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If a density of states has high-energy growth proportional to a power of energy times $e^{{\beta }_{H}E}$, its canonical Boltzmann sum fails for $\beta <{\beta }_H$. The Hagedorn temperature is the corresponding threshold. The open bosonic string mass spectrum and large-level degeneracy of an open bosonic string give ${\beta }_H=4\pi \sqrt{{\alpha }^{\prime }}$ in units with Boltzmann's constant, the speed of light and Planck's constant set to one. Polynomial prefactors affect behavior at the threshold itself; the growth alone does not distinguish a limiting temperature from a phase transition.

A physicist who coauthored Detailed Balance Limit of Efficiency of p-n Junction Solar Cells by William Shockley and Hans-Joachim Queisser.

A hard-sphere gas is a classical model of identical impenetrable spheres, each of diameter $\sigma$ and radius $\sigma /2$. Its pair potential is $u(r)=+\infty$ for centre separation $r<\sigma$ and $u(r)=0$ for $r\ge \sigma$; the convention at exact contact has no effect on the integral. It isolates the effect of particle size without an attractive potential energy.

The upper-left panel is a cross-section of a three-dimensional forbidden region: the second centre cannot enter a sphere of radius $\sigma$ around the first centre. The particles themselves have radius $\sigma /2$. Consequently, with particle volume $v_{\mathrm{s}}$ and pair excluded volume $v_{\mathrm{ex}}$,The last equality follows directly from the Mayer f-function: $f=-1$ throughout the forbidden volume and zero outside it. Confusing particle radius with contact separation changes the answer by a factor of eight; omitting the unordered-pair factor changes it by a further factor of two.

Define the packing fraction $\eta =n{v}_{\mathrm{s}}$, where $n$ is number density. Then the virial expansion beginsFor example, at the toy value $\eta =0.01$, the leading correction predicts $Z\approx 1.04$; the omitted terms still matter if accuracy beyond this approximation is needed. The classical second virial coefficient is independent of temperature because the geometry of forbidden configurations does not change when the gas is warmed. The pressure itself still scales with temperature at fixed number density.

This model makes the positive correction particularly clear: eliminating overlapping configurations changes the dependence of the available configuration space on volume, even though the allowed configurations all have zero interaction energy. The figure's other panels add a toy attractive shell and show how its Boltzmann factor competes with that geometry. The hard-sphere gas is a useful reference model, while a low-order virial expansion remains a dilute-gas approximation rather than a dense sphere packing formula.

A physical chemist whose name is used in the planetary Urey ratio.

A electrical engineer.^[ref]

The stellar temperature sequence O, B, A, F, G, K, M, with numerical subclasses approximately from hotter 0 to cooler 9 within each letter class. Spectral features follow temperature-dependent excitation, ionization and molecule survival, not a corresponding change in the bulk hydrogen abundance. A separate luminosity class describes gravity-sensitive spectral differences.

The Hasselmann stochastic climate model represents slowly responding climate variables driven by fast weather fluctuations as well as external forcing. Klaus Hasselmann's 1976 paper separates fast and slow variables, retains fluctuations lost by simple averaging, and explains why stabilizing climate feedbacks are needed to prevent indefinitely growing variability. Its general framework is broader than a single-temperature equation.^[ref]

Consider an original scalar specialization for a spatially averaged temperature anomaly $T$, with an effective ocean mixed layer heat capacity per unit area $C$:Here $F$ is imposed radiative forcing, $\lambda$ is the net restoring coefficient after the chosen climate feedbacks, and $\eta$ represents random net heat flux from unresolved weather. The anomaly energy per unit area is $CT$, so its tendency equals the net input. The weather term represents exchange with omitted degrees of freedom; it does not assert that an isolated ocean creates energy. $F$ and $\eta$ have units $\mathrm{W}{\mathrm{m}}^{-2}$, and $\lambda$ has units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

First set $\eta =0$, start from $T(0)=0$, and switch to constant $F_0>0$. The integrating factor $e^{\lambda t/C}$ givesFor $C=2.0\times 10^8\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, $\lambda =2.0\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ and $F_0=4.0\mathrm{W}{\mathrm{m}}^{-2}$, the relaxation time is $10^8\mathrm{s}$, about $3.17$ years using 365.25 days per year. At $t=\tau$, $T=2(1-e^{-1})\mathrm{K}\simeq 1.264\mathrm{K}$, with eventual warming $2\mathrm{K}$. Doubling $C$ doubles the response time but leaves this equilibrium unchanged. This separates thermal inertia from climate sensitivity.

Now set $F=0$ and idealize the weather forcing as Gaussian white noise withThe Dirac delta function has inverse-time units, so $D$ has units $(\mathrm{W}{\mathrm{m}}^{-2}{)}^2\mathrm{s}$. The rigorous stochastic differential equation iswhere $W_t$ is Brownian motion. This is an Ornstein-Uhlenbeck process. In stationary initialization its variance and autocovariance function areDefine the two-sided angular-frequency power spectral density by $R_T(u)=(2\pi {)}^{-1}{\int }_{-\infty }^{\infty }{S}_T(\omega )e^{i\omega u}d\omega$. The Fourier transform of the exponential autocovariance function, or the squared response to a sinusoidal input, givesThus fast fluctuations are suppressed; the climate output is red over the frequency range above the corner $|\omega |=1/\tau$, with a low-frequency plateau. This is a colored noise spectrum without a periodic driver. The corner is $f_c=1/(2\pi \tau )$ in cycles per second, not $1/\tau$. The continuous white input is only an approximation on times longer than weather's correlation time.

For the same $C,\lambda$ and $D=10^8(\mathrm{W}{\mathrm{m}}^{-2}{)}^2\mathrm{s}$, $v_{\infty }=0.25{\mathrm{K}}^2$ and the stationary standard deviation is $0.50\mathrm{K}$. From fixed $T(0)=0$, the variance grows as $v_{\infty }(1-e^{-2t/\tau })$, even when the forced expected value is predictable. Integrating $S_T/(2\pi )$ over all angular frequencies returns $D/(C\lambda )$, checking the normalization independently. If $\lambda =0$ the unforced variance instead grows as $2Dt/C^2$; if $\lambda <0$, deviations amplify. Neither case has the stationary spectrum above.

A red spectrum alone cannot establish attribution of recent climate change: internal variability, changing forcing and measurement effects must be distinguished. This scalar model supplies a calculable baseline for variability and persistence, not a universal spectrum for all climate timescales or a full attribution analysis.

Match an efficient, monatomic gas, fully convective star to the grey-atmosphere convection onset with thirteenth-power opacity. Its convective relation is $P=K_T{T}^{5/2}$. At the matching temperature $T_b\propto T_e$, the stellar atmosphere gives $K_T^2\propto M{L}^{-1}{T}_e^{-13}\propto M{R}^{13/2}{L}^{-17/4}$ using the Stefan–Boltzmann law. The ideal-gas relation converts this to $P=K_{\rho }{\rho }^{5/3}$ with $K_{\rho }=(\mathcal{R}/\mu {)}^{5/3}{K}_T^{-2/3}$. The polytropic mass-radius relation at index $3/2$ gives $K_{\rho }\propto G{M}^{1/3}R$, hence $K_T^2\propto M^{-1}{R}^{-3}$. Equating the two expressions proves $L^{17/4}\propto M^2{R}^{19/2}$, or $L\propto M^{8/17}{R}^{38/17}$. The entropy coefficient is constant throughout the convective envelope, but can change between models.

Heat is energy transferred across a thermodynamic system's boundary because of a temperature difference. It is measured in joules, like work and internal energy. A hot cup transfers heat to cooler surroundings; its temperature describes its state, while the transferred heat describes a process. The cup stores internal energy, rather than a substance called heat. ^[ref]

For a closed system with unchanged bulk kinetic energy and potential energy, choose $Q>0$ for heat entering and $W>0$ for work done by the system. The first law of thermodynamics givesHere $U$ is the stored internal energy. For example, a sealed piston cylinder receiving $20\mathrm{kJ}$ of heat and delivering $5\mathrm{kJ}$ of work gains $15\mathrm{kJ}$ of internal energy. A closed system retains its matter while exchanging energy; it need not be an isolated system. ^[ref]

Unlike internal energy, heat is not a state function: different paths between the same equilibrium states can transfer different $Q$, with compensating differences in work. The symbol $\delta Q$ records this dependence on the path; there is no stored quantity whose difference is always $Q$. The two paths in first law of thermodynamics illustrate this directly. Supplied heat can raise temperature as sensible heat, change a phase fraction through latent heat, or help supply expansion work. Measuring temperature alone therefore does not determine transferred heat.

For a reversible thermodynamic process, $\delta Q_{\mathrm{rev}}=TdS$, where $T$ is absolute temperature and $S$ is entropy. This identification requires reversibility: an irreversible adiabatic process has $Q=0$ and can nevertheless increase entropy. Likewise $Q=\Delta H$, with enthalpy $H$, requires the particular fixed-pressure conditions in enthalpy. These distinctions make heat useful as a precise boundary-transfer quantity in heat engines, refrigerators and heating calculations.

Historically, James Prescott Joule's paddle experiments showed that stirring water with input work could raise its temperature, helping establish conservation of energy. The lesson is still useful: a temperature rise can result from work even without incoming heat. ^[ref]

A heat bath, or thermal reservoir, is an idealized system so large that exchanging a finite amount of heat does not appreciably change its temperature.

The ideal heat bath supplies a fixed absolute temperature $T_b$ while its internal energy changes through heat transfer. A real body's suitability as a heat bath depends on its heat capacity relative to the exchange being studied, and on how quickly its internal temperature gradients disappear. Large physical size alone supplies neither condition. A finite heat bath makes the approximation quantitative; a regulated calibration bath instead uses feedback control to maintain its temperature by exchanging energy with external equipment.

For a weakly coupled system that exchanges energy but no matter with an ideal heat bath, the canonical ensemble assigns the Boltzmann distributionHere $E_n$ is a microstate's energy, $k_B$ the Boltzmann constant and $Z$ the partition function. The fixed temperature removes the need to track every exchange with the environment. A substantially changing bath temperature no longer gives this one fixed Boltzmann distribution.^[ref]

The equilibrium calculation determines where the two temperatures end, without determining how long they take. Stipulate a constant thermal conductance $G>0$ and the linear heat transfer law underlying Newton's law of cooling. With $Q$ positive from body 1 to bath 2, the ordinary differential equations areThe dot denotes a derivative with respect to time. This model assumes internal equilibration is fast compared with exchange between the bodies. Adding the two energy rates gives zero; subtracting the temperature rates givesThe temperature difference has exponential decay with time constant $\tau$. For the finite heat bath example and the stipulated $G=2\mathrm{W}{\mathrm{K}}^{-1}=2\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{K}}^{-1}$, $\tau =45\mathrm{s}$, and solving the ordinary differential equations givesAt one time constant, $T_1\simeq 315.865\mathrm{K}$, $T_2\simeq 293.793\mathrm{K}$ and $Q\simeq 3413.451\mathrm{J}$. Equal temperatures are approached as a limit at infinite time. The finite bath causes the temperature difference to shrink faster than the fixed-bath model, because both bodies change temperature.

For an ideal heat bath with the same $G$, set $T_2=T_b=290\mathrm{K}$ rather than evolving its temperature. Newton's law of cooling then yieldsChanging $G$ changes these time constants, while the final temperature is fixed by the capacities and initial conditions. The dynamical model independently checks the Second law of thermodynamics:Its integral over the relaxation reproduces the endpoint entropy change. The useful separation is between capacity, which governs how much the bath temperature shifts, and thermal conductance, which governs the exchange rate. Neither alone establishes a reliable heat bath.

Heat capacity measures the heat supplied per small temperature increase along a specified process:The subscript $X$ records the constraint, such as fixed volume or fixed pressure. The units are $\mathrm{J}{\mathrm{K}}^{-1}$. The symbol $\delta Q$ emphasizes that heat is a transfer along a path, rather than a state function with a differential $dQ$. Consequently, an unspecified heat capacity does not determine the required heat.

For a closed system of fixed composition undergoing a quasistatic process, assume that only pressure-volume work occurs, with no friction and no change in bulk kinetic energy or potential energy. The first law of thermodynamics and $H=U+pV$, where $U$ is internal energy and $H$ is enthalpy, then giveThese partial derivatives define heat capacity at constant volume and heat capacity at constant pressure, with the amount and composition also held fixed. Integrating along the selected constraint gives $Q={\int }_{T_1}^{T_2}{C}_X(T)dT$ within a single thermodynamic phase; $Q=C_X\Delta T$ requires constant $C_X$. ^[ref]

For a computed example, take a monatomic gas in the classical ideal gas regime, with amount of substance $n=1\mathrm{mol}$ and stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The equipartition theorem gives $U=3nRT/2$, and the ideal gas law gives $pV=nRT$. ThusFor the same temperature increase $\Delta T=10\mathrm{K}$, the two processes requireHere work done by the gas is positive. At fixed pressure, the extra heat supplies expansion work; the internal energy increase is the same because internal energy of an ideal gas depends only on temperature. Their final volume values differ. ^[ref]

The practical value of heat capacity is that a measured heat input and temperature response can determine it. Once its constraint and temperature dependence are known, it predicts heating requirements and supplies entropy changes through entropy from heat capacity. ^[ref]

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.

For a fixed-composition simple compressible system, a Maxwell relation gives $C_p-C_V=T(\partial p/\partial T{)}_V(\partial V/\partial T{)}_p$. At fixed volume, the differential equation of state gives $(\partial p/\partial T{)}_V={\alpha }_V/{\kappa }_T$, proving the displayed result. Positive absolute temperature and positive isothermal compressibility imply a nonnegative difference. It vanishes when thermal expansion does, so strict inequality requires a nonzero expansion coefficient. This stability argument includes nonideal internal-energy changes, unlike a purely ideal-gas expansion-work explanation.

A freely orientable classical magnetic dipole moment of fixed magnitude $m$ in a uniform magnetic field $B$ has angular canonical partition function $Z=4\pi \sinh x/x$, where $x=mB/(k_{B}T)$. Its internal energy is $U=-mB\coth x+k_{B}T$, and its fixed-field heat capacity is $C_B=k_B[1-x^2/{\sinh }^{2}x]$. At low temperature, two locally Cartesian tangent coordinates on the orientation sphere each contribute $k_{B}T/2$ by the equipartition theorem, giving $C_B\to k_B$. A discrete quantum spin in a nonzero magnetic field instead has a spectral gap and its heat capacity vanishes exponentially at sufficiently low temperature.

The heat capacity ratio compares heat capacity at constant pressure $C_p$ with heat capacity at constant volume $C_V$ at the same equilibrium state and fixed composition:Here $C_p,C_V$ refer to the whole sample, in $\mathrm{J}{\mathrm{K}}^{-1}$; molar heat capacities $C_{p,\mathrm{m}}=C_p/n$ and $C_{V,\mathrm{m}}=C_V/n$ are per mole; specific heat capacities $c_p=C_p/m$ and $c_V=C_V/m$ are per unit mass. The amount of substance $n$ or mass $m$ cancels only when both capacities use the same normalization. Thus heat capacity ratio is dimensionless and independent of sample size. For a fixed-composition ideal gas with positive finite $C_V$, Mayer's relation giveswhere $R$ is the molar gas constant. More ways to store internal energy make a given temperature increase require more heat, bringing heat capacity ratio closer to unity. ^[ref]

The equipartition theorem predicts $C_{V,\mathrm{m}}=fR/2$ when $f$ independent quadratic terms in molecular energy are thermally active and their number stays fixed; consequently $\gamma =1+2/f$. A classical monatomic gas with no appreciable electronic excitation has $f=3$, giving $C_{V,\mathrm{m}}=3R/2$ and $\gamma =5/3$. For an ideal gas of diatomic molecules with three translational and two rotational terms active, but its vibrational normal mode effectively frozen, $f=5$ gives $C_{V,\mathrm{m}}=5R/2$ and $\gamma =7/5$. A fully classical harmonic vibration adds both kinetic energy and potential energy, giving $f=7$ and $\gamma =9/7$. These are model regimes, not universal constants of every gas: rotational freezing, electronic excitation, and changing composition invalidate the simple count.^[ref]

The mechanical significance of heat capacity ratio follows from the first law of thermodynamics. For a fixed amount of ideal gas in a reversible adiabatic process, $C_V(T)dT=-pdV$. Combining this with the ideal gas law gives the local relationThe partial derivative holds at fixed entropy and composition even when heat capacity ratio varies with temperature. The finite-path formulas $p{V}^{\gamma }=\mathrm{constant}$ and $T{V}^{\gamma -1}=\mathrm{constant}$ require the additional approximation of constant heat capacity ratio. A rapid compression also needs internal equilibration to approximate this path; rapid motion alone does not ensure reversibility. The same local response gives isentropic compressibility $1/(\gamma p)$ and, for small sound disturbances in this ideal gas regime, speed of sound $a=\sqrt{\gamma p/\rho }$, where $\rho$ is mass density. Thus thermal storage controls mechanical stiffness, sound propagation and the measurements in Rüchardt experiment and Clément-Desormes experiment.

A dilute particle suspension can supply both excess mass density and heat. To first order its equation of state is $\rho /{\rho }_0=1+\gamma \varphi -\theta$, where $\varphi$ is the particle volume fraction and $\theta$ is fractional thermal expansion. Particle deposition flux reduces $\varphi$, while particle heating increases $\theta$. The competition can reverse the reduced gravity and create unstable density stratification.

A heat engine uses a repeating thermodynamic cycle to turn part of an incoming heat transfer into net outgoing work. In the two-thermal reservoir model, the hot thermal reservoir supplies heat $Q_h>0$ at fixed temperature $T_h$, and the cold thermal reservoir receives heat of magnitude $Q_c>0$ at fixed temperature $T_c<T_h$. A Stirling engine makes the repeated motion visible; the abstract energy budget applies independently of its particular mechanism.^[ref]

Choose the entire cyclic device as a closed system, with no net change in bulk kinetic energy or potential energy. Count heat entering as positive and net work delivered as positive, as in the first law of thermodynamics. Thus $Q_h$ enters with a plus sign and $Q_c$, although a positive magnitude, enters with a minus sign. Assume these two heat transfers and the net work are the only transfers across the boundary. Over one complete thermodynamic cycle, all stored energy, including internal energy $U$, returns to its starting value:The device does not accumulate heat from cycle to cycle. More precisely, heat is a transfer, while internal energy is what can be stored; even a component that temporarily stores internal energy must return to its initial state for this boundary to be cyclic.^[ref]

For engine operation, $Q_h>Q_c$ and $W>0$. Its thermal efficiency isUse net work, after subtracting any work supplied during compression or required by auxiliaries included in the chosen device. For periodic operation, average power satisfies $P_{\mathrm{out}}={\dot{Q}}_h-{\dot{Q}}_c$ and $\eta =P_{\mathrm{out}}/{\dot{Q}}_h$, with the rates averaged over complete thermodynamic cycles.^[ref]

For a stipulated model at $T_h=600\mathrm{K}$ and $T_c=300\mathrm{K}$, let $Q_h=1000\mathrm{J}$ and $Q_c=600\mathrm{J}$ per thermodynamic cycle. Then $W=400\mathrm{J}$ and $\eta =2/5=40\%$. At two thermodynamic cycles per second, these become $2000\mathrm{W}$ incoming heat transfer, $1200\mathrm{W}$ outgoing heat transfer, and $800\mathrm{W}$ net power. These are constructed values, not measurements.

Different signs describe different tasks. With the same directions of heat transfer, $Q_h=Q_c$ gives $W=0$: a device transmitting heat without net work output. If $Q_c>Q_h$, net work is supplied to that device. A refrigerator or heat pump instead extracts heat from the cold side and delivers heat to the hot side, using positive input work $W_{\mathrm{in}}$ and the balance $Q_h=Q_c+W_{\mathrm{in}}$. Their useful output is heat transfer, measured by a coefficient of performance. Finally, a one-time isothermal expansion can have $Q=W$ and $\Delta U=0$ while leaving the volume changed: it is not a complete thermodynamic cycle. A warming device also fails the cyclic assumption, and a fuel-consuming device requires an accounting of matter crossing its boundary.

For two fixed thermal reservoirs at positive thermodynamic temperatures, the Second law of thermodynamics strengthens the energy balance toEquality requires the overall operation to be a reversible thermodynamic process, including its heat transfer to the thermal reservoirs. The Carnot cycle explains how the reversible benchmark is constructed, and Carnot's theorem explains its universality and the entropy production penalty. Use kelvin values, never ratios of degree Celsius readings. The example above lies below the $50\%$ bound: its total entropy production is $-1000/600+600/300=1/3\mathrm{J}{\mathrm{K}}^{-1}$ per thermodynamic cycle.

This is useful before building a machine: warmer cooling surroundings reduce the possible thermal efficiency, even if the mechanism is unchanged. A small observed thermal efficiency does not by itself test the Second law of thermodynamics; heat leaks, friction, incomplete recovery of work, and the measurement boundary can all reduce the reported ratio. A meaningful comparison requires the actual thermal reservoir temperatures, matched heat and net work measurements, and the hypotheses of Carnot's theorem. Finite bodies whose temperatures change are treated separately there.

The practical question already appears in the 1824 Réflexions sur la puissance motrice du feu et sur les machines propres à développer cette puissance by Sadi Carnot: pp. 6–8 ask whether machines driven by heat have a fundamental limit and whether changing the working substance can improve their output. Pages 9–11 describe heating and cooling in real machines, but use the historical caloric theory, rather than the modern first law of thermodynamics balance. The historical discussion under Carnot's theorem explains that distinction.