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An isothermal expansion increases the volume of a thermodynamic system at fixed temperature. For a reversible ideal gas path, the gas delivers positive work and absorbs positive heat, with $Q=W_{\mathrm{by},\mathrm{rev}}=nRT\ln (V_f/V_i)$ and $\Delta U=0$. The heat bath supplies the expansion energy; without sufficient heat transfer, doing work instead lowers the temperature. This transfer is the hot step of a Carnot cycle, and it is why an expanding gas need not cool when its surroundings continuously replenish its internal energy.

An insulated free expansion into an evacuated compartment is different. In the idealization with no heat exchange, no external work and no container energy change, $Q=W_{\mathrm{by}}=0$. Since internal energy of an ideal gas depends only on temperature, its final equilibrium temperature equals its initial value. The transient need not have a uniform temperature or pressure, so equal endpoint temperatures do not make it a reversible isotherm. Do not apply the logarithmic isothermal ideal-gas work formula to this irreversible process. Its entropy still increases by $\Delta S=nR\ln (V_f/V_i)$; entropy change in free expansion evaluates this state function difference using a hypothetical reversible path.^[ref]

For a real gas, even equal-temperature equilibrium endpoints need not have equal internal energy. An insulated free expansion can therefore change the temperature, while a controlled isothermal expansion must provide the heat needed for both its internal energy change and its delivered work. The internal energy at constant temperature article gives the Van der Waals equation example.^[ref]

An isothermal process changes a thermodynamic system while keeping its temperature constant. It need not keep its pressure, volume or internal energy constant, and it does not generally have zero heat transfer. An adiabatic process specifies zero heat transfer instead. For example, a slowly expanding ideal gas can absorb heat from a heat bath and deliver the same amount of work while its temperature stays fixed.

For a fixed amount of substance and composition, internal energy of an ideal gas depends only on temperature. With $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work delivered to the surroundings, the first law of thermodynamics therefore givesThis equality does not determine either transfer until the path is specified. Isothermal ideal-gas work gives the reversible value; an insulated free expansion has a different value even when its equilibrium endpoints have the same temperature. The closed system convention here excludes changes in bulk kinetic energy and external potential energy.

A conducting cylinder in a large, stirred heat bath approximates an isothermal process when the piston moves slowly enough for thermal conduction to keep up. A quasistatic process keeps the gas close to internal thermodynamic equilibrium; a reversible thermodynamic process additionally requires negligible friction and vanishing driving differences in pressure and temperature. Exact equality of the gas and heat bath temperatures gives no finite heat transfer rate through a finite thermal resistance. The reversible description is a limit of arbitrarily small differences and arbitrarily slow motion, illustrated by the finite-rate isothermal piston model.^[ref]

For a real gas, internal energy at constant temperature can change with volume because molecular interaction potential energy changes. In one smooth thermodynamic phase of a fixed-composition simple compressible system,Thus an isothermal process generally has $Q=\Delta U+W_{\mathrm{by}}$, with no reason to set $\Delta U=0$. The right-hand side vanishes for the ideal gas law, but not generally for another equation of state.^[ref]

James Prescott Joule was an English physicist whose private experiments connected electricity, heat and mechanical work. His family ran a brewery in Salford; he developed an experimental laboratory while working in the family business.^[ref] The Royal Society's archive identifies his occupation as physicist and records his life as 1818–1889.^[ref]

His electrical heating experiments placed metal coils in water and measured the temperature change. Later, he used the brewery cellar to reduce ambient temperature fluctuations during friction experiments and took particular care over thermometer calibration.^[ref] These practical choices address a central difficulty of calorimetry: a small temperature rise can reflect both the intended input of energy and unwanted heat transfer to the surroundings. A quiet baseline and a known heat capacity are part of the measurement, not decoration around the formula.

The contrast between Joule heating and Joule's paddle-wheel experiment makes the connection especially clear. One supplies electrical work; the other supplies mechanical work. In each case, calorimetry tracks the resulting change of internal energy. Comparing quantitatively controlled inputs with thermal changes gives an experimental test of conservation of energy, rather than merely showing that stirring or an electric current can warm water. The paddle-wheel account describes the descending weights, vanes, thermal controls and corrections needed for that comparison.

Joule's free-expansion experiment asks a different question: how does a gas's temperature change during expansion without external work? The Joule-Thomson effect instead concerns flow through a restriction and a balance involving enthalpy. Keeping the two experiments distinct prevents a shared historical name from hiding different system boundaries and different conserved quantities.

Joule heating is the dissipative conversion of electrical work into internal energy in an electrical conductor. In a metal, the electric field supplies energy to charge carriers, which transfer it to the material. This makes an immersed resistor a useful calibrated heater and makes electrical resistance a source of losses in wiring. Wind-driven electric currents are a proposed source of heating of an inflated giant planet.

For a stationary, purely resistive Ohmic conductor, use the passive sign convention: $V$ is the terminal electric potential difference from the entry to the exit of positive electric current $I$. The absorbed electrical power and Ohm's law giveThe last two expressions assume $V=IR$ at the relevant material temperature. For constant electrical resistance and electric current over a duration $\Delta t$, the dissipated energy is $E_{\mathrm{diss}}=I^{2}R\Delta t$; otherwise use $E_{\mathrm{diss}}=\int V(t)I(t)dt$. In SI, $\mathrm{V}\mathrm{A}=\mathrm{W}=J/s$.^[ref]

Absorbed electrical power $VI$ is more general than Joule heating: a capacitor can store energy, a battery can change its chemical energy, and an electrical circuit can deliver mechanical work or emit light. Even a heated resistor need not show an indefinitely rising temperature: in a thermal steady state, its dissipated power leaves by heat transfer. With the heater inside a calorimeter boundary, electrical input crosses as work; the subsequent heat transfer to the room is a different term in the first law of thermodynamics.

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.

The Joule-Thomson effect is the temperature response of a fluid flowing from higher to lower pressure through a valve or a plug of porous medium, when the throttle is insulated and the other steady-flow energy balance assumptions hold. Its equilibrium endpoints have equal specific enthalpy. The Joule-Thomson coefficient determines whether a small drop cools or warms; cooling is not universal.

A useful contrast is between three insulated expansions. In free expansion of a gas-only closed system into a vacuum, no external work occurs and $U_2=U_1$; an ideal gas returns to its initial equilibrium temperature. In a throttle, continuous flow carries pressure work across the boundary and preserves specific enthalpy under the stated assumptions; an ideal gas again has no endpoint temperature change. In a reversible adiabatic expansion of a loaded closed system, the gas supplies work, loses internal energy and cools, even if it is an ideal gas. That process preserves entropy; throttling generally produces entropy. For a small constant-enthalpy step, $dH=TdS+Vdp=0$ gives $dS=-Vdp/T>0$ when $dp<0$ and $T,V>0$. This equilibrium comparison computes endpoint entropy differences without supplying a reversible path through the plug.

The effect is useful in refrigerators because an appropriately prepared real gas can cool while passing through a simple valve. The inlet must lie in a suitable positive-Joule-Thomson coefficient region. The Joule-Thomson inversion curve makes that requirement precise; a pressure drop by itself is not a guarantee of refrigeration.

No cyclic device can have as its sole effect the extraction of heat from one thermal reservoir and its complete conversion into net work. A putative device absorbing $Q_h>0$ and returning to its starting state with $W=Q_h$ and no other transfer would conserve energy, yet violate the Second law of thermodynamics. Rejecting heat to colder surroundings is an essential part of the ordinary two-thermal reservoir heat engine budget.^[ref]

The cyclic and sole-effect qualifications matter. A single isothermal expansion of an ideal gas can absorb heat and deliver equal work while leaving its volume changed; restoring the initial state must also be counted. Likewise, a device discharging stored internal energy or consuming fuel has another effect. Converting supplied electrical work completely to heat through Joule heating reverses the proposed conversion and is allowed. The statement also allows cyclic devices that exchange heat with one thermal reservoir while consuming net work; it forbids positive net work output with that single heat source as the only external change.

Conversely, a hypothetical complete-conversion device could take $100\mathrm{J}$ from the hot thermal reservoir and supply $100\mathrm{J}$ of work to an ideal refrigerator that extracts $100\mathrm{J}$ from the cold thermal reservoir and delivers $200\mathrm{J}$ to the hot one. At $600\mathrm{K}$ and $300\mathrm{K}$, that refrigerator is itself admissible: its reservoir entropy changes cancel. The composite would consume no external work and move a net $100\mathrm{J}$ from cold to hot, violating the Clausius statement of the second law. Together with the construction there, this proves the equivalence of the two impossibility statements.^[ref]

For the two fixed positive-temperature thermal reservoir hypotheses, Carnot's theorem gives the sharper restriction $\eta \le 1-T_c/T_h<1$. A reversible thermodynamic process reaches that bound in the ideal Carnot cycle; an ordinary device's smaller thermal efficiency reflects its entropy production and measurement boundary. An inefficient observed engine is compatible with the Second law of thermodynamics, rather than evidence that the bound has been experimentally determined.

Kinetic friction is friction during nonzero contact slip. For one-dimensional motion along a surface, choose right as positive and write $v$ for the body's contact-point velocity and $U$ for the surface's contact-point velocity, both measured in the same inertial frame. The signed relative velocity is $w=v-U$. The ideal Coulomb law for the tangential force on the body isHere $N\ge 0$ is the magnitude of the normal force, ${\mu }_k\ge 0$ is the coefficient of kinetic friction, and $\mathrm{sgn}$ is the sign function. Thus kinetic friction opposes contact slip, which can differ from the body's direction of motion. In several tangential directions the same law reads ${\mathbf{f}}_k=-{\mu }_{k}N\mathbf{w}/|\mathbf{w}|$ for a nonzero tangential relative velocity $\mathbf{w}$. The magnitude law is documented in Ling, Sanny and Moebs, OpenStax University Physics 1, §6.2: static and kinetic friction.

For a stipulated toy model, take a nonrotating box with mass $m=1\mathrm{kg}$, normal force $N=10\mathrm{N}$, coefficient of static friction ${\mu }_s=0.5$, and coefficient of kinetic friction ${\mu }_k=0.3$. These are assumed model parameters, rather than measured properties of any particular pair of materials; the prescribed normal force is consistent with horizontal support under $g=10m/{\mathrm{s}}^2$ and no other vertical forces. On stationary ground, a box already sliding right has $f_k=-3\mathrm{N}$. With an applied horizontal force $F=6\mathrm{N}$ to the right, Newton's second law gives $a=(F+f_k)/m=3m/{\mathrm{s}}^2$, for as long as rightward contact slip persists.

Direction is clearer by comparing two instantaneous states of this same model. If the box has $v=1m/s$ on stationary ground, $U=0$ and $w=+1m/s$, so kinetic friction acts left with magnitude $3\mathrm{N}$. If the box still has $v=1m/s$ but a belt beneath it has $U=2m/s$, then $w=-1m/s$: kinetic friction acts right with magnitude $3\mathrm{N}$, even while the box moves right.

In the belt case, the power delivered to the box is $P_{\mathrm{box}}=f_{k}v=+3\mathrm{W}$. With no other horizontal force, its kinetic energy therefore increases by the work-energy theorem. By Newton's third law, the box exerts $-f_k$ on the belt, so the contact power delivered to the belt is $P_{\mathrm{belt}}=-f_{k}U=-6\mathrm{W}$. Their sum is $-3\mathrm{W}$: frictional heating produces thermal internal energy at $3\mathrm{W}$. An ideal motor maintaining the prescribed belt velocity supplies $+6\mathrm{W}$ against this contact load, neglecting other losses, so its input divides into $3\mathrm{W}$ of box kinetic energy and $3\mathrm{W}$ of frictional heating. These are instantaneous rates, since the box's velocity subsequently changes. This balance follows directly from power as force times velocity and conservation of energy; the power law is given in Ling, Sanny and Moebs, OpenStax University Physics 1, §7.4: power.

When contact slip vanishes, including a transition to rolling without slipping, the nonzero-slip law no longer applies. Determine the required static friction from the no-slip condition and the equations of motion, then test its bound. Substituting $\mathrm{sgn}(0)=0$ into the sliding law would incorrectly exclude a nonzero static friction.

Latent heat is the heat absorbed or released in a first-order phase transition through its entropy jump. For an equilibrium conversion from thermodynamic phase $\alpha$ to thermodynamic phase $\beta$ at fixed pressure, evaluate both thermodynamic phases at the same temperature and pressure on their phase coexistence curve:Use one normalization throughout: $L$, $S$ and $H$ may all be per mole, giving $L$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or all per unit mass, giving $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Here $S$ is entropy and $H$ is enthalpy on that basis. Equal Gibbs free energies of the coexisting thermodynamic phases give $\Delta H=T_{\mathrm{tr}}\Delta S$. Identifying this enthalpy difference with supplied heat assumes a closed system and an equilibrium conversion with only pressure-volume work. ^[ref]

For a pure substance with positive latent heat per unit mass $L$, let $x$ be the phase fraction of the higher-enthalpy thermodynamic phase, measured by mass. While both thermodynamic phases coexist at the fixed pressure,Adding heat changes $x$ rather than the temperature; this creates the horizontal portion of a heating curve. Once $x=1$, additional heat raises the temperature according to specific heat capacity at constant pressure. Reversing the conversion releases heat. The plateau represents changing enthalpy, rather than a finite single-thermodynamic phase heat capacity satisfying $Q=C_p\Delta T$. ^[ref]

For a constructed fixed-pressure model, take $m=1\mathrm{kg}$, initial temperature $280\mathrm{K}$, transition temperature $T_{\mathrm{tr}}=300\mathrm{K}$, solid specific heat capacity at constant pressure $c_s=2\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, latent heat $L=100\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, and liquid specific heat capacity at constant pressure $c_l=3\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. These are stipulated model parameters, not empirical properties of a named material. Heating the solid to $300\mathrm{K}$ takes $40\mathrm{kJ}$ of sensible heat; converting it fully to liquid takes another $100\mathrm{kJ}$ of heat; heating the liquid to $320\mathrm{K}$ takes $60\mathrm{kJ}$. The total is $200\mathrm{kJ}$. After a total heat input of $90\mathrm{kJ}$, $x=(90-40)/100=1/2$ and the temperature is still $300\mathrm{K}$. The corresponding per-unit-mass entropy jump is $L/T_{\mathrm{tr}}=\frac{1}{3}\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$.

A continuous phase transition has no latent heat. Its heat capacity can have a jump or a singularity without a finite interval in which added heat converts coexisting thermodynamic phases at a constant temperature. A heating plateau therefore requires the stated phase coexistence conditions; it is not a feature of every phase transition. ^[ref]

A lattice energy needs a reaction, a sign convention, and a stated amount. Using the lattice-dissociation enthalpy convention for sodium chloride,The reverse lattice-formation enthalpy is negative with the same magnitude when initial and final states match. Chemistry 2e gives $769\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for this separation, per mole of sodium chloride formula units. This is about $7.97$ electronvolts per formula unit, not per individual ion. It concerns separated gaseous ions, rather than neutral chemical atoms, molten salt, or hydrated ions in water.^[ref]

Also distinguish an internal energy change from an enthalpy change: $H=U+pV$. A static point-charge calculation omits motion and thermal contributions. Reversing a reaction reverses its energy change; replacing internal energy with enthalpy additionally changes the defined quantity. The Born-Haber cycle infers a lattice-formation enthalpy by combining other enthalpies, ionization energies, and electron affinities. These imagined stages are a thermodynamic accounting path, not a claim about the actual sequence of a chemical reaction.^[ref]

For a two-dimensional Fermi gas of $N$ spin-one-half particles with Fermi energy $E_F$, the constant density of states makes the low-temperature internal energy correction and heat capacity at constant areaThus its low-temperature heat capacity obeys a power law with exponent one.

For a strong monatomic shock, the planar shock compression of a finite slab gives bulk speed $3V_s/4$ and kinetic energy $E_{\mathrm{bulk}}=9M_c{V}_s^2/32$. The initial internal energy is $E_{\mathrm{th}}=3M_{c}p/(2\rho )$. A bulk-ballistic escape estimate $E_{\mathrm{bulk}}\gtrsim W$ uses positive binding-energy magnitude $W$, not negative gravitational potential energy. With adiabatic sound speed $c_{\mathrm{ad}}^2=5p/(3\rho )$ it yields the first displayed coefficient. Using the isothermal sound speed $c_T^2=p/\rho$ instead yields $16/3\simeq 5.3$. The definitions obey ${\mathcal{M}}_T^2=(5/3){\mathcal{M}}_{\mathrm{ad}}^2$. Heating, subsequent expansion and energy loss can alter the physical escape threshold, so the coefficient is not independent of the energy budget and Mach convention.

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: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 $(\partial S/\partial P{)}_{T,N}=-\alpha V$. ^[ref]

For a simple compressible system, hold particle number $N$ and composition fixed, and allow only pressure-volume work. Write $U$ for internal energy, $H=U+PV$ for enthalpy, $F=U-TS$ for Helmholtz free energy, and $G=H-TS$ for Gibbs free energy. Their natural-variable differentials areThus the first partial derivatives, with the subscript recording the variables kept fixed, areThe symmetry of second derivatives produces all four standard Maxwell relations, including their signs:For a brief derivation of the $F$ relation, differentiate $F_T=-S$ with respect to $V$ and $F_V=-P$ with respect to $T$: $-S_V=F_{TV}=F_{VT}=-P_T$. Each other row follows in the same way. Assume continuous second partial derivatives of the relevant thermodynamic potential in the region considered. At a phase transition, the equilibrium potential may lose this smoothness; apply the identities within each smooth thermodynamic phase, and handle latent heat and phase coexistence separately. If particle numbers vary, the potential differentials acquire chemical potential terms, so the fixed-$N$ constraints must remain explicit. ^[ref]

For an ideal gas with fixed $N$, the ideal gas law gives $P=N{k}_{B}T/V$. At fixed volume, heating changes the pressure at rate $N{k}_B/V$. Consequently an isothermal expansion at positive temperature hasFor one mole with $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $V_2=2V_1$, this is $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. At $300\mathrm{K}$, a reversible isothermal expansion absorbs $Q=T\Delta S\simeq 1728.85\mathrm{J}$ and performs equal work, since its internal energy is unchanged. The entropy difference is a state function difference: the same equilibrium endpoints give the same $\Delta S$ in an insulated free expansion, although there $Q=0$ and the process is irreversible.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure, $dV=\alpha VdT$, soThus measured expansion and compression responses determine a heat capacity difference without measuring both heat capacities separately. They do not determine the absolute heat capacities or absolute entropy by themselves; an independent thermal measurement supplies that missing information. For an ideal gas, $\alpha =1/T$ and ${\kappa }_T=1/P$ reduce the result to $C_P-C_V=PV/T=N{k}_B$, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also givesIntegrating these slopes determines how a heat capacity changes with volume or pressure at fixed temperature. For an ideal gas, both right-hand sides vanish because the ideal gas law is linear in temperature at either fixed volume or fixed pressure. Consequently $C_V$ and $C_P$ depend only on temperature at fixed $N$, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's Theory of Heat, first published in 1871, presents these connections using the geometry of isothermal processes and reversible adiabatic processes. His fourth edition, page 167, explicitly relates expansion on heating at fixed pressure to heat emitted during an isothermal compression: the modern $G$ relation gives $\delta Q_{\mathrm{rev}}=TdS=-T\alpha VdP$. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^{[ref] [ref]}

Mayer's relation states that the difference between heat capacity at constant pressure and heat capacity at constant volume for a fixed-composition ideal gas isHere both capacities are total heat capacities in $\mathrm{J}{\mathrm{K}}^{-1}$, $n$ is amount of substance, $N$ is the number of particles, $R$ is the molar gas constant and $k_B$ is the Boltzmann constant. In other normalizations the same statement isThe first is a difference of molar heat capacities, in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$; the second uses specific heat capacities, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Writing $C_p-C_V=R$ with total capacities would omit the sample's amount of substance. The relation holds pointwise even when the capacities depend on temperature. ^[ref]

The proof outline is short: enthalpy is $H=U+pV=U+nRT$, and internal energy of an ideal gas is independent of volume at fixed temperature. Differentiating at fixed composition and amount of substance therefore gives $C_p=dH/dT=dU/dT+nR=C_V+nR$. It uses the ideal gas law, rather than assuming any particular number of molecular degrees of freedom. At fixed pressure, the additional heat supplies expansion work, $pdV=nRdT$. The two processes can have the same temperature endpoints and equal $\Delta U$ while requiring different heat inputs.

For one mole of the rigid-rotation diatomic ideal gas model, $C_V=5R/2$ and $C_p=7R/2$. Heating by $10\mathrm{K}$ with $R\simeq 8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ gives $Q_V\simeq 207.86\mathrm{J}$ in a rigid sealed vessel and $Q_p\simeq 291.01\mathrm{J}$ under a frictionless piston at fixed pressure. Both gain $207.86\mathrm{J}$ of internal energy, while the piston delivers approximately $83.14\mathrm{J}$ of work. This is a calculated model budget. If the capacities vary with temperature, the corresponding finite statement is stillunder the fixed-pressure and fixed-volume heating assumptions in heat capacity. Their final volume values generally differ. Combining Mayer's relation with heat capacity ratio also gives $C_V=nR/(\gamma -1)$ and $C_p=\gamma nR/(\gamma -1)$ locally; constant values cannot be extrapolated across intervals in which $\gamma (T)$ changes.

For a smooth stable single thermodynamic phase of a general simple compressible system, the broader Maxwell relation identity iswhere $\alpha$ is the coefficient of thermal expansion and ${\kappa }_T$ the isothermal compressibility, with composition fixed throughout. Substituting the ideal gas law gives $nR$; a real gas need not do so. This distinction matters near a phase transition or when interactions appreciably change the equation of state. It explains why Mayer's relation is useful: thermal measurements under different constraints become linked to a mechanical response, while the difference retains a particularly simple value for an ideal gas.^[ref]

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 iswhere $C_{\mathrm{eff}}$ is its participating heat capacity, approximately constant over the run, and $Q_{\mathrm{room}}$ 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.

Dividing two extensive quantities with the same replication factor produces an intensive quantity:Division by volume gives a density, such as mass density $\rho =m/V$ or energy density $U/V$. Division by mass gives a specific quantity, such as specific internal energy $u=U/m$ or specific volume $v=V/m$. Division by amount of substance gives a molar quantity, such as molar volume $V_m=V/n$. Their units are respectively $\mathrm{kg}{\mathrm{m}}^{-3}$, $\mathrm{J}{\mathrm{m}}^{-3}$, $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, ${\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$ and ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The denominator must be stated: “per particle”, “per mole” and “per unit mass” are different normalizations.

For a hypothetical uniform parcel with mass $2\mathrm{kg}$, volume $0.002{\mathrm{m}}^3$ and internal energy $40\mathrm{kJ}$ relative to a specified reference, $u=20\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ and $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. Three independent copies have $6\mathrm{kg}$, $0.006{\mathrm{m}}^3$ and $120\mathrm{kJ}$; both ratios remain unchanged. This permits comparison of materials without confusing the amount tested with the state tested.

If $X$ and $Y$ are additive but the parcels have different $r_i=X_i/Y_i$, their composite ratio is a weighted arithmetic mean:The denominator supplies the weights: specific internal energy uses mass weights, while mass density uses volume weights. This identity is accounting for the specified subsystems; predicting a new state after mixing also needs the relevant equation of state.

Ohm's law states that an Ohmic conductor has electric current proportional to its applied electric potential difference, with constant electrical resistance under fixed material conditions:Here $V=\varphi (A)-\varphi (B)$ is the terminal electric potential difference, and positive $I$ enters at $A$ and exits at $B$. This sign convention makes $VI$ the electrical power absorbed. The straight-line relation through the origin lets one constant predict the response at many applied electric potential differences. Merely defining $R=V/I$ at one nonzero operating point does not establish Ohm's law: a component may have a ratio that varies with applied electric potential difference. Material and temperature must be specified and held fixed; many components have a nonlinear response. ^[ref].

For the ideal resistor model $R=2\Omega$ at electric potential difference $V=5\mathrm{V}$,This power is converted to internal energy by Joule heating; a watt is a joule per second. If the model holds for time $10\mathrm{s}$, the transferred energy is $125\mathrm{J}$. Doubling electric potential difference doubles electric current and quadruples power for the same fixed $R$. The model keeps material temperature fixed even while calculating Joule heating; appreciable changes in temperature may require a changing electrical resistance. ^[ref].

In 1827, Georg Ohm published his mathematical account of electrical transport in Die galvanische Kette, mathematisch bearbeitet. On printed page 36, his relation $S=A/L$ expresses electric current in terms of total driving tension and a reduced length incorporating electrical conductivity and geometry. This is historical notation: his $A$ is not the wire cross-sectional area used here, and his $L$ is not simply the wire's physical length. The striking idea is that material response and geometry can be combined into one quantity governing an electrical circuit. ^[ref].

Choose the water, vessel and immersed paddle as the closed system, with fixed outer volume and negligible endpoint changes in bulk mechanical energy. Between common equilibrium temperatures $T_1$ and $T_2$,The effective heat capacity $C_{\mathrm{eff}}$ includes every component that participates in the common temperature change. For nearly constant $C_{\mathrm{eff}}$, the readout is $C_{\mathrm{eff}}\Delta T$. Heat transferred between water and paddle stays inside this boundary; it must not be added again as external input. The drive supplies shaft work even though the vessel has no expansion work.

For a simplified modern model, let $M$ be the total descending mass, let each of $N$ descents cover height $h$, and let $W_{\mathrm{loss},\mathrm{out}}$ be energy diverted by the external drive. With negligible remaining drive kinetic energy and elastic storage,The signed $Q_{\mathrm{room},\mathrm{out}}=-Q_{\mathrm{room},\mathrm{in}}$ may be negative if the room warms the assembly. This model omits the small endpoint and elastic corrections explicitly considered in the historical analysis. It does not equate all of the weights' lost gravitational potential energy with the water's gained internal energy.

As a proposed modern protocol, measure the falling masses and distances, calibrate the thermometer and $C_{\mathrm{eff}}$ independently, record room temperature and resting drift before and after each run, and wait for fluid motion to decay before the final equilibrium reading. Repeat with different fall budgets and run durations. A stable ratio of corrected input to corrected thermal storage supports the first law of thermodynamics over the tested range. Resting controls only estimate leakage if their thermal conditions represent the driven run; sensor lag, nonuniform temperature, unmeasured drive losses, evaporation or changing heat capacity can spoil the inference. Those are limits of the measurement, rather than evidence that heat was stored as a separate substance.

Supersonic communication. A point source moving at speed $V>c_s$ outruns its small acoustic waves. After a time $t$, the source has moved $Vt$, while a disturbance emitted at its old position has radius $c_{s}t$. The envelope of these spherical wavefronts is the Mach cone, with half-angle $\mu$ satisfyingInformation is restricted to the downstream cone, so upstream gas cannot smoothly anticipate a supersonic obstacle. Finite compressions can also steepen because their characteristics converge. Such disturbances form a shock wave, across which the fluid variables change rapidly. A perfectly uniform supersonic flow need not contain a shock: the cone expresses the communication constraint, while a disturbance or compression supplies the reason for a discontinuity.

Conserved fluxes. Work in the shock frame, take velocities normal to the front, and integrate the conservation laws through a thin steady control volume of unit area. Mass conservation gives constant flux $j=\rho u$, soThe normal momentum flux is the advected part $\rho u^2$ plus the pressure stress $p$. With negligible body-force impulse across the thin front,For an ideal gas with constant adiabatic index $\gamma$, the specific internal energy is $e=p/[(\gamma -1)\rho ]$. The energy flux iswhere $h_{\mathrm{sp}}$ is the specific enthalpy. No radiation or other heat flux escapes in this adiabatic shock model. Conservation of the energy flux, divided by the nonzero mass flux, givesThese are the Rankine-Hugoniot conditions for a perfect gas. The shock is adiabatic but irreversible: its entropy increases, so an isentropic relation cannot be imposed across the discontinuity.

Strong-shock compression. Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and neglect $p_1$ and the upstream enthalpy compared with ${\rho }_1{u}_1^2$ and $u_1^2$. Momentum conservation gives $p_2={\rho }_1{u}_1^2(1-1/\chi )$. Substitute this into energy conservation:ThusThe root $\chi =1$ describes no shock. The compressive branch of the Strong-shock Rankine-Hugoniot conditions isIt also gives $p_2=2{\rho }_1{u}_1^2/(\gamma +1)$. A monatomic gas has $\gamma =5/3$, hence $\chi =4$.

The finite slab in its initial rest frame. Denote the shock's lab speed by $V_s$. The upstream gas is initially stationary, so $u_1=V_s$ and the downstream shock-frame speed is $u_2=V_s/4$. Transforming to the initial slab frame givesThis is planar shock compression of a finite slab. In the uniform post-shock approximation, breakout occurs at $t_b=L/V_s$. The original front face has then moved $(3V_s/4)t_b=3L/4$, while the shock reaches the originally stationary far face at $L$, leaving thickness $L/4$. Equivalently, the fixed mass per area $\rho L$ becomes $4\rho L_2$. The result assumes a maintained approximately constant shock speed and no important boundary rarefaction during the crossing; an unconfined finite cloud need not remain uniformly compressed.

Escape estimate and the sound-speed convention. Let $M_c$ be the slab mass, $W>0$ its binding-energy magnitude in the star's field, and $E_{\mathrm{th}}$ its initial internal energy. Treat the potential as nearly uniform across the slab. For the bulk-ballistic escape estimate, requireThe Mach-number convention in shock unbinding estimates matters here. With the usual upstream adiabatic sound speed, $c_{\mathrm{ad}}^2=\gamma p/\rho =5p/(3\rho )$, define ${\mathcal{M}}_{\mathrm{ad}}=V_s/c_{\mathrm{ad}}$. ThenIf instead the shock speed is normalized by the isothermal sound speed, $c_T^2=p/\rho$, then ${\mathcal{M}}_T^2=(5/3){\mathcal{M}}_{\mathrm{ad}}^2$ andThe original PDF's “around five” is consistent with the latter convention or an order-of-magnitude sound-speed estimate. It is not the exact coefficient obtained with the conventional adiabatic Mach number and the stated bulk kinetic-energy criterion. Here $W=-E_{\mathrm{grav}}$ if the potential is zero at infinity; using negative gravitational energy directly on the right would be incorrect. Subsequent conversion of shock-generated internal energy to expansion, external pressure, or radiative losses would change the physical escape threshold, so this numerical factor is not universal.

The local conservation laws. The first equation is the radial Euler equation: the material acceleration $Du/Dt={\partial }_{t}u+u{\partial }_{r}u$ is caused by the pressure gradient force per unit mass, $-{\partial }_{r}p/\rho$. It neglects gravity and viscosity. The second equation is mass conservation in spherical coordinates,The term $2\rho u/r$ appears because the area of a spherical surface grows as $r^2$.

To obtain the third equation, let $e=p/[(\gamma -1)\rho ]$ be the specific internal energy of an ideal gas with constant heat capacity ratio $\gamma$. The first law of thermodynamics for a smooth adiabatically moving element isDifferentiating $e$ and collecting terms givesand henceThus the entropy parameter $p/{\rho }^{\gamma }$ is conserved along each smooth pathline; for fixed composition, specific entropy differs from $c_v\ln (p/{\rho }^{\gamma })$ only by a constant. This is local isentropic flow behind the shock. A physical shock wave itself generates entropy and must be treated with jump conditions, rather than by applying this smooth equation across the discontinuity.

The front-radius scaling. In the self-similar blast wave approximation there is no scale set by ambient pressure, cooling, gravity, or the initial source radius. Write $R\propto E^a{\rho }_0^b{t}^d$. Using $[E]=\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$ and $[{\rho }_0]=\mathsf{M}{\mathsf{L}}^{-3}$, dimensional analysis givesThus $a=1/5$, $b=-1/5$, $d=2/5$ andThe dimensionless coefficient ${\xi }_0$ depends on $\gamma$ and the resolved similarity profiles; dimensions alone do not fix it. Define the length scale $\ell (t)=(E{t}^2/{\rho }_0{)}^{1/5}$, so $\xi =r/\ell$ and the front is at the fixed coordinate $\xi ={\xi }_0$.

Kinetic energy. Substitute $\rho ={\rho }_0{f}_{\rho }(\xi )$ and $u=(r/t)f_u(\xi )$ into the volume integral. Since $r=\ell \xi$ and $dr=\ell d\xi$,But ${\rho }_0{\ell }^5/t^2=E$, so the first of the spherical blast-wave similarity energy integrals isThe conclusion requires a fixed, integrable similarity profile, as in the finite-energy blast solution. Expansion increases the swept-up mass while reducing the characteristic squared speed, leaving this product constant.

Pressure and internal energy. The derivatives of the similarity variable areWriting a prime for differentiation with respect to $\xi$, the material acceleration becomesThe radial momentum equation therefore requires a pressure gradient on the scale ${\rho }_0\ell /t^2$. Integrating over radius gives pressure similarity in a spherical blast wave:whereThe prefactor has units $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$, as required for pressure. Strictly, the momentum equation alone permits an additional pressure uniform in space and arbitrary in time. In the scale-free strong-blast solution, shock matching and the remaining thermodynamic equation fix that freedom with the same similarity scaling; any allowed constant of integration is included in $f_p$. An independently imposed external pressure would introduce another scale and invalidate the simple dimensional ansatz.

For constant $\gamma >1$, the internal energy is consequentlyIf $E$ is the total excess explosion energy, its normalization isThe two constant fractions need not be equal. No explicit solution for $f_p$ is needed to establish their time independence.