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Sensible heat is transferred heat associated with a temperature change within a single thermodynamic phase. By contrast, latent heat can change the phase fraction while the temperature stays fixed during equilibrium phase coexistence. Neither name denotes an additional stored substance: both describe how transferred heat participates in a process.

For fixed mass $m$ and composition, use the heat capacity appropriate to the heating constraint. Under the quasistatic process assumptions in heat capacity, with only pressure-volume work and unchanged bulk kinetic energy and potential energy, fixed-pressure heating givesHere $c_p$ is specific heat capacity at constant pressure; fixed-volume heating instead uses specific heat capacity at constant volume $c_V$. A cooling step has negative $Q_{\mathrm{sens}}$ with the sign convention in heat. A temperature difference has the same numerical value in units of the kelvin or degree Celsius. The integral follows by adding the small transfers $\delta Q=m{c}_p(T)dT$, rather than assuming a constant heat capacity.

As a stylized ice-to-water example for a fixed-composition closed system at fixed pressure $1\mathrm{atm}$, take $m=0.5\mathrm{kg}$, an initial temperature of $-10{}^{\circ }\mathrm{C}$, melting at $0{}^{\circ }\mathrm{C}$, and a final temperature of $20{}^{\circ }\mathrm{C}$. Assume constant specific heat capacity at constant pressure values $c_{\mathrm{ice}}=2.1\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $c_{\mathrm{water}}=4.18\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and latent heat of fusion $L_f=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. These are stipulated, rounded model values; ignore the container and heat losses, retain all the material, and assume only pressure-volume work under the quasistatic process conditions above. The three steps areOnly the first and third terms are sensible heat; the middle term is latent heat. The plateau begins after $10.5\mathrm{kJ}$ and ends after $177.5\mathrm{kJ}$ of total supplied heat. At $94\mathrm{kJ}$ total input, half the mass is liquid water and the temperature remains $0{}^{\circ }\mathrm{C}$. The equilibrium phase transition temperature depends on pressure; this example stipulates a single fixed pressure. ^[ref]

This split makes heating budgets easy to interpret: melting consumes $167/219.3\approx 76.2\%$ of the supplied heat in this model despite causing no temperature rise. Within either single thermodynamic phase, the heating curve slope is $dT/dQ=1/(m{c}_p)$; a larger heat capacity makes the slope shallower. A flat interval represents changing phase fraction, so it cannot be modeled by applying $Q=m{c}_p\Delta T$ across the phase transition. The complementary constructed example in latent heat tracks that phase fraction explicitly.

A sphere of radius $R$ settling in a stationary cylindrical tube of radius $R+d$, with $d\ll R$ and zero net through-flow, has leading speedThe neck gap is $H(z)=d+z^2/(2R)$. In the sphere frame, leading bypass flux per circumferential length is $q\simeq UR/2$; wall-motion transport is smaller by $d/R$. The pressure-driven flux through a parabolic lubrication gap gives $\Delta p=9\pi \mu UR\sqrt{2R}/(4d^{5/2})$. Pressure dominates the sphere drag, $F\simeq \pi R^2\Delta p$, and balancing $F$ against $4\pi R^3\Delta \rho g/3$ proves the speed. Tube end conditions allowing appreciable net flow change the bypass balance and can give other drag regimes.

A shock polar is the locus of possible downstream velocity vectors, or an equivalent pressure/deflection relation, as the orientation of an oblique shock varies for fixed upstream conditions. Entropy admissibility selects physical compressive shocks. The strong-shock perfect-gas velocity polar is a circle, while finite-Mach polars have a different shape.

For a horizontal field, set $C=B^2/{\mu }_0$. Static force balance gives $\delta \Pi =\rho g{\xi }_z-(\gamma p+C)\nabla \cdot \xi +Ci{k}_y{\xi }_y$. Given smooth compactly supported vertical and along-field displacements, the displayed choice makes total-pressure perturbation zero while ${\xi }_x=O(k_x^{-1})$. The nonnegative pressure and transverse-tension terms vanish as $|k_x|\to \infty$. This constructs the infimum needed for the magnetohydrodynamic energy principle; a strictly negative limit is already negative at some finite wavenumber.

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

A simple fluid has local stress determined by the history of deformation of a material element, rather than by spatial derivatives of that history. An isotropic incompressible simple fluid has an arbitrary isotropic pressure and an objective constitutive functional of the relative deformation gradients. Internal-variable equations can represent such a functional when the initial material state is fixed by the past history.

For a gravity-current nose of height $h_N$ on a small incline of slope $\theta$, the travelling front of a gravity-driven thin film satisfies $h_y=\theta (1-h_N^2/h^2)$. Close to its dry tip, $h\sim (-3\theta h_N^{2}y{)}^{1/3}$. The slope becomes order one at $h\sim \sqrt{\theta }{h}_N$, giving an $O(\sqrt{\theta }{h}_N)$ region where lubrication theory fails. The main nose length is $h_N/\theta$. Comparing capillary pressure gradients $\gamma h_N/(h_N/\theta {)}^3$ with gravitational forcing $\rho g\theta$ gives the crossover $h_N\sim \theta \sqrt{\gamma /(\rho g)}$. This concerns the main nose length; capillarity may already matter in a much smaller tip region.

Two equal circular rings of radius $R$ in parallel planes at $z=\pm L$ provide a boundary for an open film. In the ideal axisymmetric model, a stationary connected film has the catenoid profileThe catenoid existence threshold for equal rings explains why this equation can have two solutions, one solution, or none. The cancellation of principal curvatures holds for both solutions whenever they exist; the mean curvature equation alone does not select a stable minimal surface.

This ideal geometry uses $R=1$ and $L/R=0.60$. The two profiles have different waist radii $a$, although both have $H=0$. For either profile the surface area follows by integration:At this illustrated separation both profiles have greater surface area than the two disconnected disks, whose total is $2\pi R^2$. This comparison disproves a claim of global area minimization in a class allowing disconnected competitors; it does not determine the sign of the second variation for small variations of a connected film.

A reported experiment used coaxial wire rings of radius $4\mathrm{cm}$, drawn from a surfactant solution and slowly separated until collapse. The solution contained tetradecyltrimethylammonium bromide in deionized water, with glycerol added to vary viscosity. Fluorescein enabled fluorescence imaging under cyan LED arrays; a Phantom V641 camera recorded 4,000 or 5,600 frames per second. Extracted contours gave neck radius, attachment angle, and inferred surface area. They showed neck narrowing and eventual breakup, rather than indefinite persistence of a stationary catenoid. Comparing solutions across viscosities helped distinguish liquid dissipation from the surrounding air's inertia, but the ring displacement beyond the last stable state was not precisely controlled. These are collapse observations, not direct measurements of equilibrium $H$. ^[ref]

A proposed classroom measurement would instead hold two equal rings coaxial on a translation stage, leave both sides open to the same gas, and photograph the settled silhouette beside a scale. Fit $r(z)=a\cosh (z/a)$ and inspect fit residuals before inferring cancellation of the principal curvatures. Repeat at several separations and with the camera rotated about the axis to test rotational symmetry. Keep solution composition, ring wetting, and elapsed drainage time comparable, and shield the film from drafts. Film weight, drainage, evaporation, surfactant gradients, and rupture can invalidate the ideal soap film pressure balance. A silhouette fit alone neither measures the pressure jump nor tests all possible perturbations of a stable minimal surface.

For an ideal soap film with equal constant surface tension $\gamma$ on both faces, the effective tension is $\tau =2\gamma$. Choose a unit normal $N$ from the minus side to the plus side, and $\Delta p=p_{-}-p_{+}$. Translating the Young–Laplace equation to $S=-dN$ givesA single interface uses $\tau =\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma /R$. The positive-convex curvature sum in the Young–Laplace equation article is $-2H$. ^[ref]

An open film with the same resting gas on both sides has zero pressure jump. Neglecting film weight, soap film pressure balance gives $H=0$: a minimal surface. A pressurized bubble instead has nonzero mean curvature. These statements assume equilibrium, constant surface tension, and negligible normal viscous stress. Surfactant gradients or motion require the fuller normal stress balance. ^[ref]

The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2{\int }_{M}HfdA$ by the first variation of area formula. For a closed interface oriented outward, $\delta V={\int }_{M}fdA$, so stationarity of $\tau A-\Delta pV$ gives $-2\tau H-\Delta p=0$. This is a first variation calculation; testing for a stable minimal surface needs the second variation, and deciding a global minimum needs a comparison of competitors.

Linear axisymmetric compressive modes of a barotropic closure of a razor-thin disk combine radial epicyclic restoration, softened self-gravity and pressure. The off-plane razor-thin Poisson kernel supplies the exponential gravity reduction. Here $k$ is the nonnegative radial wavenumber magnitude. The full linear system also has an axisymmetric geostrophic mode.

The solubility product is the equilibrium constant for dissolution of a specified solid into its dissolved ions. For a pure bulk AB solid at the chosen temperature, pressure and standard states,The pure bulk solid's thermodynamic activity is one in this convention. Coexisting solid and chemical solution at chemical equilibrium give this equality. If all available solid dissolves, the final chemical solution may instead be undersaturated; do not impose equality merely because a salt was initially added.^[ref]

Using the thermodynamic activity definition $a_i={\gamma }_i{c}_i/c^{\circ }$, with activity coefficients ${\gamma }_i$ and reference molar concentration $c^{\circ }$, gives the independent conversionThus $K_{\mathrm{sp}}$ is dimensionless, while the conditional concentration form $K_{\mathrm{sp},c}$ has units ${\mathrm{M}}^2$ for this one-to-one model. Treating $K_{\mathrm{sp},c}$ as constant requires a justified fixed or approximately fixed activity coefficient product. Changing ionic strength can change that product even at fixed temperature; changing temperature can change the equilibrium constant itself. Specify the solid thermodynamic phase too: different crystal polymorphs need not have the same solubility product.

Solute potential is the contribution to water potential from dissolved material, with pure water assigned zero. In the nearly incompressible description, ${\Psi }_s=(RT/{\bar{V}}_w)\ln a_w$. In a dilute ideal chemical solution, ${\Psi }_s\simeq -RT{c}_{\mathrm{osm}}$, where $c_{\mathrm{osm}}$ is the osmotic concentration of dissolved particles. Thus solute potential is negative while the corresponding osmotic pressure is positive: ${\Psi }_s\simeq -\Pi$. This sign difference distinguishes a contribution to water potential from the compensating mechanical pressure.^[ref]

For transfer of the same species $X$ from a gaseous reference state into a solvent $S$, define the molar solvation free energy byHere ${\mu }^{\circ }$ is a standard chemical potential. Negative solvation free energy means that transfer into the stated solvent lowers the Gibbs free energy relative to the stated gaseous reference. Include solute-solvent interactions, making room in the solvent, and its reorganization; counting only a favorable pair interaction omits part of the transfer.^[ref]

State the temperature, gaseous reference pressure, dissolved reference concentration or other standard state convention, and species identity. Reference-state changes alter the quoted solvation free energy. For example, with an ideal dilute thermodynamic activity $a=c/c^{\circ }$, replacing $c^{\circ }$ by $\alpha c^{\circ }$ while preserving the physical chemical potential requires ${\mu }^{\circ }$ to increase by $RT\ln \alpha$. Thus the corresponding solvation free energy increases by $RT\ln \alpha$ if the gaseous reference stays fixed. This follows directly from $\mu ={\mu }^{\circ }+RT\ln a$, with $R$ the molar gas constant.

At fixed temperature and pressure, compare fully specified initial and final states usingThe entropy change is for the whole selected system, including both solute and solvent. Dispersing a solute can increase accessible arrangements while orienting nearby solvent molecules restricts others. One picture of an ordered hydration shell cannot determine the sign of the total entropy change. A negative Gibbs free energy change favors the specified process; it does not determine its reaction rate.^[ref]

An illustrative enthalpy accounting path separates solute constituents, reorganizes the solvent, and establishes their interactions:This is a chosen decomposition between compatible states, not a measured sequence. If a gaseous-ion-to-chemical solution solvation term already includes solvent reorganization, do not add that reorganization a second time. Lattice enthalpy from solution calorimetry gives the alternative Hess's law route that combines lattice separation with the complete hydration of ions term.^[ref]

The invented inputs are $+100$ for separation and $+20$ for solvent reorganization, in $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Interaction terms of $-150$ in solvent A and $-90$ in solvent B give $\Delta H_A=-30$ and $\Delta H_B=+30$. Stipulated total entropy changes of $-50$ and $+150\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ give $-T\Delta S_A=+15$ and $-T\Delta S_B=-45\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at $300\mathrm{K}$. Both selected processes therefore have $\Delta G=-15\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. This deliberately constructed comparison shows why the signs of the enthalpy and Gibbs free energy changes need not agree; it does not predict either solvent's solubility.

For incremental dissolution at a particular composition, the relevant chemical potential difference also depends on thermodynamic activities. At chemical equilibrium that driving difference vanishes. A quoted standard Gibbs free energy change and the finite change between arbitrary mixtures answer different questions; use compatible standard states and the reaction quotient when discussing solubility.

The motivating calorimetry experiment is especially instructive. A weighed potassium chloride sample dissolves in a known amount of water in a platinum-lined, shielded calorimeter; a quartz-oscillator thermometer and measured electrical heating provide the readouts. Electrical calibration before and after the reaction, comparison with a resistance thermometer, and checks on residual water and particle size constrain errors. The inspected authored article reports positive enthalpy of solution at $298.15\mathrm{K}$ for the specified dilution, despite complete dissolution. This is a combined enthalpy balance, not an isolated solvation free energy or direct single-ion hydration of ions measurement. Shield tracking and final concentration matter; the existing lattice enthalpy from solution calorimetry discussion gives the numerical result and the conversion limits.^[ref]