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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 2024 operating instructions for PHYWE's 04361-00 apparatus describe a frit between two gas chambers. A copper-coil heat exchanger first restores inlet temperature after cylinder-regulator cooling. An upstream manometer reads gauge pressure; the outlet stays at atmospheric pressure. Two NiCr-Ni thermocouples sit 2–5 mm from the frit. The digital meter's difference mode reports $0.01\mathrm{K}$ precision.^[ref]

The manual specifies at most $1\mathrm{bar}$ gauge pressure, one hour of apparatus acclimatization, half an hour of meter warm-up, about 1.5 minutes of flow settling, then 10–15 seconds of averaging. It identifies a stagnant-gas zeroing error that shifts the fitted intercept and notes a maximum nitrogen signal of about $0.3\mathrm{K}$ in this setup.^[ref]

Define $x=p_1-p_2>0$ and $y=T_1-T_2$, so a small-drop fit is $y\simeq {\mu }_{\mathrm{JT}}x+c$. A constant sensor offset $c$ leaves the slope unchanged; pressure-dependent drift or heat leakage can alter it. The observable is the differential temperature slope across the plug, not cylinder-outlet cooling. Display precision alone does not establish accuracy or prove an isenthalpic process. This low-pressure, room-temperature apparatus does not map a complete Joule-Thomson inversion curve.

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.

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 $C_{\mathrm{cal}}$, 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 $Q_{\mathrm{loss},\mathrm{out}}$, the combined enthalpy balance isHenceThe loss is positive if the room receives net heat, negative if the room supplies it. If the vessel starts at a different temperature $T_{\mathrm{cal},i}$, replace its warming term by $C_{\mathrm{cal}}(T_f-T_{\mathrm{cal},i})$; do not assume the vessel shared the water's initial temperature. For variable material properties, replace each $mc\Delta T$ by the corresponding integral $m\int c_p(T)dT$ and each vessel term by $\int C_{\mathrm{cal}}(T)dT$. The inferred constant $c_x$ then represents an average over the sample's cooling interval.^[ref]

For a synthetic no-loss example, mix $0.100\mathrm{kg}$ of hot water at $80{}^{\circ }\mathrm{C}$ with $0.200\mathrm{kg}$ of cold water at $20{}^{\circ }\mathrm{C}$ in a vessel initially at $20{}^{\circ }\mathrm{C}$. Stipulate $c_w=4180\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $C_{\mathrm{cal}}=100\mathrm{J}{\mathrm{K}}^{-1}$. The enthalpy balance givesThe hot water releases $17.337\mathrm{kJ}$: the cold water receives $15.485\mathrm{kJ}$ and the vessel $1.852\mathrm{kJ}$. Omitting the vessel predicts $40{}^{\circ }\mathrm{C}$, despite perfect insulation. These are computed numbers from stipulated properties, not laboratory observations.

For an unknown-sample example using the same cold water and vessel, stipulate $m_x=0.100\mathrm{kg}$, $T_x=80{}^{\circ }\mathrm{C}$ and $c_x=450\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With no loss, $T_f\approx 22.752{}^{\circ }\mathrm{C}$. Inverting the full balance returns $450\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; ignoring vessel storage returns about $402\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. 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.

Ocean heat content relative to a reference temperature $T_f$ is an integral through the water column. For a uniform column of depth $H$, $Q={\rho }_w{c}_{w}H(T-T_f)$ per horizontal area. Melting ice already at its fusion temperature requires its latent heat; warming cold ice first requires additional sensible heat. A surface temperature alone does not determine the column integral.

Optical-electrical heating non-equivalence means that equal absorbed optical and electrical powers do not necessarily give identical thermometer responses. Light and the heater may deposit heat in different places, producing different temperature gradients and different routes of thermal conduction to the support. Holding one thermometer reading fixed need not hold every part of the cavity at the same temperature. This differs from incomplete radiometric absorptance: one concerns where absorbed heat goes, the other how much incident light becomes absorbed heat.

A useful accounting convention iswhere the signed correction ${\delta }_{\mathrm{ne}}$ is determined by characterization of the instrument and represents the residual departure from the lumped electrical substitution heat balance. It is not a universal extra source of energy. Varying the beam location or heater location, comparing equivalent temperature excursions, and checking repeatable dark returns can test the assumptions; no single small correction proves that every other correction vanishes.

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.

Let $T_0=T_i-T_{\infty }>0$, $S_0=S_i-S_{\infty }>0$ and $R_0=\beta S_0/(\alpha T_0)$. At steady volume, the weir outflow equals the inflow $Q$. The well-mixed heat and salt balances giveWith $\theta =(T_i-T)/T_0$ and $\eta =(S_i-S)/S_0$, the flux ratio implies ${\eta }_s=q{\theta }_s$, where $q=R_F/R_0$. The actual interface ratios are $\Delta T=T_0(1-\theta )$, $\Delta S=S_0(1-\eta )$ and $R_{\rho }=R_0(1-\eta )/(1-\theta )$.

The original PDF uses $R_{\rho }^{-2}$ in the heat-flux law; the converted TeX's exponent $-3$ is a transcription error. Using the PDF yieldsSubstitute this into the heat balance to obtain the double-diffusive overflow reservoir relationsFor the usual heat-dominated positive upward buoyancy flux, $0\le R_F<1$ and $R_0>1$ give $0\le q<1$. Then the right-hand side after substituting ${\eta }_s=q{\theta }_s$ decreases from $C$ to zero as ${\theta }_s$ runs from zero to one; there is a unique physical steady root. The net upward buoyancy flux is $B_0=g\alpha F_T(1-R_F)$.

Write $\theta =T-T_e$ and use the Boussinesq approximation for air, ${\rho }_e-\rho \simeq \rho \theta /T_e$. A uniformly heated, well-mixed room has the buoyancy pressure difference $\Delta p_b=\rho gH\theta /T_e$ between its openings. Let $C_o$ denote the combined hydraulic coefficient of the two equal openings, so the effective opening area for pressure-driven ventilation givesFor example, if each orifice has conventional discharge coefficient $C_d$ in $V=C_{d}A\sqrt{2\Delta p/\rho }$, the two pressure losses add and $C_o=C_d$. The question's coefficient is the prefactor in the final heat-load law; define $c=(2C_o^2{)}^{1/3}$. This explicitly states the numerical loss-coefficient convention. Steady heat balance is $Q=\rho C_{p}V\theta$. Eliminating $\theta$ givesHere $Q$ is total heating power and $C_p$ is specific heat capacity at constant pressure. Writing $c=C_d$ without this conversion would incorrectly change the power of the discharge coefficient.

After the reduction, cooler newly heated air occupies the bottom and displaces the old warmer air upward. Use the ideal two-layer displacement ventilation model: each layer is well mixed, the upper layer remains at ${\theta }_0$, walls have negligible heat capacity, and the sharp interface is advected without exchange between the layers. Let the lower layer have depth $h$ and excess temperature $\theta$. Integrating the hydrostatic buoyancy over the full height givesThe room's excess thermal energy is $E_T=\rho C_p{A}_r[h\theta +(H-h){\theta }_0]$. The floor supplies $\lambda Q$, incoming air has zero excess temperature, and the ceiling removes $\rho C_{p}V{\theta }_0$. Thus the required heat balance isThis moving-interface energy term must include the loss of old upper-layer volume; balancing only the lower-layer temperature would miss it.

For a genuine reduction assume $0<\lambda <1$. SetThe integrated buoyancy factor is $H{\theta }_0(1-x)$, so $V/V_0=\sqrt{1-x}$. Since $Q=\rho C_p{V}_0{\theta }_0$, the thermal balance becomes the ventilation after a heating reduction equationThe material interface also moves with the upward displacement velocity, $A_r\dot{h}=V$. Therefore ${\eta }_s=\sqrt{1-x}$, and subtracting the two equations givesThe second relation is also the lower-layer energy balance: its total excess energy increases at rate $\lambda Q$. It does not assume its temperature stays constant. Since $x_s>0$ until $x=1-{\lambda }^2$, the solution remains $0\le x<1-{\lambda }^2$, and $V/V_0\ge \lambda$. Hence ${\eta }_s\ge \lambda$, so the lower layer reaches $\eta =1$ in finite time $s_H\le 1/\lambda$. Also ${\eta }_s\le 1$, so $s_H\ge 1$.

An explicit implicit solution is useful. Put $y=\sqrt{1-x}$, which decreases from $1$ toward $\lambda$. Separation givesFilling occurs when $1=1-y_H^2+\lambda s_H$, so $\lambda s_H=y_H^2$ and ${\theta }_H/{\theta }_0=y_H^2$. These relations determine the exit time and the continuous temperature at that time. For complete switch-off $\lambda =0$, the limiting solution is $y=1-s/2$, $x=\eta =s-s^2/4$ for $0\le s\le 2$, so the cold lower layer fills the room at $s_H=2$ even though its final ventilation rate is zero.

After the interface exits, the whole room is well mixed. For $z=\theta /{\theta }_0$, the natural ventilation rate is $V=V_0\sqrt{z}$, and the well-mixed ventilation temperature balance becomesEquivalently, $\rho C_p{A}_{r}H\dot{\theta }=\lambda Q-\rho C_p{C}_{o}A\sqrt{gH/T_e}{\theta }^{3/2}$. The heat-removal term is strictly increasing in $\theta \ge 0$, so every nonnegative solution tends to the unique stable equilibriumFor $\lambda =0$ the final temperature is $T_e$. The unchanged-load case $\lambda =1$ has no distinct new temperature layer and is outside the reduction scenario.

Use the well-mixed natural ventilation model for a distributed floor source. Let the building height be $H_b$, its volume be $\mathcal{V}$, its temperature excess be $\theta$, and the exterior mass density be $\rho$. Let ${\alpha }_T$ be the coefficient of thermal expansion, so reduced gravity is $g^{\prime }=g{\alpha }_T\theta$. Convert the total heat input $Q$ to the temperature-volume flux $\mathcal{H}=Q/(\rho c_p)$, where $c_p$ is specific heat capacity. If the quoted heat flux is per floor area, first multiply it by that area.

For effective opening area for pressure-driven ventilation, define the equal-opening hydraulic coefficient $K_A$ such that the signed throughflow $q$ obeys $q|q|=K_A\Pi$, with $\Pi$ the total driving pressure divided by mass density. In the effective-area convention of question 6, each opening has discharge $A\sqrt{\delta p/\rho }$, so $K_A=A^2/2$. The usual physical-area orifice law instead gives $K_A=C_d^2{A}^2$; that prefactor does not affect the stability conclusions.

Set $W=\Delta P/\rho >0$ and $G_T=g{\alpha }_T{H}_b$. Positive $q$ denotes entry at the floor and exit at the roof. hydrostatic pressure and wind combine to giveBecause the inflowing air is at exterior temperature in either direction, the heat budget isThe well-mixed ventilation temperature balance at steady state is thereforeAll roots must satisfy $\theta >0$ and their appropriate flow-direction inequality; squaring a pressure relation without imposing those conditions could add spurious branches.

The assisting case has a unique positive solution and it is stable. For opposing wind, define ${\theta }_w=W/G_T$. There is always one solution with $\theta >{\theta }_w$, a buoyancy-dominated upward flow. Below ${\theta }_w$, the heat-removal curve is $R(\theta )=\sqrt{K_A}\theta \sqrt{W-G_T\theta }$. It is zero at both endpoints, and its maximum occurs at $\theta =2{\theta }_w/3$:For $0<\mathcal{H}<R_{\max }$ there are two downward-flow roots as well as the upward-flow root. Equivalently, setting $y=G_T\theta /W$ and $\epsilon =G_T\mathcal{H}/(\sqrt{K_A}{W}^{3/2})$, the equilibrium curve isFor linear stability analysis of the opposing-wind ventilation bistability, linearize the heat budget: the eigenvalue is $-R^{\prime }(\theta )/\mathcal{V}$. The cooler downward root has $0<\theta <2{\theta }_w/3$ and $R^{\prime }>0$, so it is stable. The warmer downward root has $2{\theta }_w/3<\theta <{\theta }_w$ and $R^{\prime }<0$, so it is unstable and separates the two basins. The upward root has $R^{\prime }>0$ and is stable. At $\mathcal{H}=R_{\max }$ the two downward roots merge at a saddle-node bifurcation; above that value only the upward equilibrium remains.

For ventilation switching under wind and heating changes, increasing wind pressure or decreasing heat input both reduce $\epsilon$, so both can create the pair of wind-driven equilibria. They nevertheless have different physical and transient effects. On the hot upward branch, implicit differentiation of $\mathcal{H}=\sqrt{K_A}\theta \sqrt{G_T\theta -W}$ shows that increasing $W$ increases $\theta$, while decreasing $\mathcal{H}$ decreases $\theta$. Both reduce the upward ventilation rate, but only a wind change shifts ${\theta }_w$.

Under slow parameter variation, a state follows its current stable branch while that branch persists; the upward branch does not disappear when wind increases or heating decreases. A sufficiently large abrupt wind increase can raise ${\theta }_w$ beyond the current temperature and place that state below the new unstable threshold, causing flow reversal and cooling towards the wind-dominated equilibrium. A jump insufficient to cross its basin boundary returns to the upward equilibrium.

At fixed $W$, reducing a still-positive heat input cannot force an initially upward-flow state through ${\theta }_w$: at that boundary the temperature balance has $\dot{\theta }=\mathcal{H}/\mathcal{V}>0$. Thus heat reduction alone does not cause that reversal in this model. Conversely, decreasing wind or increasing heat from the stable downward branch eventually destroys it at the fold and forces a switch to the upward branch, giving history dependence. Wind strengthening and heat reduction have the same effect on the steady dimensionless control parameter, but not the same temperature change or switching dynamics. These stability statements use quasi-steady opening flow and a single well-mixed temperature; other stratification or airflow-inertia models require their own stability analysis.