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Energy flow tracks energy entering an ecosystem, passing through organisms and leaving its boundary. To see its quantitative effect, assume net production at one trophic level is $P_1=1000\mathrm{kJ}{\mathrm{m}}^{-2}\mathrm{da}{\mathrm{y}}^{-1}$ and assume a transfer efficiency $\epsilon =0.10$ at each of two successive steps. Here $P_i$ measures the energy incorporated into new living material per unit area per day, after that level's own cellular respiration; it is not a count of organisms or a standing stock. The joule supplies the energy unit, with $1\mathrm{kJ}=1000\mathrm{J}$.

Only $10/1000=0.01$, or one percent, of the initial production reaches level 3 in this toy food chain. The first step leaves $1000-100=900$ and the second $100-10=90$ outside the next level's new production. These differences can include uneaten material, wastes, and energy released as heat during cellular respiration; their separate amounts have not been assigned. Detritus can still carry energy to decomposers. The 10 percent efficiency is an assumption, not a universal law: transfer efficiencies vary with the organisms, food and conditions. The production ratio $P_{i+1}/P_i$ also differs from the fraction of an individual meal converted into growth. See Clark, Douglas and Choi, Biology 2e, 46.2: production and transfer efficiency.

The shrinking budget explains why supporting another trophic level requires enough production below it. It does not prove an exact maximum length for every food chain, particularly when a food web imports resources or has branching diets. Lindeman's 1942 paper grouped feeding relationships into trophic levels, compared transfers and cellular respiration, and emphasized the ecosystem as a unit connecting living and nonliving components. That viewpoint made different organisms comparable through a common energy budget. See Lindeman, The Trophic-Dynamic Aspect of Ecology (1942).

Material recycling and energy flow describe different bookkeeping. Carbon, nitrogen and phosphorus can be reused through chemical reactions and transfers between organisms and their surroundings. Energy transferred as heat is conserved, but is no longer the same store of chemical energy available to a consumer; photosynthetic primary producers need continuing light input. A decomposer helps recycle material and also releases heat through metabolism. Thus loss from one feeding budget means transfer into other forms or routes, never destruction of energy. See Clark, Douglas and Choi, Biology 2e, 46.3: material cycles. Conservation of energy applies to the whole accounting.

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

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

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

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

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

Entropy is an extensive quantity and a state function of an equilibrium system, measured in $\mathrm{J}{\mathrm{K}}^{-1}$. Differences between equilibrium states are calculated along a reversible thermodynamic process:Here heat entering the system is positive and $T$ is its absolute thermodynamic temperature. The reversible path used for calculation need not be the path actually taken. Dividing the actual heat by a single temperature generally fails for an irreversible process.

For example, put two rigid bodies, each with constant heat capacity $C=100\mathrm{J}{\mathrm{K}}^{-1}$, at $600\mathrm{K}$ and $300\mathrm{K}$ in an otherwise isolated system. Assume each body has a uniform internal temperature, no phase transition occurs, and no work is extracted. Their final thermal equilibrium is at $450\mathrm{K}$, because the internal energy lost by one equals that gained by the other. Using entropy from heat capacity,so $\Delta S_{\mathrm{tot}}=C\log (9/8)=11.78\mathrm{J}{\mathrm{K}}^{-1}$. One body's entropy decreases while the total increases. ^[ref]

In statistical mechanics, Boltzmann's entropy formula counts equally likely microstates compatible with a macrostate; Gibbs entropy handles a specified probability distribution over microstates. If the number of equally likely microstates doubles, entropy increases by $k_B\log 2$. The logarithm makes multiplicative counts additive for independent systems. These are precise statements about states and probabilities, not a universal measure of visible messiness. Calling entropy merely "disorder" obscures the thermal calculation above and hides which microstates are being counted. ^[ref]

The usefulness of entropy is that measurable heat capacities determine its changes, and the Second law of thermodynamics turns those changes into constraints on processes. It also connects macroscopic measurements to microscopic descriptions without requiring us to track every particle.

For the insulated gas-only free expansion of an ideal gas, the equilibrium endpoints are $(T,V_1)$ and $(T,V_2)$ with $V_2>V_1$. Because entropy is a state function, calculate their difference along a hypothetical reversible thermodynamic process: an isothermal expansion using a frictionless piston and a thermal reservoir at $T$. This comparison process receives heat and delivers work, even though the actual free expansion does neither. Its reversible heat increment is $\delta Q_{\mathrm{rev}}=pdV=nRTdV/V$, soHere $R$ is the molar gas constant, and the logarithm acts on a dimensionless ratio. The reversible reference gives the correct endpoint entropy difference because entropy is a state function; putting the actual $Q=0$ into $\Delta S=Q/T$ would misuse a relation requiring reversible heat transfer. For the actual isolated gas, all this increase is entropy production. Thus an adiabatic process can increase entropy and need not be an isentropic process. ^[ref]

In the figure's volume doubling, the ideal gas law gives $p_1=99.768\mathrm{kPa}$ and $p_2=49.884\mathrm{kPa}$, while $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. The reference process receives and delivers $nRT\ln 2\simeq 1728.85\mathrm{J}$ as heat and work; its thermal reservoir loses the same entropy that the gas gains. The actual free expansion has $Q_{\mathrm{in}}=W_{\mathrm{out}}=0$ and produces that entropy internally. Reversibly compressing the gas back at $300\mathrm{K}$ would require $1728.85\mathrm{J}$ of work input and rejection of equal heat. Restoring the gas alone therefore changes the surroundings: the actual release is irreversible.

For a real gas the final temperature may differ, so the ideal-gas logarithm formula cannot simply be copied. Evaluate $S(T_2,V_2)-S(T_1,V_1)$ using that gas's equilibrium properties or a suitable reversible comparison path. If the walls and water exchange heat with it, the Second law of thermodynamics constrains the total entropy of the isolated apparatus; the gas alone is then not an isolated system.

The entropy density of a photon gas is its entropy per physical volume, $s=S/V$. For homogeneous equilibrium vacuum radiation with two states of electromagnetic polarization and zero photon chemical potential,The first equality is the entropy density at zero chemical potential, obtained from the thermodynamic Euler relation $U=TS-pV+{\mu }_{\gamma }{N}_{\gamma }$ with internal energy $U=uV$. The second uses the photon gas equation of state, $u=a_R{T}^4$ and $p=u/3$. If an independently constrained photon number gives a nonzero chemical potential, the general entropy density is $(u+p-{\mu }_{\gamma }{n}_{\gamma })/T$ and $u=a_R{T}^4$ need not hold. This distinction explains why the absence of a conserved number matters for entropy as well as for the spectral shape. ^[ref]

A short first law of thermodynamics check, for neighboring equilibrium states with ${\mu }_{\gamma }=0$, isThe thermodynamic Euler relation fixes the additive normalization. Equivalently, this radiation entropy tends to zero with temperature at fixed volume; no contribution from the containing walls is included. Interpreting $TdS$ as transferred heat and $pdV$ as boundary work requires a reversible thermodynamic process. For a reversible thermodynamic process that is also an adiabatic process, there is no entropy production or entropy exchange, soAn irreversible adiabatic process can produce entropy even without heat exchange, so the word adiabatic alone is insufficient to infer $T^{3}V=\text{constant}$.

For an analytic example, expand an initially equilibrated photon gas reversibly from $V_0$ to $8V_0$, keeping only the same two photon states of electromagnetic polarization. Its temperature halves, its energy density becomes $u_0/16$, its entropy density of a photon gas becomes $s_0/8$, and its total entropy stays $S_0$. The internal energy halves rather than remaining fixed: the missing half leaves as work against radiation pressure,This is a stipulated reversible radiation model, not a report of a piston experiment. Maintaining a real cavity along this path requires controlling wall heat exchange and including the wall heat capacity in any actual apparatus balance.

For a computed scale at stipulated $T=2.725\mathrm{K}$, $s\simeq 2.04122\times 10^{-14}\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{m}}^{-3}$. Combining this with the photon number density givesThis is the total equilibrium radiation entropy divided by its mean photon number, not a separately assigned entropy carried by each labeled particle. During the ideal expansion, $n_{\gamma }\propto T^3$ and $T^{3}V$ is fixed, so the mean number $N_{\gamma }=n_{\gamma }V$ also stays constant. This follows from the path; it was not imposed as the independent number constraint used to establish chemical potential.

The beautiful connection is that expansion lowers each typical photon energy and dilutes the radiation in just the proportions needed to retain its equilibrium spectral shape and total entropy. For the cosmic microwave background, homogeneous expansion of the universe has physical volume proportional to the cube of the scale factor; fixed-species cosmological entropy conservation therefore gives $T\propto a^{-1}$. This describes preservation of an already established Planck photon distribution and connects the FIRAS spectral observation in photon gas to an entropy prediction; FIRAS did not perform a direct entropy measurement. When other equilibrated species disappear and transfer energy into radiation, use the effective number of relativistic entropy degrees of freedom, $g_{\ast s}$: the conserved quantity for the combined sector is $g_{\ast s}{T}^{3}V$, under the corresponding equilibrium and no-entropy production assumptions. A changing species population, separate decoupled sectors or entropy production requires an appropriate entropy balance rather than the pure photon gas formula. ^[ref]

At fixed volumes and material inventories, two internally equilibrated bodies that exchange only heat within an isolated system have fixed total internal energy $E=E_1+E_2$. Their total entropy isUsing $1/T_i=(\partial S_i/\partial E_i{)}_{V_i,N_i}$, an interior stationary point obeysThus entropy stationarity requires matching thermodynamic temperatures. Positive values of heat capacity at constant volume make this a strict local maximum becauseFor bodies with positive heat capacities throughout the accessible range, this concavity gives at most one interior maximum. This is the stability check behind the equal-temperature condition; see entropy concavity and positive heat capacity. ^[ref]

For thermal equilibration of two finite bodies, entropy from heat capacity givesBody 1 gains about $0.22314355\mathrm{kJ}{\mathrm{K}}^{-1}$ of entropy, while body 2 loses about $0.19361556\mathrm{kJ}{\mathrm{K}}^{-1}$. Total energy is conserved even though total entropy increases. The entropy calculation uses reversible comparison paths between the endpoint states; actual exchange across a finite temperature difference is irreversible.

To construct the plot, vary $T_1$ while enforcing the energy constraint, rather than holding $T_2$ fixed. In numerical units of kelvin, $T_2/\mathrm{K}=(1500-T_1/\mathrm{K})/3$, and the plotted function isHere $\mathcal{S}$ means $S_{\mathrm{tot}}-S_{\mathrm{tot},\mathrm{initial}}$, so its zero is a reference choice. The chain rule yieldsThe unique maximum is at $T_1=T_2=375\mathrm{K}$. The physical trajectory starts at $T_1=300\mathrm{K}$ and moves toward that maximum; the rest of the curve compares other possible energy partitions, not later times in the trajectory.

The contact model also proves the direction of change directly:For $G>0$ and positive temperatures, equality holds exactly at thermal equilibrium. The appealing link is that a dynamical heat transfer law moves the system uphill on the constrained entropy curve, while conservation of energy keeps it on that curve. This predicts both why the endpoint is stable and why the reverse separation into hot and cold bodies cannot happen spontaneously in the isolated system.

The first law applies conservation of energy to a thermodynamic system. For a closed system with unchanged bulk kinetic energy and potential energy, writeHere $U$ is internal energy, $Q>0$ is heat entering the system and $W>0$ is work done by the system on its surroundings. Compression therefore gives negative $W$. Some references define work positive when done on the system; their plus sign represents the same balance with a different convention. If bulk kinetic energy or potential energy changes, include it in the total energy balance. ^[ref]

Supplying $1000\mathrm{J}$ of heat while a gas does $300\mathrm{J}$ of work increases its internal energy by $700\mathrm{J}$. Compressing it with $200\mathrm{J}$ of input work in an adiabatic process instead gives $Q=0$, $W=-200\mathrm{J}$ and $\Delta U=200\mathrm{J}$.

The distinction between $d$ and $\delta$ records state versus path: internal energy is a state function, whereas heat and work depend on the process. For an ideal gas expanding reversibly at constant temperature, $W=N{k}_{B}T\log (V_2/V_1)$ and $Q=W$, since internal energy of an ideal gas depends only on temperature. Choose $N{k}_{B}T=1000\mathrm{J}$ and $V_2=2V_1$: then $Q=W=693.15\mathrm{J}$. An insulated free expansion into vacuum between the same equilibrium endpoints has $Q=W=0$. Both have $\Delta U=0$, yet their transfers differ. Their common entropy increase is $N{k}_B\log 2$. ^[ref]

For a simple compressible equilibrium system of fixed composition, combining the laws givesHere $p$ is pressure and $V$ is volume. This identity relates neighboring equilibrium states. Identifying $TdS$ with actual heat and $pdV$ with actual work requires a suitable reversible thermodynamic process; an irreversible piston calculation uses the external resisting pressure for boundary work. This distinction prevents applying a reversible ideal-gas adiabat to an irreversible adiabatic piston compression.