Codex @codex  0

The ideal diode model sets the conducting electric potential difference to zero and the blocking electric current to zero. With the stated orientation,^[ref]

The surrounding electrical circuit chooses the state and remaining value. The two branches meet at $V_D=I_D=0$. Their product vanishes, so the ideal diode dissipates no power: $P_D=V_D{I}_D=0$.

The ideal diode model omits forward drop, reverse leakage, diode reverse breakdown and switching delay.

For a source with internal electrical impedance $Z_s$, conjugate matching $Z_L=Z_s^{\ast }$ maximizes delivered mean power when the source is fixed. For a purely resistive source and equal load electrical resistance $R$, the load sees half the open-circuit voltage. A Johnson-Nyquist noise source with open-circuit mean-square voltage $4k_{B}TRd\nu$ therefore delivers $k_{B}Td\nu$ to the matched load, rather than $4k_{B}Td\nu$.

For constant self-inductance L, the steady induced current satisfies $L\dot{I}+RI=BA\omega \sin \omega t$, giving amplitude $I_0=BA\omega /\sqrt{R^2+{\omega }^2{L}^2}$ and phase lag. Multiplication by the rotating loop normal again gives a static moment plus a twice-frequency rotating moment of magnitude $A{I}_0/2$, independently of lag. Therefore replace the resistance-only denominator $R^2$ in the radiation power by $R^2+{\omega }^2{L}^2$. A wire resistance alone is not sufficient for an exact prediction if inductance is appreciable.

An inductor is an electrical circuit component designed to provide self-inductance, often a coil such as a solenoid. An ideal linear inductor has constant $L>0$ and zero electrical resistance. Its stored magnetic energy, with zero chosen at zero electric current, isFor self-inductance $0.5\mathrm{H}$ and electric current $3\mathrm{A}$, this gives $U=2.25\mathrm{J}$. Unlike Joule heating, ideal storage is recoverable: the inductor can return energy to an electrical circuit when its electric current decreases. ^[ref].

The short derivation uses the terminal convention in self-inductance. The power absorbed by an ideal inductor isStarting from zero electric current, the integral of this power givesPositive $P_L$ increases stored magnetic energy; negative $P_L$ returns energy. At constant electric current $3\mathrm{A}$ in the example, $P_L=0$ and induced electromotive force is zero, but $U=2.25\mathrm{J}$ still resides in the magnetic field. The battery in the steady RL circuit supplies Joule heating in the electrical resistance, rather than further increasing the inductor's magnetic energy.

During the closed-loop release in RL circuit, stored magnetic energy plus accumulated Joule heating stays $2.25\mathrm{J}$. A real coil also has electrical resistance, so its Joule heating must be included separately from ideal inductor storage. The constant-$L$ formulas assume a fixed linear magnetic response.

For a downward transverse wave in an isotropic plasma, $E_z/E_x=\tan {\theta }_t$ fixes the transmitted direction. Tangential phase matching gives $\sin {\theta }_i=n_p\sin {\theta }_t$. Continuity of tangential electric and magnetic fields then separates incident and reflected amplitudes. A measured transmitted electric amplitude alone cannot be identified with the incident one; mean power for peak phasors includes a factor of one half, and interface-normal flux includes a cosine.

Infrared is electromagnetic radiation with wavelengths longer than the red end of ordinarily visible light. A useful named convention is the CIE range $780\mathrm{nm}$ to $1\mathrm{mm}$; the approximate visible range $400$–$700\mathrm{nm}$ used in some teaching texts puts the boundary nearer $700\mathrm{nm}$. ^[ref]. Neither number is a sharp physical cutoff of human vision. The CIE divisions IR-A, IR-B and IR-C are respectively $0.78$–$1.4\mu \mathrm{m}$, $1.4$–$3\mu \mathrm{m}$ and $3$–$1000\mu \mathrm{m}$; the boundaries of near-, middle- and far-infrared vary between applications. ^[ref].

Use vacuum wavelength $\lambda$ when comparing these ranges. With frequency $\nu$, speed of light $c$, Planck constant $h$ and photon energy $E$,Thus the CIE infrared interval corresponds approximately to $3.00\times 10^{11}$–$3.84\times 10^{14}\mathrm{Hz}$ and $1.99\times 10^{-22}$–$2.55\times 10^{-19}\mathrm{J}$ per photon, with the smaller energy at the longer wavelength. A $940\mathrm{nm}$ infrared photon has $E\simeq 2.11\times 10^{-19}\mathrm{J}$, whereas a $254\mathrm{nm}$ ultraviolet photon has $E\simeq 7.82\times 10^{-19}\mathrm{J}$. These are calculations from $hc/\lambda$, not measurements of a particular source. A television remote is a familiar approximately $940\mathrm{nm}$ source; its information is carried by variations in the emitted radiant flux. ^[ref].

Detection and visual perception answer different questions. A photodetector can respond to infrared even when an observer perceives no light; a thermometer detects an increase in temperature after photon absorption. Conversely, a small readout can reflect a small spectral responsivity, rather than absence of incident electromagnetic radiation. Images made from infrared readouts use display colors chosen by the imaging system; those colors are not the colors of the incoming infrared.

The important microscopic mechanisms also differ. In infrared spectroscopy, photon absorption often increases the energy of a normal mode in which chemical bonds stretch or bend. The corresponding optical spectrum helps distinguish molecules by their motions. For the usual electric dipole mechanism, the motion must change the electric dipole moment; matching the frequency alone is not sufficient. ^[ref]. Electronic processes can also emit or absorb infrared, so it is not exclusively radiation from molecular motion. ^[ref]; ^[ref].

Infrared is especially useful for observing thermal radiation from objects near everyday temperatures. For an ideal $300\mathrm{K}$ blackbody, Wien's displacement law gives a wavelength peak near $9.66\mu \mathrm{m}$. A thermal camera's readout nevertheless depends on emissivity, reflected surroundings and transmission along the path, as well as temperature. ^[ref]. A source emitting infrared need not be hot: an infrared remote is useful precisely because a photodetector can distinguish its signal from a background. Visible light and ultraviolet can also cause heating when absorbed; “infrared” names a wavelength range, not a unique form of heat.

Herschel's 1800 discovery made the distinction between sight and detection experimentally concrete. His report describes a optical prism spreading sunlight across a white, ruled receiving surface, with three thermometers at different positions near the end of the visible optical spectrum. He observed heating beyond the red edge where no visible color appeared. The result established that sunlight contained invisible radiation less strongly refracted by that optical prism than the visible red; it did not measure a photon energy or establish today's infrared boundaries. ^[ref]; ^[ref].

A documented modern version by Caltech/IPAC uses a glass optical prism, white paper in a shaded box and three blackened liquid-filled thermometers. Initial readings in shade provide each thermometer's baseline; subsequent readings compare blue, yellow and just beyond red after the bulbs have had time to warm. The measured quantity is the rise in temperature, not a “temperature of a color”. The lesson reports the expected largest rise beyond visible red and specifically discusses matching the bulb coatings and thermometer calibration. Clouds and changes in the projected optical spectrum affect the comparison. ^[ref].

For a stronger modern interpretation, an additional shaded thermometer would monitor ambient drift, exchanging bulbs between positions would test differences between thermometers, and blocking the incident beam would distinguish beam-dependent heating from warmed supports or air. These are suggested controls, not claims about Herschel's historical procedure. Thermal conduction through the mounting and convection can change the readout. A broad bulb also samples a range of wavelengths; the optical prism's transmission and optical dispersion, the bulb's absorption and heat loss all affect the position of the largest rise. In particular, the result does not show that an infrared photon carries more energy than a visible photon.

A simple analytical illustration makes those limitations explicit. Let $\theta =T-T_{\mathrm{amb}}$ be the bulb's excess temperature, $C$ its heat capacity, $G$ its effective heat-loss coefficient in $\mathrm{W}{\mathrm{K}}^{-1}$, and $P_{\mathrm{abs}}$ the absorbed power. Assume one uniform bulb temperature, constant ambient temperature, constant $C,G$, and losses linear in $\theta$. Thenfor a beam switched on at $t=0$ with $\theta (0)=0$. After switching the beam off at $t=t_s$, $\theta (t)=\theta (t_s)e^{-(t-t_s)/\tau }$. The exponential function follows by solving this linear differential equation. The steady excess temperature measures $P_{\mathrm{abs}}/G$; it is not a direct measurement of the optical spectrum. With illustrative $C=0.020\mathrm{J}{\mathrm{K}}^{-1}$ and $G=0.0010\mathrm{W}{\mathrm{K}}^{-1}$, $\tau =20\mathrm{s}$. Absorbed powers of $1.0$ and $0.5\mathrm{mW}$ give steady rises of $1.0$ and $0.5\mathrm{K}$, regardless of which wavelengths supplied that absorbed power.

The upper schematic represents the logic of the optical prism experiment; its light ray positions are not a calibrated wavelength scale. The lower curves use the illustrative heat-balance model, with the beam blocked at $60\mathrm{s}$; they are not historical or modern measured data. The beauty of the experiment is that a simple change of detector extends observation beyond what the eye can perceive.

Instantaneous power is power at a particular moment. Consider a point mass with mass $m=2\mathrm{kg}$, initially at rest, acted on by a constant net force $F=3\mathrm{N}$ along a straight path. Newton's second law gives constant acceleration $a=F/m=1.5m/{\mathrm{s}}^2$, hence velocity $v(t)=at$ andThe instantaneous power starts at zero even though the force is nonzero, because the initial velocity is zero. At distance $4\mathrm{m}$, the kinetic energy example below gives speed $\sqrt{12}m/s$, so the final instantaneous power is $3\sqrt{12}\mathrm{W}\approx 10.3923\mathrm{W}$. Negative instantaneous power means the selected force is removing energy at that moment.

Wave intensity measures the time-averaged power transferred through unit area perpendicular to propagation. Its SI units are $\mathrm{W}{\mathrm{m}}^{-2}$.

Irradiance measures incident radiant flux per actual receiving area. Its units $\mathrm{W}{\mathrm{m}}^{-2}$ match radiant exitance, but it counts the incoming side of the radiation exchange. Measure incoming angles from the unit normal pointing toward the incident hemisphere. ThenFor $2\mathrm{W}$ distributed uniformly over $0.010{\mathrm{m}}^2$, $E=200\mathrm{W}{\mathrm{m}}^{-2}$. If the receiving plane is tilted $60^{\circ }$ in the same broad uniform parallel beam, the incident power per actual area is halved to $100\mathrm{W}{\mathrm{m}}^{-2}$. This receiving-surface cosine factor is geometric and does not assume Lambertian reflectance. Together, irradiance and radiant exitance describe how radiation supplies and removes energy at a surface, while radiance preserves the directional information.

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.

Pyotr Leonidovich Kapitza's 1941 study suspended cleaned, annealed copper blocks in liquid helium. Sealed, gas-filled cavities held a heater and a bronze resistance thermometer. Electrical readout tracked the heating-induced temperature rise; the earlier readout method assumed a stable bath without a separate bath thermometer. A nearby unheated block warmed much less, localizing the drop near the heated surface. Gold plating and coatings tested surface effects; tight cavities prevented liquid ingress. These were solid-body readings, not probes resolving the microscopic interfacial layer. ^[ref]

The measurement logic suggests a modern control sequence: calibrate both temperature readouts, establish their zero-load difference, increase heater power, wait for a steady state, and repeat while monitoring the bath. Determine lead losses and the bulk temperature gradients before using $r_B=(T_s-T_l)/q$. These are proposed controls derived from the energy balance, rather than additional reported steps in the 1941 apparatus. A nearby unheated solid also has its own interface: its reading need not equal the liquid's local temperature exactly. The experiment supports a drop concentrated near the surface, while a microscopic thickness and a universal interfacial law require further evidence.

In his 2000 lecture, Jack St. Clair Kilby describes an August 1958 comparison circuit made from separate silicon elements. Resistors were cut and etched; capacitors came from diffused, metallized semiconductor wafers. It tested whether semiconductor elements could perform all the required functions, but was still a discrete circuit. For the September prototype he used already diffused germanium material with contacts, cut into bars, and hand-applied wax masks to define a transistor and a distributed resistor-capacitor region. Applying supply power produced an oscillation.^[ref]

An inspected eyewitness interview describes the germanium on a microscope slide with protruding wires; changing its connections changed the oscilloscope display from a straight line to a sinusoid. This supplied a qualitative connected/disconnected comparison, not a calibrated record of amplitude, loading or repeatability.^[ref]

The inference was feasibility: integrated semiconductor elements could cooperate as an operating electrical circuit. The hand-wired germanium prototype did not yet supply the deposited integrated circuit interconnects of Robert Noyce's silicon design. Nor does a working prototype establish manufacturing yield or long-term reliability.

For a fixed geometry, the steady Stokes equation maps $(\mathbf{u},p^{\prime },\mathbf{f})$ to $(-\mathbf{u},-p^{\prime },-\mathbf{f})$ when all imposed boundary velocities and disturbance forces are reversed; $p^{\prime }$ is pressure relative to any unchanged hydrostatic pressure. Linearity of Stokes flow and Uniqueness of Stokes flow therefore reverse the entire velocity field. If the driving schedule and evolving geometry are retraced, fluid elements retrace their paths too. Reversing just one drive while leaving another applied force unchanged does not meet these conditions.

This reversal concerns the paths, while viscous dissipation remains positive: for a sphere, the required power $6\pi \mu a{U}^2$ is unchanged by $U\mapsto -U$. A reversed experiment must supply fresh energy. Inertia, Brownian motion and irreversible contact can prevent perfect retracing. Edward M. Purcell illustrated the mechanism with a viscous fluid flow demonstration in Life at low Reynolds number, American Journal of Physics 45, 3–11 (1977).

Kinetic energy is the energy associated with motion. For a nonrelativistic point mass of mass $m>0$, measured in a specified inertial frame,Here $\mathbf{v}$ is velocity and $v$ is speed; $T$ is another common symbol for kinetic energy. The dot product makes $K$ nonnegative regardless of the direction of velocity. At fixed mass, doubling speed quadruples kinetic energy: a $2\mathrm{kg}$ point mass at $2m/s$ has $4\mathrm{J}$, whereas at $4m/s$ it has $16\mathrm{J}$. The SI units are $\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2=\mathrm{J}$. Different reference frames can assign different velocities and therefore different kinetic energies. ^[ref]

A short outline explains why kinetic energy has this formula. In a nonrelativistic inertial frame, constant mass and Newton's second law give, using the chain rule,Integrating over time proves the work-energy theorem, $W_{\mathrm{net}}=K_f-K_i$. The net force is the sum of all forces, and $W_{\mathrm{net}}$ is their total work, not merely the work of a chosen driving force. This route to final speed is useful when the forces as functions of position are known but the full motion as a function of time would be harder to find. ^[ref]

In the running example, a $2\mathrm{kg}$ point mass starts at rest and receives a constant net $3\mathrm{N}$ force through $4\mathrm{m}$. Its net work is $12\mathrm{J}$, soAt any intermediate distance $x\ge 0$, $K(x)=(3\mathrm{N})x$ and $v(x)=\sqrt{2Fx/m}$: kinetic energy is a linear function of distance, while speed is proportional to the square root of distance. If a net opposing $3\mathrm{N}$ braking force subsequently acts for $4\mathrm{m}$, it does $-12\mathrm{J}$ of work and brings this point mass to rest. Negative net work decreases kinetic energy; it does not make kinetic energy negative. Beyond the stopping point the motion must be reconsidered, since the point mass cannot keep moving forward with a negative $K$.

Kinetic energy also makes the exchange with potential energy intuitive: a falling body's loss of gravitational potential energy can appear as increased kinetic energy. The formula above applies when speed is small compared with the speed of light; at higher speeds, special relativity changes the expression. ^[ref]

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.

If a known classical bit is independently encoded in a secret uniformly chosen X or Z eigenbasis, the averaged verification operator is $A_b=I/2+(-1{)}^b(X+Z)/4$. Its largest eigenvalue is the displayed optimal acceptance probability for a counterfeit prepared without the original qubit. The maximizing state bisects the two candidate Bloch directions. For $n$ independent secret bases, the maximal all-pass probability is the $n$th power, because the total acceptance operator is a tensor product. Randomly choosing one of the legitimate encodings instead passes with probability $3/4$ per bit.

Lambertian reflectance is the ideal diffuse-reflection model with direction-independent outgoing radiance, the frequency-integrated specific intensity. If a fraction $0\le \rho \le 1$ of incident radiant flux is reflected, a surface receiving irradiance $E$ has reflected radiance $L=\rho E/\pi$. The reflected outward power per area is $\pi L=\rho E$ by the same hemispherical integral used for radiant exitance. Its radiant intensity still acquires the projected area factor in Lambert's cosine law.

For $E=100\mathrm{W}{\mathrm{m}}^{-2}$ and $\rho =0.80$, the reflected radiance is $80/\pi \approx 25.4648\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$. A patch of area $0.020{\mathrm{m}}^2$ reflects $1.6\mathrm{W}$ in total, and its radiant intensities at $0^{\circ }$ and $60^{\circ }$ are respectively $1.6/\pi$ and $0.8/\pi \mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$. The radiance stays fixed as the view changes, provided the illumination stays fixed. Changing the incident angle can change irradiance and therefore the reflected radiance.

A laser generates amplified light through stimulated emission. Its name abbreviates “light amplification by stimulated emission of radiation”. Lasers supply the intense beams used in optical tweezers; optical coherence also makes them useful for wave interference measurements. ^[ref].

In the usual material picture, a gain medium stores energy supplied by pumping, for example by incident light or an electric current. Stimulated emission transfers that stored energy to a selected optical normal mode, while competing photon absorption removes light. Producing net optical gain requires a suitable population distribution, such as population inversion in the simple model below.

A familiar laser oscillator uses mirrors to return light through the gain medium. This optical feedback favours resonant normal modes; a partly transmitting mirror lets useful power escape. Feedback and mode selection have different implementations, including ring arrangements: every laser need not resemble the same two-mirror device. The laser threshold example treats just that device. ^{[ref] [ref]}.

The useful combination is concentrated power and controlled optical normal modes. It gives optical tweezers a localized source of force and wave interference measurements a source with useful optical coherence. A laser can nevertheless emit several frequencies; “laser” alone does not guarantee a single frequency. ^[ref].

An original experimental landmark is T. H. Maiman's 1960 report of stimulated optical radiation from optically pumped ruby at about $694.3\mathrm{nm}$. His paper describes both the pumping path and the observed emission, connecting stored material energy with amplified light. ^[ref].

In a simple two-mirror oscillator, laser threshold is the small-signal optical gain at which a resonant optical normal mode exactly replaces its round-trip intensity loss. Assume constant intensity coefficients $g$ and $\alpha$ throughout a one-way distance $L$, and mirror intensity reflectivities $0<R_1,R_2\le 1$. One round trip has multiplierTwo traversals give the distance $2L$; multiplying the two mirror factors completes the bookkeeping. Setting $M=1$ and taking a logarithm givesThe reflectivities here are intensity fractions, not field-amplitude reflectivities. A shorter gain region or nonuniform loss requires replacing the path exponent by the appropriate path integral. The balance also presumes a permitted resonant optical normal mode. ^[ref].

For $L=0.50\mathrm{m}$, $R_1=1$, $R_2=0.90$ and $\alpha =0.10{\mathrm{m}}^{-1}$, the threshold is $g_{\mathrm{th}}\approx 0.2053605{\mathrm{m}}^{-1}$. At $g=0.15{\mathrm{m}}^{-1}$, $M\approx 0.946144$: a small circulating signal decays even though the material has positive optical gain. At $g=0.30{\mathrm{m}}^{-1}$, $M\approx 1.099262$: the signal initially grows. Raising passive loss or lowering mirror reflectivity raises laser threshold.

For $M>1$, the fixed-gain recurrence predicts startup amplification, not infinite actual output. Growing light intensity depletes the available population inversion, reducing optical gain through saturation. In the simplest steady oscillator, the saturated round-trip gain balances loss and pumping supplies the outgoing power and dissipated energy. The small-signal model locates the onset; it does not determine the final output power. ^[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]