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Photosynthesis captures light energy and transforms it into chemical forms that a biological cell can use. In plants, algae and cyanobacteria, this supports carbon fixation: incorporation of carbon from carbon dioxide into chemical compounds used to build living material. Capturing light and incorporating carbon are distinct processes coupled through metabolism. See OpenStax Biology 2e, 8.1: overview of photosynthesis.

For oxygenic photosynthesis, a simplified net equation writes a six-carbon sugar, glucose, as the representative product:This is net stoichiometry, not a one-step mechanism. A worked bookkeeping check counts six carbon chemical atoms, twelve hydrogen chemical atoms, and eighteen oxygen chemical atoms on each side: the product side has six oxygen chemical atoms in the sugar and twelve in molecular oxygen. The light label supplies energy, not additional chemical atoms.

Crucially, the released molecular oxygen comes from water, not carbon dioxide. Cancelling water made and consumed in different steps hides this distinction in the net equation. The Calvin cycle supplies a net three-carbon triose phosphate, such as glyceraldehyde 3-phosphate, for subsequent sugar synthesis; it does not directly export glucose. See OpenStax Biology 2e, 8.2: light-dependent reactions and OpenStax Biology 2e, 8.3: using light energy to make organic molecules.

The useful achievement is connecting incoming light to both usable energy and new biological material. ATP is an immediate energy currency; NADPH supplies electrons for synthesis; carbon dioxide supplies the carbon incorporated into carbohydrates. These roles are complementary. Conservation of energy still holds: photosynthesis transforms energy, never creates it, and some absorbed energy is dissipated as heat. A plant also carries out cellular respiration, including in biological cells that perform photosynthesis. Gas exchange reflects both processes. Oxygenic photosynthesis remains central to interpreting terrestrial molecular oxygen as an atmospheric biosignature gas. See OpenStax Biology 2e, 8.3: photosynthesis and respiration.

The study of physical properties and processes in the oceans, including fluid flow, waves and transport of heat.

Low-temperature solid helium-3 can have greater entropy than the liquid through disordered nuclear spins, enabling compression cooling. ^[ref]

The sign follows from the Clapeyron equation. Let $S_l,S_s$ be molar entropies and $V_l,V_s$ molar volumes of liquid and solid, respectively, on their phase coexistence curve. Equality of their chemical potentials, with $d\mu =-SdT+VdP$, givesWith $S_s>S_l$ and $V_l>V_s$, the slope is negative. Reversibly freezing one mole at fixed temperature requires heat into the sample,Here $L_{s\to l}$ uses the melting convention for latent heat of fusion. This is the opposite sign from ordinary ice melting. The inequalities specify the regime; freezing does not universally cool a substance. In an approximately isolated sample, the growing higher-entropy phase draws on the sample's thermal energy instead of an external heater. Irreversible losses and incoming heat oppose the cooling.

An original deliberately local model makes the cooling quantitative. Fix a total amount $n$ of material, let $x$ be the mole fraction of solid, and stipulate molar entropiesThe constants $a$ and $b$ have units of molar entropy per temperature and molar entropy, respectively. Ignore their pressure dependence and the cell's own heat capacity. This is a local thermodynamics model, not an extrapolation to zero temperature. A constant positive $b$ cannot remain valid once the solid's spins order.

For a reversible thermodynamic process with no heat exchange, total entropy is constant. Differentiating giveswhen $b>aT$ and $x<1$. Define $T_0=b/a$; solving between two states givesIn particular, stipulate $x_0=0$, $T_i=T_0/4$ and $x=1/8$. ThenThe last expression is the local reversible heat requirement per mole frozen at the final state, divided by $b{T}_0$; it is not the heat exchanged during the isolated compression. That compression exchanges no heat. This distinction separates the latent-heat sign from the actual boundary heat budget.

The utility of the model is its explicit condition: $S_s>S_l$ makes $dT/dx<0$, while positive total heat capacity supplies a stable thermal response. More generally, for molar entropies depending only on temperature, let $C_l,C_s$ be the molar heat capacities. Then $dT/dx=-T(S_s-S_l)/[(1-x)C_l+x{C}_s]$. Incoming heat or entropy production adds terms to the balance and can overwhelm the ideal cooling. Neither the numerical ratios above nor the straight entropy lines are measurements of helium-3.

A small longitudinal electric field drives conductor current in a lossy coaxial line. Its cross product with the azimuthal magnetic field sends energy radially into both conductors, where it becomes heat; forward power decreases with distance.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.

Radiometric absorptance is the fraction of incident radiant flux converted to heat in the specified receiver: $P_{\mathrm{abs}}=a{P}_{\mathrm{in}}$, with $0\le a\le 1$ for a passive receiver. Specify the entrance boundary, wavelength, incidence angle and receiver temperature. Radiation reflected or transmitted out of that receiver is not absorbed power. A cavity can improve radiometric absorptance by giving light repeated opportunities for wave absorption; a dark coating alone does not establish $a=1$.

For broadband light, spectral radiant flux matters: $P_{\mathrm{abs}}=\int a(\lambda ){\Phi }_{e,\lambda }(\lambda )d\lambda$. A single effective radiometric absorptance therefore depends on the incident optical spectrum as well as on the receiver. If the stated incident boundary lies before a window of known power transmission ${\tau }_w$, the receiver absorbs $a{\tau }_w{P}_{\mathrm{before}}$ when other optical losses are absent. Window loss and receiver radiometric absorptance are separate corrections.

Consider a fixed, homogeneous, nonabsorbing, nonmagnetic sphere of radius $a$ in vacuum. Let $m$ be its relative refractive index, a fixed real number; in vacuum this equals the sphere's refractive index. Treat $m$ as independent of wavelength over the range being compared, neglecting optical dispersion. Air can be approximated by vacuum here because its refractive index is close to one. These assumptions describe a toy particle, rather than a literal model of an individual air molecule.

The Rayleigh scattering cross-section $\sigma$ is an effective area: the total scattered power is $P_{\mathrm{sca}}=I_{\mathrm{inc}}\sigma$, where $I_{\mathrm{inc}}$ is the incident light intensity. For $ka\ll 1$ its leading approximation isThe units are area, since $k^4{a}^6$ has units ${\mathrm{m}}^{-4}{\mathrm{m}}^6={\mathrm{m}}^2$. This Rayleigh scattering cross-section measures scattering, not conversion of the incident electromagnetic radiation into heat. The definition follows ^[ref].

To see the mechanism, the nearly uniform applied electric field produces an electric dipole moment with peak magnitudewhere $E_0$ is the incident electric field peak wave amplitude and ${ϵ}_0$ is vacuum permittivity. This is the sphere's response in electrostatics, with its permittivity divided by vacuum permittivity equal to $m^2$ under the nonmagnetic assumption. ^[ref]. The averaged electric dipole radiation power is $P_{\mathrm{sca}}={\omega }^4{p}_0^2/(12\pi {ϵ}_0{c}^3)$, with angular frequency $\omega =ck$ and speed of light $c$. Dividing by $I_{\mathrm{inc}}={ϵ}_{0}c{E}_0^2/2$ gives the displayed Rayleigh scattering cross-section. Thus doubling radius multiplies the electric dipole moment by eight and the scattered power by $64$, provided the larger sphere still satisfies $ka\ll 1$. ^[ref].

Loss of summer sea ice need not be irreversible at fixed external forcing. Open water can release excess heat in autumn, and thin ice grows faster than thick ice during winter because it offers less resistance to thermal conduction. Continued climate change can nevertheless shift the long-term ice cover downward. A one-summer perturbation and persistent forcing are different physical experiments.

A motivating real procedure is NIST's ambient thermometer verification. Its apparatus is a deep beaker of water, a calibrated reference thermometer, a test thermometer, and a magnetic stirrer with any heater switched off. Let the water approach room temperature, stir gently, immerse both sensing tips well below the surface, and wait before reading. NIST specifies about $20\mathrm{cm}$ of water, $10$–$15\mathrm{cm}$ immersion where possible, and an initial five-minute wait. Record test then reference, repeat in reverse order, and apply the reference's certificate correction. Subtract corrected reference from test: the difference estimates the test instrument's error at this bath temperature. Agreement between the two reading orders checks repeatability. ^[ref]

For example, suppose a test reads $24.7{}^{\circ }\mathrm{C}$, a reference reads $24.4{}^{\circ }\mathrm{C}$, and its certificate says to add $0.1{}^{\circ }\mathrm{C}$. The corrected reference is $24.5{}^{\circ }\mathrm{C}$, so the test error is $+0.2{}^{\circ }\mathrm{C}$: it reads high. These are illustrative readings. This verification tests a previously calibrated instrument at one point; it does not establish the two-point response used in two-point thermometer calibration example.

Additional controls follow from the thermal equilibrium premise. Keep tips away from walls and bottom, interchange probe positions to test a temperature gradient, and check that longer waits and modest immersion changes leave agreement unchanged within the required tolerance. A flat trace alone can be a steady state maintained by competing heat transfer paths. Five minutes is a procedural starting point, not a universal equilibration time. Excessive stirring can supply work and warm the bath, while probe stems can exchange heat with the room. These controls test whether both probes sample the same state; neither one-point agreement nor this experiment proves the zeroth law of thermodynamics for every material and condition.

A refrigerator removes heat from the region we want to keep cold and releases heat to warmer surroundings. Here we consider a work-driven cyclic device: its useful output is the cold-side extraction $Q_c$, and its input is net work $W$. A food compartment is a familiar example; the same cooling objective applies when removing heat from a room. The coefficient of performance measures extraction per unit work, rather than the fraction of input energy converted into output. ^[ref]

Choose the complete machine as the thermodynamic system. Let $Q_c>0$ be heat entering from the colder side, $Q_h>0$ be heat delivered to the hotter side, and $W>0$ be net work supplied, all over the same complete cycle. Assume these are its only energy transfers, no matter crosses the outer boundary, and there is no net change in stored energy. Returning to the initial state gives $\Delta U=0$, so the first law of thermodynamics readsThe heat leaving the warmer side includes both the extracted heat and the supplied work. For instance, stipulated transfers $Q_c=3\mathrm{kJ}$ and $W=1\mathrm{kJ}$ require $Q_h=4\mathrm{kJ}$; the cooling coefficient of performance is $3$. This is an illustrative budget, not measured performance. At steady operation the same balance holds for average transfer rates, ${\dot{Q}}_h={\dot{Q}}_c+P_{\mathrm{in}}$, where $P_{\mathrm{in}}$ is input power.

This budget explains a useful counterexample: operating a refrigerator with its door open inside an otherwise isolated room cannot cool the room as a whole. If both heat transfers occur within the room and the electrical input comes from outside, the room gains $Q_h-Q_c=W$ each cycle. Locally cold air does not imply net cooling. The Second law of thermodynamics permits cold-to-hot transfer because work is supplied; it forbids a cyclic device whose sole effect is that transfer. See the Clausius statement of the second law.

For a two-component mixture, let $x$ be component A's mole fraction in the liquid and $y$ its mole fraction in the coexisting vapor. At interior compositions, define the relative volatility of A with respect to B by the ratio of composition odds:Thus $\alpha >1$ means A is enriched in the vapor, while $\alpha =1$ means $y=x$. Solving this definition givesHere a constant relative volatility is a stipulated model of vapor-liquid equilibrium, not a measured property of a named chemical solution. Real relative volatility can vary with composition, temperature, and pressure. The relation specifies no absolute temperature scale: even fixing $\alpha =2$ does not determine a boiling-temperature curve or a heat requirement.

A probe's sheath and leads carry heat between the sensing region and the surroundings through thermal conduction. In a stable, sufficiently uniform heat bath, compare the indicated temperature as insertion depth increases, allowing settling each time. A plateau supports adequate immersion for that apparatus and required uncertainty; it does not prove that every depth or medium works. Keep the sensing element and reference thermometer at comparable depths, and check bath gradients independently, so moving the probe does not merely sample a different temperature. The documented platinum resistance thermometer comparison calibration includes these controls.

Glycolysis converts glucose into two pyruvate molecules, with a net two ATP and reduced NADH; it does not directly consume molecular oxygen. Pyruvate processing and the citric acid cycle, also called the TCA cycle, release carbon dioxide and supply reduced carriers. Thus releasing carbon dioxide and consuming molecular oxygen occur at different stages.^[ref]

An electron transport chain reoxidizes carriers such as NADH. Coupling some of its redox steps to proton transport establishes an electrochemical gradient; ATP synthase can use that gradient to form ATP from adenosine diphosphate and inorganic phosphate. This is oxidative phosphorylation. A separate route, substrate-level phosphorylation, transfers a phosphate group from a reacting substrate to adenosine diphosphate. In typical eukaryotic biological cells, the respiratory chain occupies the inner membrane of a mitochondrion.^[ref]

The useful achievement is controlled transfer: fuel oxidation can replenish an ATP supply that supports active transport and synthesis. ATP is a reusable chemical carrier, not newly created energy; ATP hydrolysis and the coupled cellular process have their own net free energy balance. Some fuel energy is dissipated as heat.^[ref]

There is no universal yield of $36$ or $38$ ATP per glucose. Actual ATP recovery depends on carrier transfer routes, coupling and diversion of intermediates to other metabolism. In particular, an oxygen uptake rate cannot be converted into an ATP production rate by assuming one fixed multiplier.^[ref]

At fixed temperature, a photon gas has constant energy density $u$ and radiation pressure $p=u/3$. An increase $\Delta V$ in volume requires both $\Delta U=u\Delta V$ and expansion work $p\Delta V$, giving absorbed heat $4u\Delta V/3$.

Postmortem muscular stiffening associated with loss of sufficient ATP to maintain normal detachment of actin–myosin cross-bridges. It is distinct from freezing and heat coagulation.

Roberta Rudnick is a geologist who investigates the composition and origin of continental crust. Much of the lower continental crust is inaccessible to direct sampling, so its composition must be inferred by combining geochemistry, petrology and seismology. Her 1995 synthesis with David M. Fountain linked laboratory measurements of elastic wave speeds in rocks to seismic refraction surveys, using granulites to constrain the composition of the deep continental crust.^[ref]

Their reconstruction distinguished a compositionally heterogeneous, on average mafic lower continental crust from an intermediate-composition middle continental crust. The distinction matters for Earth's heat budget: the abundance of potassium, thorium and uranium, and hence heat production from radioactive decay, varies between rock types and with depth. Seismic wave speeds alone are not a unique chemical analysis; different rocks can have overlapping elastic properties, making independent constraints from geochemistry essential.^[ref]

Research author of Nature and composition of the continental crust: A lower crustal perspective by Roberta L. Rudnick and David M. Fountain.

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.

In Shearwood and Sloan's implementation, piston-mounted magnets induce an electromotive force in a stationary coil through Faraday's law of induction; an oscilloscope records the signal. They observed faster oscillations at smaller trapped volume values and stronger damping at small cylinder volume values. The signal tracks velocity locally, so its amplitude is not directly piston displacement. Calibration also identified inlet dead space beyond the graduated cylinder volume. ^[ref]

Repeat at smaller initial amplitudes to test the linear approximation, and compare several $V_0$ values. Heat transfer weakens the restoring force toward the isothermal process limit, biasing the inference toward $1$. Check leakage through resting-piston drift and before-and-after trapped volume readings. For viscous damping with envelope $e^{-t/\tau }$, the measured angular frequency obeys ${\omega }_0^2={\omega }_d^2+{\tau }^{-2}$; this correction does not remove thermal leakage. Internal equilibration must be faster than an oscillation period, with heat transfer and leakage slow over a run. Large amplitude or sticking friction require a richer model. ^[ref]

Salt fingering is double-diffusive convection in which narrow rising and sinking fluid structures develop because heat and dissolved salt diffuse at different rates. Warm, salty water above colder, fresher water can have a stable total mass density stratification: the stabilizing temperature difference outweighs the destabilizing salinity difference.^[ref]

A small upper parcel displaced downward enters colder surroundings. It loses heat relatively quickly while retaining more of its salt, becoming denser and continuing downward. A fresher parcel displaced upward can gain heat while remaining relatively fresh and light. The distinction between thermal diffusivity and solutal diffusivity supplies the feedback; stability of the initial combined mass density profile does not exclude this mechanism. The density ratio records the competing contributions, while the salt-finger instability criterion specifies the parameter regime.

A laboratory analog can replace heat and salt by two dissolved substances. David Osmond describes a 1820-by-80-by-120 millimetre tank with salt and sugar solutions entering at opposite ends and an overflow at the centre. The supplied solutions had matching mass density, but salt diffused faster than sugar. The salt-rich plume could lose salt faster than it gained sugar, producing a lighter core and denser surroundings. Fingering and diffusive layers appeared in different regions, showing that identical source densities need not remain dynamically equivalent when composition diffuses.^[ref]

In this analog, salt plays the faster-diffusing role played by heat in the ocean, while sugar plays the slower-diffusing role of oceanic salt. Keeping that correspondence explicit prevents reversing the physical explanation.