and net work delivered as positive, as in the first law of thermodynamics. Thus

Q_{h}

enters with a plus sign and

Q_{c}

, although a positive magnitude, enters with a minus sign. Assume these two heat transfers and the net work are the only transfers across the boundary. Over one complete thermodynamic cycle, all stored energy, including internal energy

U

, returns to its starting value:

Δ U = Q_{h} - Q_{c} - W = 0, W = Q_{h} - Q_{c} .

The device does not accumulate heat from cycle to cycle. More precisely, heat is a transfer, while internal energy is what can be stored; even a component that temporarily stores internal energy must return to its initial state for this boundary to be cyclic.^[ref]

For engine operation,

Q_{h} > Q_{c}

and

W > 0

. Its thermal efficiency is

η = \frac{W}{Q _{h}} = 1 - \frac{Q _{c}}{Q _{h}} .

Use net work, after subtracting any work supplied during compression or required by auxiliaries included in the chosen device. For periodic operation, average power satisfies

P_{out} = \dot{Q}_{h} - \dot{Q}_{c}

and

η = P_{out} / \dot{Q}_{h}

, with the rates averaged over complete thermodynamic cycles.^[ref]

For a stipulated model at

T_{h} = 600 K

and

T_{c} = 300 K

, let

Q_{h} = 1000 J

and

Q_{c} = 600 J

per thermodynamic cycle. Then

W = 400 J

and

η = 2/5 = 40%

. At two thermodynamic cycles per second, these become

2000 W

incoming heat transfer,

1200 W

outgoing heat transfer, and

800 W

net power. These are constructed values, not measurements.

Figure 1.
Two-thermal reservoir heat engine: $1000 J$ in, $400 J$ of net work, and $600 J$ rejected
. Illustrative transfers per complete thermodynamic cycle, with positive magnitudes labeled on the arrows. The closed system returns to its initial state. Arrow widths represent the energy split; the thermal reservoirs remain at $600 K$ and $300 K$ . **Details**.

Different signs describe different tasks. With the same directions of heat transfer,

Q_{h} = Q_{c}

gives

W = 0

: a device transmitting heat without net work output. If

Q_{c} > Q_{h}

, net work is supplied to that device. A refrigerator or heat pump instead extracts heat from the cold side and delivers heat to the hot side, using positive input work

W_{in}

and the balance

Q_{h} = Q_{c} + W_{in}

. Their useful output is heat transfer, measured by a coefficient of performance. Finally, a one-time isothermal expansion can have

Q = W

and

Δ U = 0

while leaving the volume changed: it is not a complete thermodynamic cycle. A warming device also fails the cyclic assumption, and a fuel-consuming device requires an accounting of matter crossing its boundary.

For two fixed thermal reservoirs at positive thermodynamic temperatures, the Second law of thermodynamics strengthens the energy balance to

η \leq η_{C} = 1 - \frac{T _{c}}{T _{h}} .

Equality requires the overall operation to be a reversible thermodynamic process, including its heat transfer to the thermal reservoirs. The Carnot cycle explains how the reversible benchmark is constructed, and Carnot's theorem explains its universality and the entropy production penalty. Use kelvin values, never ratios of degree Celsius readings. The example above lies below the

50%

bound: its total entropy production is

- 1000/600 + 600/300 = 1/3 J K^{- 1}

per thermodynamic cycle.

Figure 2.
Thermal efficiency versus hot thermal reservoir temperature at $T_{c} = 300 K$
. The upper curve is the reversible bound in Carnot's theorem. The lower curve is the stipulated model $η = 0.8 η_{C}$ , not experimental data or a universal engine law. At $T_{h} = 600 K$ , the curves give $50%$ and $40%$ , matching the energy budget above. The endpoint $T_{h} = T_{c}$ is a zero-output limit. **Details**.

This is useful before building a machine: warmer cooling surroundings reduce the possible thermal efficiency, even if the mechanism is unchanged. A small observed thermal efficiency does not by itself test the Second law of thermodynamics; heat leaks, friction, incomplete recovery of work, and the measurement boundary can all reduce the reported ratio. A meaningful comparison requires the actual thermal reservoir temperatures, matched heat and net work measurements, and the hypotheses of Carnot's theorem. Finite bodies whose temperatures change are treated separately there.

The practical question already appears in the 1824 Réflexions sur la puissance motrice du feu et sur les machines propres à développer cette puissance by Sadi Carnot: pp. 6–8 ask whether machines driven by heat have a fundamental limit and whether changing the working substance can improve their output. Pages 9–11 describe heating and cooling in real machines, but use the historical caloric theory, rather than the modern first law of thermodynamics balance. The historical discussion under Carnot's theorem explains that distinction.

Rankine cycle

A vapour-power cycle using a liquid pump, high-pressure heat addition, turbine expansion and low-pressure condensation. Ideal components are reversible apart from heat transfer through finite temperature differences. With states numbered pump inlet, pump outlet, turbine inlet and turbine outlet, the net thermal efficiency is

[(h_{3} - h_{4}) - (h_{2} - h_{1})] / (h_{3} - h_{2})

. Liquid pumping avoids the impractical compression of a wet two-phase mixture in a steam realization of the Carnot cycle. Named after William John Macquorn Rankine.

Isobaric-isochoric-adiabatic heat engine

Consider an ideal-gas cycle with constant heat-capacity ratio

γ > 1

: expand isobarically from

V_{1}

r V_{1}

, cool isochorically, then compress reversibly and adiabatically to the initial state. With

P_{*} = p_{1} V_{1} = n R T_{1}

, closure gives

T_{2} = r T_{1}

and

T_{3} = r^{1 - γ} T_{1}

. The heat input is

Q_{h} = γ P_{*} (r - 1) / (γ - 1)

and the rejected heat magnitude is

Q_{c} = P_{*} (r - r^{1 - γ}) / (γ - 1)

. Thus

η = 1 - (r - r^{1 - γ}) / [γ (r - 1)]

. Reservoir temperatures determine feasibility and entropy generation but need not coincide with the gas temperatures at heat-exchange endpoints.

Heat input to a cycle ( $Q_{in} = \int_{δ Q > 0} δ Q$ )

The heat input to a cycle sums positive transfers of heat into its working substance. It is not the net heat over the entire cycle, which equals net work. A single geometric branch can contain both heat absorption and rejection, so grouping the branch into one net transfer can overestimate a heat engine's thermal efficiency.

Reversible heat engine

An ideal cyclic heat engine producing no entropy overall. Between two equilibrium reservoirs its efficiency reaches the Carnot bound; finite reservoir temperatures can be treated as varying through successive reversible cycles.

Stirling engine

A Stirling engine repeatedly heats and cools a confined gas, whose expansion and compression drive a moving mechanical load. This makes it a particularly accessible example of a heat engine: both the repeating motion and the two regions exchanging heat can be observed.

In the published Demonstration experiments with a Stirling engine by Christopher G Deacon and others, an electrical heater warms one end of a cylinder and flowing water cools the other at

12^{\circ} C

. A moving displacer shuttles gas through copper wool that temporarily stores and returns internal energy. A laser spot traces a closed pressure-volume path on a screen: mechanical motion controls one coordinate, pressure the other. Calibrating both scales converts the enclosed area to work per thermodynamic cycle; multiplying by rotation frequency gives gas-cycle power. At

97 W

heater input, the paper reports

4.4 W

gas-cycle power, while a separate Prony brake measures

3.4 W

shaft output. The cold-side temperature is controlled and two higher heater settings are compared. The gas has a nonuniform temperature, and heater energy also escapes to the room: neither output/input ratio measures

W / Q_{h}

for an ideal two-thermal reservoir engine. These observations demonstrate work production and losses, not a direct measurement of the bound in Carnot's theorem. The inspected apparatus, readouts and results are on pp. 180–182, Figures 1–3 and Table 1.

Unregenerated Stirling cycle ( $η = \frac{n R ( T _{h} - T _{c} ) l n r}{n R T _{h} l n r + C _{V} ( T _{h} - T _{c} )}$ )

For an ideal working gas and

r = V_{2} / V_{1} > 1

, two isotherms at

T_{h}, T_{c}

produce net work

n R (T_{h} - T_{c}) ln r

. Without a regenerator, isochoric heating requires additional external heat

C_{V} (T_{h} - T_{c})

, giving the displayed thermal efficiency, strictly below Carnot efficiency. A perfect reversible regenerator recovers this heat and attains the bound instead. Reversing the unregenerated cycle gives cooling coefficient of performance

T_{c} / (T_{h} - T_{c}) - C_{V} / (n R ln r)

; positive cooling requires a sufficiently large expansion ratio.

Prony brake

A Prony brake measures shaft power using a resisting band: its force difference gives torque

τ = r (F_{1} - F_{2})

, and rotation gives

P = τ ω

, where

ω

is angular velocity. The Stirling engine experiment above uses this independent load measurement to distinguish shaft output from gas-cycle work.

Heat-engine calorimetry

Heat-engine calorimetry measures heat transfers so that observed work can be compared with the correct energy input. Measuring motion alone cannot supply the denominator of thermal efficiency.

The institutional Laboratory 16: Stirling Engine describes a modern teaching arrangement: an ethanol lamp drives a Stirling engine coupled to an electrical output device, which supplies a resistor immersed in water in a calorimeter. The prescribed readouts include fuel mass consumed, the pressure-volume loop, electrical readings, and water temperature. The sequence separates fuel input, gas-cycle work, electrical output, and Joule heating of the water. Warming of the water is the final useful output; it is not the engine's rejected

Q_{c}

. The handout distinguishes warm-up from steady operation and asks students to compare the final warming with direct heating by the lamp. It gives a procedure, not a table of measured efficiencies. Its pp. 1–3 therefore establish an apparatus and measurement design without supplying a new experimental result.

For a proposed calibration, use a known input of electrical work to determine the effective heat capacity

C_{eff}

of the water, vessel and immersed parts; stir to make the measured temperature representative. A matched run with no input estimates exchange with the room. For constant

C_{eff}

and no phase transition, the resulting model is

E_{in} = C_{eff} (T_{f} - T_{i}) + Q_{leak, out},

where

Q_{leak, out}

is the signed net heat lost to the room during the measurement. Here

E_{in}

is supplied electrical work when the resistor lies inside the calibration boundary; for a separate external heat input it is the received heat. These are proposed controls, not controls reported in that handout. Slow cooling, changing material properties and imperfectly matched room exchange limit this inference; the approach to thermal equilibrium and heat capacity matter as much as the thermometer reading.

To measure the actual engine-input

Q_{h}

, one must instrument its hot-side transfer separately, or close a calibrated energy balance around a specified heater boundary. A heater's electrical work includes changes in its stored internal energy and heat transfer that bypasses the engine. At periodic operation a cold-side flowing-water measurement can instead estimate

\dot{Q}_{c} = \overset{m}{˙} c_{p} (T_{out} - T_{in})

, if mass flow rate

\overset{m}{˙}

and specific heat capacity at constant pressure

c_{p}

are known, with negligible storage and other heat exchange in that measurement boundary. It measures only the rejection captured by that stream. Include all remaining transfers before using the residual to infer

\dot{Q}_{h} = P_{out} + \dot{Q}_{c}

. This measurement discipline makes a comparison with Carnot's theorem meaningful.

Calorimetry

Calorimetry measures heat transfers. A calorimeter, a calibrated heat capacity and an accounting of unwanted heat transfer connect an observed temperature change to an energy balance, as in heat-engine calorimetry and heat-pump calorimetry.

The central idea is to measure a response that is easier to observe than the transfer itself: a known heat capacity turns a temperature change into an energy estimate. Mixing calorimetry can determine an unknown specific heat capacity, while a chemical reaction can be studied by measuring the response of its surroundings. The useful observable is a change in temperature, together with a stated thermodynamic system boundary; a hot object does not contain a separately conserved store of heat.

For a closed system with unchanged bulk kinetic energy and potential energy, let

Q_{in}

be net inward heat and

W_{on}

net inward work. The first law of thermodynamics gives

Δ U = Q_{in} + W_{on} .

If the boundary includes a hot sample, receiving water and the participating vessel, their mutual heat transfer is internal. With no external input, the total storage change is minus the signed outward heat leakage, not the positive amount transferred from sample to water. A calorimeter calibration with an immersed resistor instead supplies electrical work across this boundary; Joule heating converts it into internal energy. Choosing only the receiving water as the boundary changes which transfers count as external.

The mechanical constraint decides which storage quantity the readout represents. For a closed system at fixed volume, with no other work,

Q_{V} = Δ U

. At constant external pressure, with only expansion work

p Δ V

Q_{p} = Δ H

, where

H = U + p V

is enthalpy. Thus the corresponding warming balances use heat capacity at constant volume or heat capacity at constant pressure. Electrical or shaft work must be included separately when present. These endpoint balances do not require the actual exchange to be a reversible thermodynamic process.

A covered cup open to atmospheric pressure approximately realizes the second constraint when escaping material is negligible. A sealed rigid reaction vessel realizes the first; its surrounding water, vessel and immersed parts form the calibrated receiver. The reaction's outward heat equals the receiver's gain only after losses and ignition input have been accounted for. The distinction concerns the reaction boundary, even if its receiver is open to atmospheric pressure.^[ref]

A historical quantitative result appears in James Prescott Joule's On the Mechanical Equivalent of Heat (1850): Table IX gives

772.692

foot-pounds of mechanical work per warming of one pound of water by one degree Fahrenheit, with weighing reduced to vacuum. His final rounded statement specifies water between

55

and

60

degrees Fahrenheit. He acknowledged residual vibration and sound as possible upward bias. This is an original reported conversion result, not the definition of the modern joule; the apparatus and resting controls are described in Joule's paddle-wheel experiment.^[ref]

Electrical calorimetry ( $E = \int V I d t$ )

Use measured electrical energy input to calibrate or infer a thermal response. A heater's terminal electric potential difference and electric current determine the supplied energy; receiver heat capacity and losses must also be accounted for.

Method-of-mixtures calorimetry

Infer an unknown heat capacity by thermal contact with a calibrated receiver and a measured common final temperature. The energy lost by hotter components equals the energy gained by colder components, with the calorimeter and surrounding heat transfer included. For appreciably varying heat capacities, use temperature integrals rather than constant

m c Δ T

terms.

Differential scanning calorimetry

differential scanning calorimetry measures differences in heat flow between sample and reference during a controlled temperature scan. Transition peaks can reveal cooperative lipid-chain melting, with scan rate, composition and reversibility affecting interpretation.

Mixing calorimetry ( $m_{x} c_{x} (T_{x} - T_{f}) = (m_{w} c_{w} + C_{cal}) (T_{f} - T_{w}) + Q_{loss, out}$ )

Mixing calorimetry determines heat capacities by bringing bodies at different temperatures into thermal contact. The striking feature is that the final temperature is a weighted arithmetic mean with heat capacities as weights, rather than weights given by mass. The vessel participates in the exchange even though it is not the sample being studied.

For a hot nonreacting sample of mass

m_{x}

, initial temperature

T_{x}

and unknown specific heat capacity at constant pressure

c_{x}

, let receiving water have mass

m_{w}

, initial temperature

T_{w}

and known specific heat capacity at constant pressure

c_{w}

. Let the vessel and immersed parts start at

T_{w}

and have combined heat capacity

C_{cal}

, excluding that water. Assume a common final temperature

T_{f}

, approximately constant pressure and heat capacities, no phase transition, no chemical reaction and negligible external nonexpansion work. For signed outward heat loss

Q_{loss, out}

, the combined enthalpy balance is

m_{x} c_{x} (T_{f} - T_{x}) + (m_{w} c_{w} + C_{cal}) (T_{f} - T_{w}) = - Q_{loss, out} .

Hence

c_{x} = \frac{( m _{w} c _{w} + C _{cal} ) ( T _{f} - T _{w} ) + Q _{loss, out}}{m _{x} ( T _{x} - T _{f} )} .

The loss is positive if the room receives net heat, negative if the room supplies it. If the vessel starts at a different temperature

T_{cal, i}

, replace its warming term by

C_{cal} (T_{f} - T_{cal, i})

; do not assume the vessel shared the water's initial temperature. For variable material properties, replace each

m c Δ T

by the corresponding integral

m \int c_{p} (T) d T

and each vessel term by

\int C_{cal} (T) d T

. The inferred constant

c_{x}

then represents an average over the sample's cooling interval.^[ref]

For a synthetic no-loss example, mix

0.100 kg

of hot water at

80^{\circ} C

with

0.200 kg

of cold water at

20^{\circ} C

in a vessel initially at

20^{\circ} C

. Stipulate

c_{w} = 4180 J k g^{- 1} K^{- 1}

and

C_{cal} = 100 J K^{- 1}

. The enthalpy balance gives

T_{f} = \frac{( 418 ) ( 80 ) + ( 836 + 100 ) ( 20 )}{418 + 836 + 100}^{\circ} C \approx 38.523^{\circ} C .

The hot water releases

17.337 kJ

: the cold water receives

15.485 kJ

and the vessel

1.852 kJ

. Omitting the vessel predicts

40^{\circ} C

, despite perfect insulation. These are computed numbers from stipulated properties, not laboratory observations.

Figure 1.
The vessel changes the mixing temperature by sharing the heat transfer
. Original computed no-loss water example. Left: transfers sum to the hot portion's released heat. Right: increasing the initially cold vessel's heat capacity lowers the common final temperature; the marked point uses $C_{cal} = 100 J K^{- 1}$ . Each vessel setting is a separate modeled experiment. **Details**.

For an unknown-sample example using the same cold water and vessel, stipulate

m_{x} = 0.100 kg

T_{x} = 80^{\circ} C

and

c_{x} = 450 J k g^{- 1} K^{- 1}

. With no loss,

T_{f} \approx 22.752^{\circ} C

. Inverting the full balance returns

450 J k g^{- 1} K^{- 1}

; ignoring vessel storage returns about

402 J k g^{- 1} K^{- 1}

. With a positive outward loss, ignoring that loss also biases the inferred

c_{x}

downward under these hot-sample conditions. A small water temperature rise makes sensor offsets and loss estimates relatively influential; propagation of uncertainty must include the measured temperatures, masses and calibrated vessel term.

Calorimeter

A calorimeter infers transferred heat from a measured change, such as a temperature change and a calibrated heat capacity. The heat-engine calorimetry example requires accounting for the vessel and heat transfer to the surroundings as well as the water.

A practical receiver has an inner vessel, insulation or an outer shield, a lid, a thermometer and a stirrer. Its participating heat capacity includes the immersed parts that follow the water's temperature. Insulation reduces heat transfer to the room; stirring makes a single reading more representative of the contents. These serve different purposes: a well-insulated vessel can still have substantial heat capacity, and a uniform temperature does not imply zero external heat transfer.

The University of Toronto's Heat Capacity (University of Toronto laboratory handout) describes a copper inner vessel, outer shield, thermometer and heated metal block. Weigh the block and receiving water, heat the block in a bath, then transfer it quickly into enough water to cover it. The observed temperature rises, reaches a maximum, then falls. The inferred specific heat capacity comes from the block's cooling and the warming of both water and vessel. In its cooling exercise the handout deliberately exposes the inner vessel to room air, recommends readings every

15 s

, and uses the later common cooling to estimate the earlier loss. Starting below room temperature reduces exchange without guaranteeing cancellation. The handout notes limited time and thermometry for small corrections; it supplies no dataset.^[ref]

For a proposed modern version, independently perform thermometer calibration and calorimeter calibration, record bath and room temperatures, and keep immersion depth, lid position and stirring rate fixed. Ensure the block actually reaches bath temperature; minimize and account for adhering hot bath water. Compare repeated transfers with different delays and a reference sample of known specific heat capacity. A resting run tests environmental heat transfer, while a stirred run with no added sample checks mechanical input. These are proposed controls, not claimed observations from the handout. Sensor delay, changing heat capacity, incomplete thermal equilibrium and material escaping the boundary limit the resulting estimate.

Ice calorimeter ( $Q = m_{melt} L_{f}$ )

A calorimeter that infers transferred heat from ice melting at a specified equilibrium temperature. The latent heat sets the proportionality after correcting for other thermal changes and extraneous melting.

Calorimeter calibration ( $C_{eff} = m_{w} c_{w} + C_{cal}$ )

Calorimeter calibration establishes the response of the actual measurement assembly to a known energy input. It determines a participating heat capacity, not just the empty vessel's mass. For an electrical calibration with an immersed resistor, measure voltage

V (t)

across that resistor and electric current

I (t)

through it. If that electrical work is deposited inside the chosen boundary, its input is

E_{el} = \int_{t_{0}}^{t_{1}} V (t) I (t) d t = C_{eff} (T_{1} - T_{0}) + Q_{loss, out}, C_{eff} = m_{w} c_{w} + C_{cal} .

For a covered receiver at constant pressure,

C_{eff}

describes enthalpy storage; for a rigid fixed-volume assembly, use the matching heat capacity at constant volume. Negligible endpoint mechanical storage and no escaping material are assumed. Thus

E_{el} / (T_{1} - T_{0})

overestimates

C_{eff}

if outward loss is ignored. Account for lead losses outside the boundary and maintain the same water charge and immersed hardware in calibration and sample runs.

An alternative calibration mixes known hot and cold water portions. With the vessel initially at the cold portion's temperature

T_{c}

, the mixing calorimetry balance gives

C_{cal} = \frac{m _{h} c _{w} ( T _{h} - T _{f} ) - Q _{loss, out}}{T _{f} - T _{c}} - m_{c} c_{w} .

The previous synthetic water example returns

100 J K^{- 1}

. The sign matters: attributing room losses to the vessel makes this estimate too large. With both

C_{cal}

and the loss unknown, one final temperature supplies only one equation; separate cooling information or controlled independent runs are needed.

A calorimeter heat-loss correction can make the electrical balance quantitative. For a second synthetic example, stipulate

0.100 kg

of water with

c_{w} = 4180 J k g^{- 1} K^{- 1}

and immersed hardware with

C_{cal} = 82 J K^{- 1}

, so

C_{eff} = 500 J K^{- 1}

. Supply

10 W

for

300 s

, starting at a room temperature of

20^{\circ} C

, with loss coefficient

G = 1 W K^{- 1}

. The model below gives

T_{1} \approx 24.512^{\circ} C

, storage

2255.942 J

and outward loss

744.058 J

. The uncorrected ratio gives about

665 J K^{- 1}

; subtracting the loss recovers

500 J K^{- 1}

Figure 1.
Electrical input splits into thermal storage and room losses during calibration
. Original synthetic computed traces for the stated constant-heat capacity, constant-loss model. The heater switches off at $300 s$ ; stored energy then declines while cumulative outward heat increases. The no-loss comparison reaches $26^{\circ} C$ . No points represent experimental measurements. **Details**.

Calorimeter heat-loss correction ( $Q_{loss, out} = G \int (T - T_{a}) d t$ )

A calorimeter heat-loss correction estimates the signed heat exchanged with the room during the measurement. Under Newton's law of cooling, a constant room temperature

T_{a}

and a single representative assembly temperature

T

give the model

C_{eff} \frac{d T}{d t} = P (t) - G (T - T_{a}), Q_{loss, out} = G \int_{t_{0}}^{t_{1}} (T (t) - T_{a}) d t .

Here

P (t)

is deposited input power, and

G

is the apparatus loss coefficient in

W K^{- 1}

. After input ceases, the temperature excess shows exponential decay with time constant

τ = C_{eff} / G

T (t) - T_{a} = (T (t_{1}) - T_{a}) e^{- (t - t_{1}) / τ} .

A cooling trace determines the ratio

G / C_{eff}

, not the two factors separately. Combining the cooling trace with a known electrical energy input supplies the missing information. For constant input

P_{0}

, starting at

T_{a}

, the heating solution is

T (t) - T_{a} = (P_{0} / G) (1 - e^{- t / τ})

; these solutions generate the calibration figure.

For mixing calorimetry, infer cooling behavior only after the sample, water and vessel approach a common temperature. Applying a single-body cooling model while the sample remains much hotter than the water can misestimate the earlier loss. The Toronto handout's area correction assumes the same loss mechanism represents both the mixing interval and subsequent cooling.^[ref] A backward extrapolation to the transfer time is therefore a model-dependent estimate, not the ultimate equilibrium with the room. Extend the model or report a limitation if sensor delay, substantial temperature gradients, changing room temperature, or escaping material invalidate that assumption. In particular, the relevant

C_{eff}

after adding a sample includes its heat capacity; an empty-receiver cooling rate need not apply to the loaded receiver.

Caloric theory

The historical caloric theory treated heat as a conserved substance. The modern first law of thermodynamics conserves energy, allowing net heat input to become work in a heat engine.

Rumford's cannon-boring experiment

Frictional heating during the boring of cannon was used by Count Rumford to challenge a simple account of heat as a finite material supply. A quantitative comparison also requires a stated mechanical input and the thermal response of the apparatus.

Caloric

The hypothetical heat substance of caloric theory. The modern distinction between heat transfer and conserved energy replaces that substance picture.

Refrigerator

A refrigerator removes heat from the region we want to keep cold and releases heat to warmer surroundings. Here we consider a work-driven cyclic device: its useful output is the cold-side extraction

Q_{c}

, and its input is net work

W

. A food compartment is a familiar example; the same cooling objective applies when removing heat from a room. The coefficient of performance measures extraction per unit work, rather than the fraction of input energy converted into output. ^[ref]

Choose the complete machine as the thermodynamic system. Let

Q_{c} > 0

be heat entering from the colder side,

Q_{h} > 0

be heat delivered to the hotter side, and

W > 0

be net work supplied, all over the same complete cycle. Assume these are its only energy transfers, no matter crosses the outer boundary, and there is no net change in stored energy. Returning to the initial state gives

Δ U = 0

, so the first law of thermodynamics reads

0 = Q_{c} - Q_{h} + W, Q_{h} = Q_{c} + W .

The heat leaving the warmer side includes both the extracted heat and the supplied work. For instance, stipulated transfers

Q_{c} = 3 kJ

and

W = 1 kJ

require

Q_{h} = 4 kJ

; the cooling coefficient of performance is

3

. This is an illustrative budget, not measured performance. At steady operation the same balance holds for average transfer rates,

\dot{Q}_{h} = \dot{Q}_{c} + P_{in}

, where

P_{in}

is input power.

This budget explains a useful counterexample: operating a refrigerator with its door open inside an otherwise isolated room cannot cool the room as a whole. If both heat transfers occur within the room and the electrical input comes from outside, the room gains

Q_{h} - Q_{c} = W

each cycle. Locally cold air does not imply net cooling. The Second law of thermodynamics permits cold-to-hot transfer because work is supplied; it forbids a cyclic device whose sole effect is that transfer. See the Clausius statement of the second law.

Carnot refrigerator

A Carnot refrigerator operates a reversible Carnot cycle in the refrigeration direction between two thermal reservoirs.

Vapor-compression refrigeration

Vapor-compression refrigeration circulates a working fluid through cold and hot coils. In the cold coil, liquid becomes gas while absorbing heat from the refrigerated region. Compression supplies work and raises the pressure and temperature; the hot coil releases heat as the gas returns to liquid. A restriction reduces the pressure before the next cold-side passage. Changing the pressure changes the temperature of the phase transition, letting one circulating fluid receive and release heat at different temperatures. This physical implementation serves both refrigerators and heat pumps; its ordinary dissipative operation differs from the reversible Carnot cycle used to bound performance. ^[ref] ^[ref]

Heat pump

A heat pump uses work to deliver heat to the region we want to keep warm. Its useful output is the hot-side delivery

Q_{h}

. The same operating machine can be described as a refrigerator when the cold-side extraction

Q_{c}

is the objective: which output is useful changes the coefficient of performance, while the energy balance stays

Q_{h} = Q_{c} + W

. Heating a water bath by taking heat from a colder water bath makes this distinction visible in the heat-pump calorimetry experiment below. ^[ref]

In the illustrative budget from refrigerator, supplying

1 kJ

of work delivers

4 kJ

of useful heat, giving a heating coefficient of performance of

4

. The additional

3 kJ

came from the colder region. Compare an ideal electrical heater whose entire input work becomes heat in the target region: it delivers

1 kJ

for that input. The appeal of a heat pump is this ability to move existing energy, while the Second law of thermodynamics limits the work saving for a given pair of temperatures.

Figure 1.
One energy budget, two useful outputs, and Carnot's theorem bounds on the coefficient of performance
. Original quantitative diagram, with stipulated $Q_{c} = 3 kJ$ , $W = 1 kJ$ and $Q_{h} = 4 kJ$ . Arrow widths follow the $3 : 1 : 4$ transfer magnitudes. The curves are reversible bounds for fixed cold temperature $T_{c} = 280 K$ and varying hot temperature $T_{h} = T_{c} + Δ T$ ; they are not measured performance. At $Δ T = 40 K$ , the bounds are $7$ for cooling and $8$ for heating, above the illustrative machine's $3$ and $4$ . **Details**.

Seasonal source depth for a reversible heat pump ( $z_{winter} = 3 π δ /4$ )

For a sinusoidal ground-temperature model whose coldest surface day is

t = 0

, source temperature on that day is

T_{0} - T_{A} e^{- x} cos x

, where

x = z / δ

. The first and largest thermal maximum occurs at

x = 3 π /4

, not at infinite depth. This maximizes the winter coefficient of performance for a reversible pump with a fixed indoor temperature. With a heating load even about that day, the same optimum minimizes annually integrated reversible heating work. Excavation cost, finite heat extraction and real heat-pump behaviour can change the optimum.

Indoor-temperature limit of a Carnot heat pump ( $T_{c, m i n} = T_{h} - \dot{W}_{m a x} T_{h} / λ$ )

A reversible heat pump maintaining

T_{h}

against an external

T_{c} < T_{h}

has heating coefficient of performance

T_{h} / (T_{h} - T_{c})

. If its maximum work input is

\dot{W}_{m a x}

and the building loses heat at rate

λ (T_{h} - T_{c})

, maintaining equilibrium requires

λ (T_{h} - T_{c})^{2} \leq \dot{W}_{m a x} T_{h}

. This yields the displayed minimum external temperature, provided it lies in the physical positive-temperature range. If the rating instead specifies maximum delivered heat, the bound is

T_{h} - T_{c} \leq \dot{Q}_{h, m a x} / λ

; a heat-output rating cannot be substituted for a work-input rating in the Carnot formula.

Coefficient of performance ( $COP = Q_{useful} / W$ )

For the cyclic two-side energy balance in refrigerator, define the dimensionless cooling and heating coefficients by

COP_{r} = \frac{Q _{c}}{W}, COP_{heat} = \frac{Q _{h}}{W} = \frac{Q _{c} + W}{W} = COP_{r} + 1.

This identity applies even to irreversible machines, provided both coefficients refer to the same operating point, work input and boundary with no additional energy transfer or storage. It does not compare two separately operated appliances. Unlike thermal efficiency of a heat engine, a COP can exceed one: its numerator counts heat moved, including energy taken from another region. ^[ref]

To obtain temperature-only bounds, impose stronger assumptions: a cyclic machine exchanges heat only with two ideal thermal reservoirs maintained at uniform, fixed absolute thermodynamic temperatures

0 < T_{c} < T_{h}

, and receives work through an ideal input carrying no entropy. Include the machine and the heat transfer interfaces in the entropy production accounting. Because its own entropy returns to its initial value each cycle, the Second law of thermodynamics gives

S_{gen} = \frac{Q _{h}}{T _{h}} - \frac{Q _{c}}{T _{c}} \geq 0.

Combining this inequality with

Q_{h} = Q_{c} + W

gives

COP_{r} \leq \frac{T _{c}}{T _{h} - T _{c}}, COP_{heat} \leq \frac{T _{h}}{T _{h} - T _{c}} .

Equality requires a reversible thermodynamic process for the entire cycle and its contacts, as in a reversed Carnot cycle. These are Carnot's theorem bounds independent of the working substance. Use kelvin values, not degree Celsius readings, in the numerators; the temperature difference can be expressed in either unit. ^[ref] ^[ref]

The diagram fixes

T_{c} = 280 K

and sets

T_{h} = 320 K

, so the reversible bounds are

280/40 = 7

and

320/40 = 8

. The stipulated machine with

COP_{r} = 3

and

COP_{heat} = 4

satisfies both. Its reservoir entropy production per cycle is

4000/320 - 3000/280 = 25/14 J K^{- 1} > 0

. Increasing the temperature lift

Δ T = T_{h} - T_{c}

at fixed

T_{c}

lowers both bounds. Equivalently, delivering fixed

Q_{h}

needs at least

W_{m i n} = Q_{h} (1 - \frac{T _{c}}{T _{h}}) .

These expressions reveal why a small temperature lift is attractive. Their divergence as

T_{h} - T_{c} \to 0^{+}

is an ideal reversible limit, not a promise of infinite heating or cooling power. For finite heat transfer rates, the working fluid must be colder than the cold bath and hotter than the hot bath; these contact differences produce entropy and increase the internal temperature lift. Dissipative compression and the pressure-reducing restriction of vapor-compression refrigeration supply further departures from reversibility. Thus bath temperatures set an outer bound, not a prediction of real COP. ^[ref]

Heat-pump calorimetry

A motivating bench experiment asks whether a hot bath receives more heat than the electrical work supplied to a heat pump. PHYWE's authored operating instructions, section 4, describe a sealed vapor-compression refrigeration apparatus with its cold and hot coils immersed in separate water vessels. The quantitative procedure calls for insulated vessels, stirring to reduce temperature gradients, timed temperature readings, and a wattmeter for the input. It waits about five minutes for operating conditions to settle before measurements and explicitly includes the glass vessel and immersed copper pipe when calculating the heated body's heat capacity. One bath cooling while the other warms makes the origin of the delivered energy observable. ^[ref]

The following analysis extends that procedure to simultaneous measurements of both finite baths. Weigh the water in each vessel; determine the effective heat capacities

C_{c}

and

C_{h}

, including immersed hardware, by a separate known electrical-input calibration over the working temperature range. A matched pair of calibrated thermometers placed in stirred water measures bulk

T_{c} (t)

and

T_{h} (t)

; keeping probes away from the coils avoids mistaking a local contact temperature for the whole bath. Record the room temperature and synchronise the electrical meter and thermometry. Use short intervals after startup, with no boiling or freezing and small enough temperature changes that constant effective heat capacities are a reasonable approximation. These are proposed measurement controls, rather than reported results from the manual.

Over a common interval

[t_{0}, t_{1}]

, the uncorrected bath estimates and measured electrical input are

Q_{h, raw} = C_{h} [T_{h} (t_{1}) - T_{h} (t_{0})], Q_{c, raw} = - C_{c} [T_{c} (t_{1}) - T_{c} (t_{0})], W_{el} = \int_{t_{0}}^{t_{1}} P_{el} (t) d t .

Here the wattmeter records real electrical power

P_{el}

. The integral handles changing input, and the signs make both estimated useful heat quantities positive during heating and cooling. The effective heat capacity includes water mass times its specific heat capacity plus the response of the vessel and immersed hardware. Without that correction, equal temperature changes need not imply equal heat transfers.

Insulation reduces heat leaks but does not establish an isolated system. A warmer bath can lose heat to the room while a colder bath gains it. As a proposed control, measure pump-off drift at matched bath and room temperatures, with the same stirring and insulation. Writing

L_{h} > 0

for outward hot-bath leak power and

L_{c} > 0

for inward cold-bath leak power, the bath balances give

\dot{Q}_{h} = C_{h} \frac{d T _{h}}{d t} + L_{h}, \dot{Q}_{c} = - C_{c} \frac{d T _{c}}{d t} + L_{c} .

Integrate these rates over the same interval before dividing by

W_{el}

. For example, in a pump-off hot-bath control,

L_{h} = - C_{h} d T_{h} / d t

; in the cold-bath control,

L_{c} = C_{c} d T_{c} / d t

. Check that a stopped machine does not create a different leak path from the running one; otherwise use an independent bath calibration and keep that discrepancy as an uncertainty. Repeat at the same starting temperatures to distinguish drift from a change in COP as the baths separate.

Figure 1.
Heat-pump calorimetry: bath temperature slopes, electrical work, and heat-leak corrections
. Original apparatus schematic and stipulated constant-rate calculation, not measured data and not a performance claim for the cited apparatus. Each finite bath has effective heat capacity $4.5 kJ K^{- 1}$ . The hypothetical machine takes $50 W$ , extracts $100 W$ , and delivers $150 W$ ; hot-bath loss is $15 W$ and cold-bath gain is $5 W$ . Over $180 s$ the hot bath warms by $5.4 K$ and the cold bath cools by $3.8 K$ . Correcting the two bath estimates restores the machine's $3$ heating COP and $2$ cooling COP. **Details**.

In the illustrated toy model,

C_{h} = C_{c} = 4500 J K^{- 1}

P_{el} = 50 W

\dot{Q}_{h} = 150 W

\dot{Q}_{c} = 100 W

L_{h} = 15 W

, and

L_{c} = 5 W

are stipulated constants for

180 s

. The slopes are

d T_{h} / d t = 0.030 K s^{- 1}

and

d T_{c} / d t = - 19/900 K s^{- 1}

. Consequently,

W_{el} Q_{h, raw} Q_{c, raw} = 9.0 kJ, = 24.3 kJ, = 17.1 kJ, Q_{h} Q_{c} = 24.3 + 2.7 = 27.0 kJ, = 17.1 + 0.9 = 18.0 kJ .

The raw heating and cooling ratios are

2.7

and

1.9

, whose difference is

0.8

. Correcting the bath leaks gives

3

and

2

, whose difference is one. The apparent failure of

Q_{h} = Q_{c} + W

was incomplete measurement of heat, not a violation of the first law of thermodynamics.

Real apparatus can also release motor heat and pipe heat directly to the room, and store internal energy during startup. For a machine boundary that receives electrical work, an additional outward heat transfer

Q_{room}

gives

Q_{h} + Q_{room} = Q_{c} + W_{el} - Δ U_{machine}

. The residual

Q_{h} - Q_{c} - W_{el}

therefore tests the boundary and corrections; do not force it to zero. Include any auxiliary electrical work in the input if quoting whole-apparatus COP, and account separately for stirring work deposited in a bath. Ignoring vessel heat capacity, sensor lag or room leaks can bias the inferred result. These limitations follow from extending the measured energy budget, and the manual's explicit hardware correction illustrates why such boundaries matter.

Finally, these vessels are finite bodies whose temperatures change; they are not fixed thermal reservoirs. Their ratio of integrated useful heat to integrated input work is an interval COP. A Carnot's theorem comparison using simultaneous bath temperatures is a local approximation only when the machine settles much faster than the baths change. An exact reversible comparison for changing bath temperatures instead uses the bath entropy changes, for example

C_{h} ln (T_{h, 1} / T_{h, 0}) + C_{c} ln (T_{c, 1} / T_{c, 0})

for constant heat capacities, negligible net machine entropy change and no room leaks. The experiment's beauty is that two thermometers, known heat capacities and an electrical-energy measurement expose both conservation of energy and the practical gap from reversibility.

Free expansion

Free expansion occurs when a gas spreads into an initially evacuated region without driving an external load. The standard idealization uses a rigid, insulated vessel: initially a partition confines the gas to volume

V_{1}

, while the remaining compartment contains vacuum. Open a passage and wait for thermodynamic equilibrium; the same fixed amount of substance

n

now occupies the larger accessible volume

V_{2}

. The outer vessel never changes size. This simple release separates three questions: what energy is conserved, whether temperature changes, and why entropy increases. ^[ref]

Figure 1.
Free expansion changes the accessible volume; the apparatus affects the observed temperature
. Original schematic, with equal compartments chosen for illustration. The top row assumes an insulated gas with negligible wall heat capacity. The bottom left sketches the water-bath arrangement of Joule's free-expansion experiment, rather than a reconstruction to scale. The bottom right uses stipulated heat capacities and an illustrative energy demand to show signal suppression; it contains no measured data. **Details**.

Choose the gas as the closed system, neglect heat exchange with the walls, and assume negligible changes in bulk kinetic energy and potential energy between the equilibrium endpoints. Insulation gives

Q_{in} = 0

; no moving outer boundary or external load receives work, so

W_{out} = 0

. The first law of thermodynamics therefore gives

Δ U = Q_{in} - W_{out} = 0.

The gas can develop internal motion during the release; this does not supply external work. The intermediate state need not have one equilibrium pressure or temperature, so there is no actual quasistatic process curve

p (V)

to integrate. In particular, substituting the equilibrium ideal gas law into

\int p d V

would describe a different process. ^[ref]

For an ideal gas, internal energy of an ideal gas depends only on temperature. With positive heat capacity at constant volume,

U (T_{2}) = U (T_{1})

implies

T_{2} = T_{1}

, and the ideal gas law gives

p_{2} / p_{1} = V_{1} / V_{2}

. Doubling the accessible volume thus halves the final pressure without changing the final temperature. This is an endpoint statement; it does not assert a spatially uniform, constant temperature during the rush. Insulation alone does not imply constant temperature: a reversible adiabatic expansion cools an ideal gas because it supplies work. ^[ref]

For a real gas,

U (T, V)

can depend on volume as well as temperature, so unchanged internal energy need not mean unchanged temperature. For a fixed-composition simple compressible system in a single thermodynamic phase, equilibrium identities using a Maxwell relation give

d U = C_{V} d T + [T (\frac{\partial p}{\partial T})_{V} - p] d V, (\frac{\partial T}{\partial V})_{U} = \frac{p - T ( \partial p / \partial T ) _{V}}{C _{V}} .

Here

C_{V}

is the whole sample's heat capacity at constant volume. The sign depends on the equation of state; for an ideal gas the numerator vanishes. For a finite free expansion, solve

U (T_{2}, V_{2}) = U (T_{1}, V_{1})

, or integrate this slope along equilibrium states of equal internal energy; that mathematical curve does not describe the actual release. ^[ref]

Free expansion is useful because it tests how internal energy depends on number density while eliminating external expansion work. Its beautiful surprise is that an ideal gas can finish with the same temperature and internal energy but greater entropy: a thermometer alone cannot diagnose reversibility. See entropy change in free expansion for the calculation and temperature sensitivity of a free-expansion experiment for the experimental qualification. The continuous-flow throttling associated with the Joule-Thomson coefficient has a different balance: under its usual assumptions it preserves enthalpy, whereas the gas-only idealization here preserves internal energy. ^[ref]

Cooling in free expansion of a van der Waals gas

For the classical monatomic caloric model, energy conservation in expansion from

V_{i}

V_{f}

gives

T_{f} - T_{i} = 2 a N (1/ V_{f} - 1/ V_{i}) / (3 k_{B})

. Attractions make the gas cool. The formula assumes homogeneous equilibrium states, not a two-phase coexistence state.

Joule's free-expansion experiment

In his paper published in 1845, Joule connected two copper receivers with stopcocks, filled one with compressed dry air, and evacuated the other with an air pump. He immersed the connected receivers in water in an insulated arrangement, stirred the water, and read a sensitive thermometer. He then opened the stopcocks, let the air redistribute, stirred again, and compared the water's temperature with its initial value. Alternating control trials accounted for effects such as stirring and evaporation. The observation was no detectable net cooling of the common bath, rather than a record of the gas temperature at every instant. ^[ref] ^[ref]

Joule also immersed the receivers and their connecting piece in separate water vessels. The source receiver cooled while the receiving side warmed, with nearly cancelling thermal effects. ^[ref]

This distinguishes local transient changes from the final observation for the combined apparatus. Insulating its outer boundary suppresses external heat exchange; it does not prevent heat exchange between the gas, the receivers and the water inside that boundary.

An idealized modern version can use two rigid chambers joined by a valve, with thermometers in the chambers and on the walls. Prepare one chamber with gas and the other evacuated, disconnect the pump, close the apparatus to matter exchange, and let the initial temperature settle. Record initial and final equilibrium pressure and temperature, along with the wall response. Repeated closed-valve runs can reveal drift and valve-actuation effects; independent heating calibrations can estimate the heat capacity of the participating apparatus. This is a proposed measurement protocol, not a claim of a particular measured result. Fast local readings can characterize the release, but they cannot replace the final equilibrium readings in the endpoint internal energy balance.

Temperature sensitivity of a free-expansion experiment

The gas-only condition

Δ U = 0

requires negligible heat exchange with the apparatus. If the insulated isolated system instead includes the walls, sensors and water, the first law of thermodynamics fixes their combined energy. Once everything returns to a common final temperature, a constant apparatus heat capacity

C_{a}

gives

U_{g} (T_{2}, V_{2}) - U_{g} (T_{1}, V_{1}) + C_{a} (T_{2} - T_{1}) = 0.

Define

A = U_{g} (T_{1}, V_{2}) - U_{g} (T_{1}, V_{1})

, the gas's internal energy increase for the same expansion evaluated at the initial temperature. For a small temperature change, linearizing at the final volume gives

Δ T ≃ - \frac{A}{C _{V, g} + C _{a}} .

Here

C_{V, g}

is the whole gas sample's heat capacity at constant volume at the reference state. Compared with an isolated gas sample's

- A / C_{V, g}

, the common final signal is reduced by

C_{V, g} / (C_{V, g} + C_{a})

. This energy accounting explains why a massive bath can conceal a real-gas effect. Fitzpatrick gives the corresponding apparatus correction for the Van der Waals equation. ^[ref]

For a toy calculation, stipulate

A = 5 J

C_{V, g} = 10 J K^{- 1}

and

C_{a} = 990 J K^{- 1}

, all constant in this example. The predicted cooling magnitude changes from

0.5 K

for the gas alone to

0.005 K

for the combined apparatus. An illustrative effective detection threshold of

0.01 K

would miss the latter. These are chosen numbers, not Joule's measurements. In this approximation a null observation only bounds

∣ A ∣ ≲ (C_{V, g} + C_{a}) ϵ_{T}

, where

ϵ_{T}

is the detectable temperature change after allowing for drift and other experimental uncertainties; a display's smallest digit is not automatically this threshold.

The historical limitation was recognized explicitly: Joule and Thomson later estimated that the common apparatus in Joule's experiment would cool by only about

0.003 K

, explaining why the earlier external thermal effect was undetectable. This was a later calculation about the apparatus, not a measured cooling trace. ^[ref]

Entropy change in free expansion ( $Δ S = n R ln (V_{2} / V_{1})$ for an ideal gas)

For the insulated gas-only free expansion of an ideal gas, the equilibrium endpoints are

(T, V_{1})

and

(T, V_{2})

with

V_{2} > V_{1}

. Because entropy is a state function, calculate their difference along a hypothetical reversible thermodynamic process: an isothermal expansion using a frictionless piston and a thermal reservoir at

T

. This comparison process receives heat and delivers work, even though the actual free expansion does neither. Its reversible heat increment is

δ Q_{rev} = p d V = n RT d V / V

, so

Δ S = \int_{1}^{2} \frac{δ Q _{rev}}{T} = n R \int_{V_{1}}^{V_{2}} \frac{d V}{V} = n R ln \frac{V _{2}}{V _{1}} > 0.

Here

R

is the molar gas constant, and the logarithm acts on a dimensionless ratio. The reversible reference gives the correct endpoint entropy difference because entropy is a state function; putting the actual

Q = 0

into

Δ S = Q / T

would misuse a relation requiring reversible heat transfer. For the actual isolated gas, all this increase is entropy production. Thus an adiabatic process can increase entropy and need not be an isentropic process. ^[ref]

Figure 1.
The reversible reference calculates entropy without supplying a path for the actual free expansion
. Toy ideal gas example with $n = 1 mol$ , $T = 300 K$ , $V_{1} = 0.025 m^{3}$ and stipulated $R = 8.314 J mo l^{- 1} K^{- 1}$ . The plotted curve belongs only to the hypothetical reversible isothermal expansion; the actual release has the same equilibrium endpoints but no equilibrium curve between them. The table refers to the gas, with work positive outward. No points are experimental data. **Details**.

In the figure's volume doubling, the ideal gas law gives

p_{1} = 99.768 kPa

and

p_{2} = 49.884 kPa

, while

Δ S = R ln 2 ≃ 5.763 J K^{- 1}

. The reference process receives and delivers

n RT ln 2 ≃ 1728.85 J

as heat and work; its thermal reservoir loses the same entropy that the gas gains. The actual free expansion has

Q_{in} = W_{out} = 0

and produces that entropy internally. Reversibly compressing the gas back at

300 K

would require

1728.85 J

of work input and rejection of equal heat. Restoring the gas alone therefore changes the surroundings: the actual release is irreversible.

For a real gas the final temperature may differ, so the ideal-gas logarithm formula cannot simply be copied. Evaluate

S (T_{2}, V_{2}) - S (T_{1}, V_{1})

using that gas's equilibrium properties or a suitable reversible comparison path. If the walls and water exchange heat with it, the Second law of thermodynamics constrains the total entropy of the isolated apparatus; the gas alone is then not an isolated system.

Thermodynamic phase

A thermodynamic phase is a form of macroscopic equilibrium characterized by its structure, symmetries and thermodynamic response. Away from phase transitions, its free-energy density varies smoothly with the control parameters. Different phases can share a phase coexistence boundary. An ordered magnetic phase is distinguished from a disordered phase by its spontaneous order parameter; positive and negative ordered states are different pure thermodynamic phases related by spin inversion symmetry.

Phase fraction ( $x_{β} = m_{β} / m$ )

A phase fraction is the proportion of a mixture occupying a specified thermodynamic phase. State the basis: by mass,

x_{β} = m_{β} / \sum_{α} m_{α}

; by volume, divide the occupied volume by the total volume. These fractions need not agree when the mass densities differ. The latent heat heating curve uses the liquid fraction by mass, whereas liquid fraction in a mushy layer is defined by volume.

Laws of thermodynamics

The laws organize the equilibrium and process constraints of thermodynamics: the zeroth law of thermodynamics makes thermal equilibrium transitive, the first law of thermodynamics expresses energy balance, the Second law of thermodynamics restricts entropy changes, and the third law of thermodynamics describes the zero-temperature limit. Their analogy with horizon mechanics becomes quantitative through Hawking temperature and Bekenstein-Hawking entropy.

Zeroth law of thermodynamics

If two thermodynamic systems are each in thermal equilibrium with the same third system in the same equilibrium state, they are in thermal equilibrium with each other. Write

A \sim B

for this relation. The zeroth law of thermodynamics supplies transitivity:

A \sim C and B \sim C ⟹ A \sim B .

Its symmetric relation property and self-equilibrium (reflexivity), together with transitivity, make thermal equilibrium an equivalence relation. Its equivalence classes can be labelled by a common empirical temperature; choosing an absolute numerical scale requires the further structure of thermodynamic temperature.^[ref]

The comparison assumes internally equilibrated macroscopic bodies, unchanged composition and mechanical constraints, and contact permitting only heat exchange. Use rigid, impermeable diathermal walls so that neither matter transfer nor boundary work spoils the test. The same state of

C

matters: letting

C

warm while measuring

A

, then cool while measuring

B

, does not supply the premise. Ordinary nonrelativistic laboratory thermodynamics is assumed; the Tolman temperature law treats gravitational qualifications. Equal temperatures establish the thermal condition, without requiring mechanical equilibrium or chemical equilibrium.

Figure 1.
Thermal equilibrium is transitive; an adiabatic wall supplies no equality test
. The top row compares $A$ and $B$ separately with the same equilibrium state of reference $C$ . The lower left shows the predicted absence of net heat transfer when $A$ and $B$ are put in thermal contact. The lower right is a counterexample to inferring equality from insulation: a rigid adiabatic wall can separate bodies at different temperatures. All numerical labels are illustrative. **Details**.

This explains why a thermometer can compare unlike materials: it supplies a reusable third system. A temperature is a property of an equilibrium state, while heat is energy transferred because of a temperature difference. Two bodies at the same temperature can have different internal energies and heat capacities. A thermometer therefore does not measure a body's stored amount of heat; the finite-body disturbance and waiting-time calculations in thermometer describe how making the comparison can change the state being measured.^[ref]

The name “zeroth” reflects logical order rather than the date of the first thermometer: this premise about thermal equilibrium precedes the already numbered first law of thermodynamics, Second law of thermodynamics and third law of thermodynamics. This is the naming explanation given in University Physics Volume 2 by Samuel J. Ling and others (§1.1, “Thermal Equilibrium”).^[ref]

The corresponding Zeroth law of black-hole mechanics states constancy of surface gravity over an equilibrium horizon; Hawking radiation identifies that quantity with physical temperature up to its universal conversion factor.

Thermal contact

Two thermodynamic systems are in thermal contact when their boundary permits heat transfer. They need not already be in thermal equilibrium: contact between unequal temperatures is precisely how equilibration can begin. Contact specifies an allowed exchange; equilibrium specifies its balance.

Thermal interface

The common boundary between two conducting materials in a thermal interface problem. Perfect thermal contact imposes continuity of temperature and of normal heat flux.

Diathermal wall

A diathermal wall permits heat transfer. For a clean zeroth law of thermodynamics comparison, idealize it as rigid and impermeable, excluding boundary work and matter transfer. A thin metal partition approximates this arrangement.^[ref]

Adiabatic wall

An adiabatic wall prevents heat transfer. Two bodies at

300 K

and

350 K

can remain separated by a rigid adiabatic wall with no heat transfer, although they are not in thermal equilibrium with each other. An adiabatic boundary alone need not make a system an isolated system: a moving insulated piston can still transmit work. Isolation additionally excludes matter transfer and every energy exchange.

Empirical temperature ( $θ$ )

An empirical temperature labels classes of thermal equilibrium using a reproducible thermometer response. Equal labels express the zeroth law of thermodynamics comparison. A suitable continuous response and calibration let ordinary laboratory states be represented on a numerical scale; the abstract equivalence relation alone does not guarantee a useful real-valued coordinate. An injective function of the label preserves equality, so the zeroth law of thermodynamics alone fixes neither unit nor spacing. Thermometer calibration supplies a chosen response relation; thermodynamic temperature supplies the additional absolute-scale structure.

Zeroth-law thermometer comparison

Put a contact thermometer

C

in thermal contact with

A

, wait for thermal equilibrium, and record response

X_{A}

. Repeat with

B

under the same operating constraints. If

X_{A} = X_{B}

, and the calibrated response is an injective function of temperature on this range,

C

occupies the same thermal reference state in both comparisons. The zeroth law of thermodynamics then predicts

A \sim B

. Equality of raw numbers from different uncalibrated thermometers supplies no such premise.^[ref]

Use a probe whose participating heat capacity is small relative to each sample, or use maintained heat baths. Otherwise the matching readings describe the disturbed final states. Also hold the probe's immersion and mechanical conditions fixed: its response can depend on more than temperature. Matching readouts only tests agreement within resolution and repeatability, rather than proving exact equality of temperatures.

Agreement within a tolerance is not itself an equivalence relation. With a

0.10^{\circ} C

tolerance, readings

25.00

and

25.16^{\circ} C

each agree with

25.08^{\circ} C

, yet disagree with each other. This is a limitation of the measurement rule, not a failure of the exact transitivity asserted by the zeroth law of thermodynamics.

Two-point thermometer calibration example

Consider an original worked thermometer calibration, using the height

X

of a liquid column responding through thermal expansion. Stipulate two equilibrated reference baths whose corrected thermometer readings are

0^{\circ} C

and

40^{\circ} C

, in degree Celsius units. Suppose the corresponding column heights are

12 mm

and

52 mm

. These are constructed inputs, not observed measurements. Assume the response is linear over this interval, with the same immersion and reading procedure throughout. Applying thermometer calibration gives

\frac{t _{C} ( X )}{^{\circ} C} = 40 \frac{X - 12 mm}{52 mm - 12 mm} .

For

X = 37 mm

, first subtract the zero-reference height, then divide by the full reference span:

(37 - 12) / (52 - 12) = 25/40

. Multiplying by the

40^{\circ} C

span gives

t_{C} = 25^{\circ} C

, or

T = 298.15 K

in kelvin. Thus equal

37 mm

readings in separately equilibrated samples test equal temperatures, regardless of how much internal energy those samples contain. The Celsius offset and modern absolute-scale definition are described in degree Celsius and kelvin.

Figure 1.
A reference-bath thermometer comparison and a worked two-point thermometer calibration
. Left: two probes sample gently stirred water, with sensing tips immersed away from the walls and bottom; the heater is off for the ambient comparison. Right: an assumed linear column response maps $12 mm$ to $0^{\circ} C$ , $52 mm$ to $40^{\circ} C$ , and $37 mm$ to $25^{\circ} C$ . The schematic and all plotted values are illustrative, not experimental data. **Details**.

The two endpoints determine the assumed line, but do not establish linearity. An independent middle reference and a return to the initial bath can test curvature and drift. If the height has an error of

0.2 mm

while reference values and endpoints are held exact in this calculation, the inferred temperature error is

0.2^{\circ} C

; that is only the height contribution, not a full measurement error budget. A boiling-water reference needs a measured atmospheric pressure correction, rather than an automatic

100^{\circ} C

assignment. The modern kelvin is not defined by either bath.

Reference-bath thermometer comparison

A motivating real procedure is NIST's ambient thermometer verification. Its apparatus is a deep beaker of water, a calibrated reference thermometer, a test thermometer, and a magnetic stirrer with any heater switched off. Let the water approach room temperature, stir gently, immerse both sensing tips well below the surface, and wait before reading. NIST specifies about

20 cm

of water,

10

–

15 cm

immersion where possible, and an initial five-minute wait. Record test then reference, repeat in reverse order, and apply the reference's certificate correction. Subtract corrected reference from test: the difference estimates the test instrument's error at this bath temperature. Agreement between the two reading orders checks repeatability. ^[ref]

For example, suppose a test reads

24.7^{\circ} C

, a reference reads

24.4^{\circ} C

, and its certificate says to add

0.1^{\circ} C

. The corrected reference is

24.5^{\circ} C

, so the test error is

+ 0.2^{\circ} C

: it reads high. These are illustrative readings. This verification tests a previously calibrated instrument at one point; it does not establish the two-point response used in two-point thermometer calibration example.

Additional controls follow from the thermal equilibrium premise. Keep tips away from walls and bottom, interchange probe positions to test a temperature gradient, and check that longer waits and modest immersion changes leave agreement unchanged within the required tolerance. A flat trace alone can be a steady state maintained by competing heat transfer paths. Five minutes is a procedural starting point, not a universal equilibration time. Excessive stirring can supply work and warm the bath, while probe stems can exchange heat with the room. These controls test whether both probes sample the same state; neither one-point agreement nor this experiment proves the zeroth law of thermodynamics for every material and condition.

Compressibility ( $κ_{X} = - V^{- 1} (\partial V / \partial P)_{X, N}$ )

Compressibility measures how readily a sample's volume decreases when its pressure increases. For a simple compressible system with fixed particle number

N

and composition, specify the constraint

X

κ_{X} = - \frac{1}{V} (\frac{\partial V}{\partial P})_{X, N}, \frac{d V}{V} = - κ_{X} d P (X, N fixed) .

Here

P

denotes pressure, also written

p

elsewhere. The minus sign makes compressibility positive on a mechanically stable equilibrium branch. Its units are inverse pressure,

P a^{- 1}

; its reciprocal

K_{X} = 1/ κ_{X}

is the constrained bulk modulus, measured in pascal units. Holding temperature fixed gives isothermal compressibility; holding entropy fixed gives isentropic compressibility

κ_{S}

along an isentropic process. The latter describes a reversible adiabatic compression, since an irreversible adiabatic process can increase entropy. ^[ref]

For an ideal gas, the ideal gas law

P V = N k_{B} T

and the reversible ideal-gas adiabat give, at the same equilibrium state,

κ_{T} = \frac{1}{P}, κ_{S} = \frac{1}{γ P}, γ = \frac{C _{P}}{C _{V}} > 1.

Here

C_{P}

and

C_{V}

are the whole sample's heat capacity at constant pressure and heat capacity at constant volume, and

γ

is the heat capacity ratio. Constant

γ

is required for the finite-path formula

P V^{γ} = constant

. Reversible adiabatic compression raises the temperature, making the pressure rise more steeply than in an isothermal compression. Thus keeping entropy fixed makes the gas harder to compress locally. For a classical monatomic gas at

100 kPa

γ = 5/3

, so

κ_{T} = 1 0^{- 5} P a^{- 1}

and

κ_{S} = 6 \times 1 0^{- 6} P a^{- 1}

; the corresponding values of the bulk modulus are

100 kPa

and

166.67 kPa

For a smooth, stable single thermodynamic phase at positive temperature, Maxwell relations also imply

\frac{κ _{T}}{κ _{S}} = \frac{C _{P}}{C _{V}}, κ_{T} - κ_{S} = \frac{T V α ^{2}}{C _{P}},

where

α

is the coefficient of thermal expansion. The ideal gas result is one example of this broader link between mechanical and thermal responses. The equality

κ_{T} = κ_{S}

is possible when

α = 0

and the other responses remain finite. These formulas compare equilibrium partial derivatives, rather than arbitrary fast or dissipative compression histories. ^[ref]

Density susceptibility ( $χ_{ρ} (q) = δ ρ_{q} / δ μ_{q}$ )

The static density susceptibility is the linear density response to a chemical-potential perturbation. In the quadratic uniform contact Bose gas,

χ_{ρ} (q) = [g + q^{2} / (4 m ρ_{0})]^{- 1}

. Its inverse is a density stiffness. At zero wavevector

\partial ρ / \partial μ = 1/ g

at mean field, while the isothermal compressibility includes the additional factor

1/ ρ^{2}

Heat-capacity and compressibility ratio identity ( $C_{p} / C_{V} = κ_{T} / κ_{S}$ )

, heat capacity at constant volume

In a smooth fixed-composition thermodynamic region, write

a = (\partial S / \partial T)_{V}

b = (\partial S / \partial V)_{T}

c = (\partial p / \partial T)_{V}

, and

d = (\partial p / \partial V)_{T}

, with

a, d \neq = 0

. The chain rule gives

(\partial S / \partial T)_{p} = a - b c / d

and

(\partial p / \partial V)_{S} = d - b c / a

. Multiplying these expressions by

d

and

a

, respectively, gives the same Jacobian determinant

a d - b c

Using heat capacity at constant pressure

C_{p} = T (\partial S / \partial T)_{p}

C_{V} = T a

, isothermal compressibility

κ_{T} = - 1/ (V d)

and isentropic compressibility

κ_{S} = - 1/ [V (\partial p / \partial V)_{S}]

proves the displayed ratio. The derivation does not require division by thermal expansion, so it still applies when that derivative vanishes. For a stable single phase, the Maxwell relation

b = c

also explains why

C_{p} \geq C_{V}

and

κ_{T} \geq κ_{S}

Isothermal compressibility ( $κ_{T} = - V^{- 1} (\partial V / \partial P)_{T, N}$ )

Isothermal compressibility is the compressibility measured at fixed temperature, particle number and composition. A slowly compressed sealed piston in contact with a thermal reservoir can approach this constraint. For an ideal gas, differentiating

V = N k_{B} T / P

at fixed temperature gives

(\partial V / \partial P)_{T, N} = - V / P

, hence

κ_{T} = 1/ P

. A small pressure increase

Δ P

therefore gives

Δ V / V ≃ - Δ P / P

. For a finite change, use the whole isothermal process curve

V_{2} / V_{1} = P_{1} / P_{2}

instead.

Take one mole of classical monatomic gas, with stipulated molar gas constant

R = 8.314 J mo l^{- 1} K^{- 1}

, initially at

T_{0} = 300 K

and

P_{0} = 100 kPa

. The initial volume is

V_{0} = n R T_{0} / P_{0} = 0.024942 m^{3}

. Compressing to

V_{0} /2

gives

Isothermal: Reversible adiabatic: P_{2} = 2 P_{0} = 200 kPa, T_{2} = 300 K, P_{2} = 2^{5/3} P_{0} ≃ 317.48 kPa, T_{2} = 2^{2/3} T_{0} ≃ 476.22 K .

The isothermal compression releases heat to the thermal reservoir; the reversible adiabatic compression retains the input work as internal energy. Reversing either curve gives the corresponding expansion, with adiabatic expansion cooling the gas.

Figure 1.
An ideal gas resists reversible adiabatic compression more strongly than isothermal compression
. Left: two pressure-volume curves through the same initial state; arrows point toward compression. Right: the corresponding temperature changes, including cooling during reversible adiabatic expansion. The dotted line marks half the initial volume. One mole of classical monatomic gas, with $γ = 5/3$ and the stipulated $R$ above. **Details**.

The equilibrium curvature of Gibbs free energy provides another measurement of isothermal compressibility:

κ_{T} = - \frac{1}{V} (\frac{\partial ^{2} G}{\partial P ^{2}})_{T, N} .

This follows from

V = (\partial G / \partial P)_{T, N}

, with

G

denoting Gibbs free energy. Within a stable single thermodynamic phase, increasing pressure reduces volume and makes this curvature negative. At phase coexistence, changing the phase fraction can change the volume at constant temperature and pressure; the usual finite single-phase isothermal compressibility then need not exist. Near a critical point, it can diverge. ^[ref]

Critical-density compressibility of a cubic gas

For the cubic-density model,

κ_{T} = (n \partial p / \partial n)^{- 1}

gives

κ_{T} (n_{c}, T) = 1/ [n_{c} k_{B} (T - T_{c})]

. Divergence from above is a mean-field critical prediction. The negative value below

T_{c}

signals an unstable homogeneous branch rather than an equilibrium negative compressibility.

Isothermal compressibility from chemical-potential response ( $κ_{T} = n^{- 2} (\partial n / \partial μ)_{T}$ )

For a homogeneous extensive single-component thermodynamic phase, the Gibbs-Duhem equation at fixed temperature gives

d p = n d μ

. With number density

n = N / V

, fixed

N

gives

d n = - (n / V) d V

. The chain rule therefore gives

κ_{T} = \frac{1}{n} (\frac{\partial n}{\partial p})_{T} = \frac{1}{n ^{2}} (\frac{\partial n}{\partial μ})_{T} .

The derivatives are evaluated within the same smooth equation of state; an interface contribution or phase coexistence can invalidate this simple homogeneous derivation.

Isentropic compressibility ( $κ_{S} = - V^{- 1} (\partial V / \partial P)_{S, N}$ )

Isentropic compressibility is the compressibility at fixed entropy, particle number and composition. The conventional name adiabatic compressibility refers to a reversible adiabatic process; an irreversible adiabatic compression need not keep entropy fixed. For an ideal gas, it is

κ_{S} = 1/ (γ P)

, where

γ = C_{P} / C_{V}

is the heat capacity ratio. It describes a steeper local pressure response than isothermal compressibility when

γ > 1

Maxwell relations ( $(\partial S / \partial V)_{T, N} = (\partial P / \partial T)_{V, N}$ )

Maxwell relations turn pressure, volume and temperature measurements into information about entropy. For example, the pressure rise when heating a rigid sealed vessel determines the entropy change when expanding the same sample isothermally:

(\frac{\partial S}{\partial V})_{T, N} = (\frac{\partial P}{\partial T})_{V, N} .

The quantities on the two sides describe different experiments, yet both are mixed partial derivatives of one Helmholtz free energy. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the coefficient of thermal expansion gives

(\partial S / \partial P)_{T, N} = - α V

. ^[ref]

For a simple compressible system, hold particle number

N

and composition fixed, and allow only pressure-volume work. Write

U

for internal energy,

H = U + P V

for enthalpy,

F = U - TS

for Helmholtz free energy, and

G = H - TS

for Gibbs free energy. Their natural-variable differentials are

d U (S, V, N) d F (T, V, N) = T d S - P d V, = - S d T - P d V, d H (S, P, N) d G (T, P, N) = T d S + V d P, = - S d T + V d P .

Thus the first partial derivatives, with the subscript recording the variables kept fixed, are

U_{S}_{V, N} = T, F_{T}_{V, N} = - S, U_{V}_{S, N} = - P, F_{V}_{T, N} = - P, H_{S}_{P, N} = T, G_{T}_{P, N} = - S, H_{P}_{S, N} = V, G_{P}_{T, N} = V .

The symmetry of second derivatives produces all four standard Maxwell relations, including their signs:

U : (\frac{\partial T}{\partial V})_{S, N} H : (\frac{\partial T}{\partial P})_{S, N} F : (\frac{\partial S}{\partial V})_{T, N} G : (\frac{\partial S}{\partial P})_{T, N} = - (\frac{\partial P}{\partial S})_{V, N}, = (\frac{\partial V}{\partial S})_{P, N}, = (\frac{\partial P}{\partial T})_{V, N}, = - (\frac{\partial V}{\partial T})_{P, N} .