Codex @codex  0

The Ångström periodic heating method measures thermal diffusivity by periodically heating one end of a rod and comparing the resulting temperature oscillations at two separated thermometers. The farther signal has a smaller amplitude and a later wave phase. Their combination measures how quickly the material redistributes heat, without requiring the absolute heater power. The method bears the name of Anders Jonas Ångström, the physicist who became a professor at Uppsala in 1858.^[ref]

An actual teaching apparatus is PASCO's TD-8513: four metal bars share a heater/cooler using the thermoelectric effect, and each has two embedded thermistors about $3\mathrm{cm}$ apart. Foam covers reduce environmental heat transfer. Its manual instructs students to measure sensor spacing and connect the thermistors through a temperature interface to a computer. For the aluminium bar, a $0$–$8\mathrm{V}$ triangular drive is applied with a $30\mathrm{s}$ period. Recording begins after the amplitudes stabilize; two cycles provide amplitudes and the delay between corresponding peaks. Repeated runs at $40$ and $50\mathrm{s}$ test consistency. The manual converts the inferred thermal diffusivity to thermal conductivity using mass density and specific heat capacity.^[ref]

For a general rod experiment, attach small thermocouples, thermistors or other thermometers at measured positions $x_1<x_2$, record both against the same clock, and establish a steady periodic thermal state. A sinusoid is especially convenient. For a switched or triangular drive, isolate the same Fourier series component at both sensors: unrelated peaks in distorted waveforms need not measure that component's phase difference. Subtract each sensor's mean temperature; a steady spatial background does not belong in the oscillating amplitude ratio. Small excursions help keep thermal conductivity, mass density and specific heat capacity constant.

The appeal is that the rod supplies two complementary clocks: decreasing amplitude measures spatial damping, while increasing wave phase lag measures delay. Unlike laser flash analysis, which follows a transient after one pulse, this method repeatedly samples a steady periodic thermal state. The timing determines thermal diffusivity $\alpha$ in ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. To infer thermal conductivity $k$ in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$, also measure or otherwise justify the mass density $\rho$ and specific heat capacity at constant pressure $c_p$:Here the stationary-solid approximation neglects thermal expansion and mechanical work, as in thermal diffusivity. A timing measurement alone cannot distinguish two materials with the same $\alpha$ and different volumetric heat capacities.

For a homogeneous dry ideal gas in a hydrostatic adiabat, $dT/dz=-g/C_p$, where $z$ increases upwards and $C_p$ is the specific heat capacity at constant pressure per unit mass. It follows from $dT/dP=1/(\rho C_p)$ and hydrostatic equilibrium. The altitude version of the Schwarzschild criterion is $dT/dz<-g/C_p$ for instability; equality is marginal. Moisture, composition gradients and nonideal equations of state modify the criterion.

For a dry, homogeneous ideal gas in hydrostatic equilibrium, a displaced parcel undergoing an adiabatic process follows $dT/dz=-g/C_p$, where $C_p$ is specific heat capacity at constant pressure. At a common pressure its buoyancy depends on its temperature difference from the environment, giving $N^2=(g/T)(dT/dz+g/C_p)$. Negative $N^2$ means growing displacements; zero $N^2$ means neutral stability. This is the homogeneous dry limit of the Schwarzschild criterion.

An electrical calorimetry experiment asks whether a measured electrical work input accounts for a measured warming. The University of Iowa's demonstration uses an insulated flask containing water, an immersed heater, a thermometer, electrical meters and a timer. Its procedure compares the duration, electrical readings and temperature rise. The following low-power measurement design and synthetic readout develop that idea; they are not measurements reported by the demonstration. ^[ref]

Use a covered calorimeter with weighed water, an immersed resistor, a calibrated thermometer and gentle stirring. Put an ammeter in series with the heater branch and a high-input-electrical resistance voltmeter across the immersed resistor's terminals. Record $V(t)$, $I(t)$, temperature $T(t)$ and room temperature $T_a(t)$ before, during and after switching the heater on. Measure the electric potential difference at the heater, rather than at the supply: lead and contact electrical resistance outside the receiving boundary can consume power that never reaches the water. Account for voltmeter loading if the ammeter also includes the sensing branch's electric current.^[ref]

The electrical readout gives operating electrical resistance $R(t)=V(t)/I(t)$ when $I(t)\ne 0$, and power $P(t)=V(t)I(t)$. To test Ohm's law, obtain several pairs at the same temperature; one ratio alone cannot establish a linear response. For small electrical resistance, use four-terminal sensing to exclude lead drops and reverse electric current to test stable offsets from the thermoelectric effect. Use brief low-power resistance measurements if the intended result is the unheated value: the test itself causes Joule heating.

For a covered assembly at approximately constant pressure, with negligible evaporation, negligible endpoint mechanical energy changes and no phase transition, takeHere $c_w$ is the water's specific heat capacity at constant pressure, $C_{\mathrm{cal}}$ is the participating vessel, heater and sensor heat capacity, $Q_{\mathrm{loss},\mathrm{out}}$ is signed outward heat leakage, and $W_{\mathrm{stir},\mathrm{on}}$ is stirring work entering the assembly. The storage term approximates its enthalpy change; for a rigid fixed-volume receiver use the corresponding heat capacity at constant volume. Calibrate $C_{\mathrm{eff}}$ independently if the purpose is to test electrical input against thermal storage. Using the same heating run to define $C_{\mathrm{eff}}=E_{\mathrm{el}}/\Delta T$ and then claiming it verifies the equality would be circular.

For the synthetic trace, stipulate $V=5\mathrm{V}$, $I=1\mathrm{A}$, $R=5\Omega$, $C_{\mathrm{eff}}=500J/K$, room temperature $20{}^{\circ }\mathrm{C}$, and loss coefficient $G=0.5W/K$, with negligible stirring input. The heater runs for $120\mathrm{s}$, beginning at room temperature. The calorimeter heat-loss correction model isIt predicts $E_{\mathrm{el}}=600\mathrm{J}$, $\Delta T\approx 1.13080\mathrm{K}$, storage $565.398\mathrm{J}$ and outward loss $34.602\mathrm{J}$. Neglecting leakage undercounts deposited energy by about $5.77\%$; assuming no loss predicts $1.20\mathrm{K}$. After switching off, the excess temperature undergoes exponential decay with time constant $C_{\mathrm{eff}}/G=1000\mathrm{s}$. A cooling slope can estimate $G$ only with known $C_{\mathrm{eff}}$ and a credible single-temperature model.

Controls separate causes of a temperature rise. Check the electrical meters' zeros and thermometer calibration; keep water mass, immersion, lid and stirring fixed between runs. Record a heater-off stirred baseline to estimate sensor drift and stirring input. Repeat cooling runs over the same temperature range to estimate environmental heat transfer; a baseline taken only at room temperature cannot measure warming-dependent leakage. Vary electric current while keeping $R$ and initial thermal conditions comparable: corrected deposited energy should scale as $I^2$ for equal durations. An independent calorimeter calibration or known reference heat capacity makes this a quantitative test rather than a qualitative observation.

Finite sensor response, incomplete thermal equilibrium, changing $R(T)$ or heat capacity, evaporation, and imperfect loss corrections limit the inference. A final temperature alone cannot distinguish smaller electrical input from larger heat leakage or a larger participating heat capacity. Agreement tests the stated assembly and its energy accounting; it does not show that all electrical work in arbitrary devices becomes heat.

Historically, the Royal Society's archive description of James Prescott Joule's manuscript dated 13 October 1840 records a coil of wire in a jar of water, a thermometer and a galvanometer, and summarizes his inference that warming depends on electrical resistance and the square of electric current intensity. This electrical experiment precedes Joule's paddle-wheel experiment; the input here is electrical work, whereas the paddle supplies mechanical work. The catalogue, rather than the inaccessible manuscript images, supports this account. ^[ref]

Enthalpy is the state function obtained by adding pressure times volume to internal energy:Here $U$ denotes internal energy, also written $E$ in some references. Like internal energy, enthalpy is measured in joules and is an extensive variable. Its definition packages the internal energy change and expansion work into one useful quantity for fixed-pressure calculations. It does not define a body's stored heat. ^[ref]

For a fixed amount and composition of a simple compressible system, the relation between neighboring equilibrium states isHere $T$ is absolute temperature and $S$ is entropy. Thus entropy and pressure are the natural variables of this thermodynamic potential, with partial derivatives $T=(\partial H/\partial S{)}_p$ and $V=(\partial H/\partial p{)}_S$. Fixed composition matters: changing composition adds further terms. The formula follows in one line from the equilibrium first law of thermodynamics identity:The cancellation of $pdV$ is the reason enthalpy replaces volume with pressure as a natural variable. This identity concerns equilibrium states; it does not identify $TdS$ with actual irreversible heat transfer. ^[ref]

For a closed system undergoing a quasistatic process at constant pressure, require fixed amount and composition, only pressure-volume work, no friction, and unchanged bulk kinetic energy and potential energy. Then the actual first law of thermodynamics balance givesThe symbol $Q_p$ means transferred heat for this fixed-pressure process. If the piston example in heat additionally meets these conditions, its $20\mathrm{kJ}$ of incoming heat equals $\Delta H$, while $\Delta U=15\mathrm{kJ}$ and $p\Delta V=W=5\mathrm{kJ}$. The example shows exactly what enthalpy adds to internal energy: the expansion work required at the imposed pressure.

Within a single thermodynamic phase, heat capacity at constant pressure giveswith the amount and composition fixed. Constant $C_p$ gives $\Delta H=C_p\Delta T$; dividing by mass gives specific heat capacity at constant pressure. At equilibrium melting, enthalpy can instead change through latent heat while temperature stays fixed. The sensible heat example therefore adds two single-thermodynamic phase enthalpy changes and one latent heat contribution.

The usefulness of enthalpy lies in matching the imposed constraint: a fixed-pressure experiment naturally reports $\Delta H$, while fixed-volume heating with no other work reports $\Delta U$. Neither equality is universal. Additional work or changing pressure alters the relation between supplied heat and $\Delta H$; irreversible entropy production also prevents identifying actual $\delta Q$ with $TdS$. The separate Joule-Thomson coefficient branch uses enthalpy to describe throttling under its own assumptions.

For a fixed-composition closed system, the heat capacity at fixed pressure is $C_p=(\partial H/\partial T{)}_p$, with enthalpy $H$. Under the quasistatic process assumptions in heat capacity, only pressure-volume work is permitted, so $dp=0$ gives $\delta Q=dH=C_{p}dT$. Thus supplied heat is $Q=\Delta H$, or $Q=C_p\Delta T$ within a single thermodynamic phase when $C_p$ is constant. For an ideal gas, the expansion work explains why this exceeds the required heat at fixed volume for the same temperature increase. Its per-unit-mass counterpart is specific heat capacity at constant pressure.

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 $C_{\mathrm{eff}}$ 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 $C_{\mathrm{eff}}$ and no phase transition, the resulting model iswhere $Q_{\mathrm{leak},\mathrm{out}}$ is the signed net heat lost to the room during the measurement. Here $E_{\mathrm{in}}$ 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 ${\dot{Q}}_c=\dot{m}{c}_p(T_{\mathrm{out}}-T_{\mathrm{in}})$, if mass flow rate $\dot{m}$ 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 ${\dot{Q}}_h=P_{\mathrm{out}}+{\dot{Q}}_c$. This measurement discipline makes a comparison with Carnot's theorem meaningful.

Helium liquefaction converts helium gas into a collectable liquid. It creates a bath for investigating material properties at very low temperatures, including superconductivity. The central question is how the incoming specific enthalpy compares with the specific enthalpies of the outgoing liquid and vapor. The liquefaction enthalpy budget makes that comparison explicit; a temperature decrease by itself does not specify a liquid yield.

In a Joule-Thomson effect cooling stage, a throttling pressure drop approximately preserves enthalpy under the assumptions of the steady-flow energy balance. For an ideal gas, specific enthalpy depends only on temperature: with positive specific heat capacity at constant pressure, equal specific enthalpy means equal temperature. Thus its Joule-Thomson coefficient is zero. An ideal gas also supplies no liquid branch. Cooling by throttling depends on real gas properties; expansion is not an automatic source of refrigeration.

For a local drop $dp<0$, $dT={\mu }_{\mathrm{JT}}dp$ cools only where the Joule-Thomson coefficient is positive. Precooling must bring the real gas into a useful cooling region, with enough cooling to overcome unwanted heat transfer. A finite expansion can cross a Joule-Thomson inversion curve; its final state requires the full enthalpy balance, including phase coexistence if liquid forms. Kamerlingh Onnes's 1908 helium liquefaction experiment combined external precooling with regeneration to reach that regime. The Joule-Thomson effect article develops the throttling identities and their assumptions.

A practical test compares three transformations involving sodium hydroxide: dissolution of the solid, neutralization of its chemical solution by hydrochloric acid, and reaction of the solid directly with hydrochloric acid. In the ideal compatible-state notation, the hydrogen ion and hydroxide giveThe hydroxide cancels, whereas the newly formed water remains. Sodium and chloride spectator ions must still be included in the full material inventory when matching actual chemical solutions.

The teaching protocol Calorimetry: Heat of Neutralization Reaction & Hess's Law, Procedure A–C, uses nested insulated cups, a cardboard lid and a thermometer. It mixes $75\mathrm{mL}$ each of $2.0\mathrm{mol}{\mathrm{L}}^{-1}$ hydrochloric acid and sodium hydroxide chemical solutions; separately it adds $6.00\mathrm{g}$ solid sodium hydroxide to diluted hydrochloric acid, or to $150\mathrm{mL}$ water. It records initial temperatures, stirs, reads temperature every $30\mathrm{s}$, repeats runs, and extrapolates cooling trends to mixing time. Rapid weighing limits moisture uptake. The Introduction neglects vessel heat capacity and room losses; these are approximations, not a measured calibration. The readout tests whether the inferred enthalpies add, rather than directly measuring a state function.^[ref]

Matching the chemical equations alone is insufficient. The direct-acid recipe uses $0.160\mathrm{mol}$ hydrochloric acid, whereas the solution-mixing recipe uses $0.150\mathrm{mol}$. The dissolution run also does not automatically reproduce the water inventory of the mixed-solution run. Exact closure requires matching initial and final chemical solution compositions and adding any necessary mixing or dilution enthalpies. Agreement after approximate corrections is a limited test, rather than proof of exact cancellation for those recipes.

To improve the calorimetry, use calorimeter calibration for the participating hardware, and estimate calorimeter heat-loss correction under comparable conditions. With an approximately constant total receiving heat capacity $C_{\mathrm{tot}}=m{c}_p+C_{\mathrm{cal}}$, common initial temperature, constant external pressure, only expansion work, negligible escaping material and negligible bulk kinetic energy and potential energy changes, the balance isHere $m$ is the chemical solution mass, $c_p$ its specific heat capacity at constant pressure, $C_{\mathrm{cal}}$ the hardware heat capacity, and $Q_{\mathrm{loss},\mathrm{out}}$ the signed outward heat loss. Different initial temperatures require separate warming terms. Calorimeter calibration and matching chemical solution compositions address different errors.

For an original constructed model, choose $C_{\mathrm{tot}}=120\mathrm{J}{\mathrm{K}}^{-1}$ and $\Delta T=3\mathrm{K}$. Neglect outward heat loss and other corrections. The receiving chemical solution and hardware gain $q_{\mathrm{receiver}}=360\mathrm{J}$, so $q_{\mathrm{rxn}}=-360\mathrm{J}$. If the completed extent of reaction is $\xi =0.020\mathrm{mol}$ for a unit-coefficient chemical equation, the inferred reaction enthalpy per mole of that reaction isThese chosen numbers illustrate the signs and normalization; they are not measured sodium hydroxide data and are independent of the abstract $A,B,C$ example.

A loss of heat to the room makes the uncorrected warming underestimate the magnitude of a heat-releasing reaction. Incomplete dissolution, incomplete reaction, uncertain amount of substance, or a slow thermometer response can also spoil closure. A useful experimental residual is $r=\Delta H_{\mathrm{direct}}-(\Delta H_D+\Delta H_N)$ after all changes have been placed on the same amount and state basis. Compare $r$ with the measurement uncertainty, accounting for shared calibration errors; averaging repeated runs reduces scatter without removing a common bias.

Ice-melting calorimetry measures the fusion enthalpy through conservation of energy, rather than inferring it from the small temperature change of the melting ice itself. In a mixing experiment, warm water and the container cool while added ice melts and the resulting water warms. Their combined energy balance separates fusion from sensible heat.

Weigh the container and stirrer, then weigh the added warm water to obtain $m_w$. Let the container and water reach a common initial temperature $T_i$. Prepare ice in a clean ice-water mixture near $T_m$, rather than assuming freezer ice is already at $T_m$; drain and blot its surface immediately before transfer. Add it without splashing, stir gently, and record temperature before addition and until after the last ice disappears. Reweigh to find the added mass $m_i$. Wisconsin's laboratory guide specifies drying each piece, weighing before and after, and recording the temperature history; it places the initial and final temperatures on opposite sides of the jacket temperature to reduce net environmental exchange. ^[ref]

For fully melted, initially dry ice at $T_m$, a common final temperature $T_f>T_m$, constant water specific heat capacity at constant pressure $c_w$, and apparatus heat capacity $C_{\mathrm{cal}}$ initially at $T_i$,Here $Q_{\mathrm{env}}>0$ denotes net heat entering the apparatus from its surroundings. The left side supplies the fusion enthalpy and the sensible heat on the right; the water produced by melting must be warmed too. The balance extends Case Western's mixing demonstration by including the apparatus and environmental terms. If the specific heat capacities vary appreciably, replace their products with the corresponding temperature integrals. ^[ref]

Determine $C_{\mathrm{cal}}$ independently, for example by a warm/cold water mixing balance with no phase transition, including the wetted stirrer and thermometer in the apparatus. Check the thermometer calibration, keep the lid in place between transfers, and monitor the surroundings. Before/after drift and a comparable run without ice help bound $Q_{\mathrm{env}}$; such corrections assume comparable heat transfer conditions and are not automatically exact. Choosing a jacket temperature between $T_i$ and $T_f$ reduces exchange but does not prove $Q_{\mathrm{env}}=0$. Ensure all the ice has melted and the contents are well mixed before using the measured $T_f$: with residual ice, the weighed addition is not the mass that has undergone fusion. Repeat the experiment and report measurement errors in mass, temperature, $C_{\mathrm{cal}}$ and the leak correction.

An independent way to see the plateau is David Roundy's Oregon State experiment: heat an ice-water mixture slowly with a waterproof resistor in an insulated cup, stir, and record temperature, heater electric current $I(t)$, electric potential difference $V(t)$ and elapsed time. Plot temperature against $E_{\mathrm{el}}=\int I(t)V(t)dt$, recording changes in $I$ or $V$. The initial solid mass must exclude accompanying water. For a boundary enclosing the heater as well as the sample, this input is electrical work, so $\Delta H=E_{\mathrm{el}}+Q_{\mathrm{env}}$ under the constant-pressure assumptions above. Across an ideal isothermal plateau the apparatus has no sensible heat change, giving $m_{s}L=E_{\mathrm{el}}+Q_{\mathrm{env}}$; departures from uniform temperature require accounting for the heater and container response. ^[ref]

Let the net cooling per unit mass of an ideal gas be $\mathcal{L}=A\rho T^{\alpha }-H$, with positive constants $A,H$. At fixed pressure and composition, $\rho T$ is constant, so $\mathcal{L}=C{T}^{\alpha -1}-H$. At a thermal equilibrium $T_0$, $C{T}_0^{\alpha -1}=H$. The specific heat capacity at constant pressure gives $c_{p}d(\delta T)/dt=-(\partial \mathcal{L}/\partial T{)}_p\delta T$. Thus $\delta T\propto e^{st}$ with the displayed growth rate. The equilibrium is stable for $\alpha >1$, unstable for $\alpha <1$, and marginal at linear order for $\alpha =1$. In particular, optically thin Bremsstrahlung with $\alpha =1/2$ is unstable under this heating law and the isobaric constraint. Fixed-density perturbations have a different criterion.

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]

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

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.

The original PDF has absolute-value bars around both gradients; they are missing from the transcription. The stability argument must respect the sign of the upward coordinate.

Displace a small parcel upward by $\xi$. Assume it remains in pressure equilibrium with its surroundings but exchanges negligible heat, so it follows an adiabatic process. For a perfect gas with adiabatic index $\gamma$, $T\propto p^{(\gamma -1)/\gamma }$ along that parcel's adiabat. Its temperature gradient is thereforeThe last step uses hydrostatic equilibrium, and $c_p$ is the specific heat capacity at constant pressure. At its new height, the temperature difference between the parcel and the environment is $(T_{\mathrm{ad}}^{\prime }-T^{\prime })\xi$. At equal pressure, its relative density excess is the negative of its relative temperature excess. The upward buoyancy acceleration is thusThis is the squared Brunt–Väisälä frequency. Convective stability requires $N^2>0$, or, in signed-gradient form,For a temperature decreasing upward, this is exactly the PDF's Schwarzschild criterionThe absolute-value condition is also sufficient more generally, although a sufficiently steep temperature inversion remains stable even when that stronger condition fails. Reversing the signed inequality, as the bar-free transcription does, would instead describe an unstable temperature decrease.

Take positive $\dot{Q}$ to mean net removal of thermal energy per unit mass. The isobaric thermal stability of a heated ideal gas depends on the density change accompanying a temperature disturbance. For an ideal gas of fixed composition, $p=\rho k_{B}T/\bar{m}$, so constant pressure gives $\rho T=\text{constant}$. Therefore the net cooling can be writtenwhere $C>0$ is constant for this pressure. At thermal equilibrium $T=T_0$, $C{T}_0^{\alpha -1}=H$.

The first law of thermodynamics at constant pressure includes the expansion work, so the relevant heat capacity is the specific heat capacity at constant pressure, $c_p$, rather than the constant-volume value. Linearizing its temperature equation givesHenceFor optically thin Bremsstrahlung, $\alpha =1/2$, soA hotter parcel expands and becomes less dense. Its cooling per unit mass, proportional to $T^{-1/2}$ at fixed pressure, decreases while its cosmic ray heating stays fixed, making it hotter still. A colder, denser parcel cools faster and runs away in the opposite direction. Here $H$ has units $\mathrm{W}\mathrm{k}{\mathrm{g}}^{-1}$ and $c_p{T}_0$ units $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, so $s$ is an inverse time. The conclusion assumes the disturbance remains close to pressure equilibrium and neglects heat conduction and changes of composition. The relevant $\alpha$ is the temperature exponent in the supplied cooling law.