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(a) Treat the lake as a heat bath at fixed temperature $T_0=288.15\mathrm{K}$ and the block's heat capacity as constant. Its initial temperature is $T_i=473.15\mathrm{K}$, and $C=mc=355\mathrm{J}{\mathrm{K}}^{-1}$. Although the cooling is irreversible, entropy is a state function, so evaluate the block's change along an imagined reversible thermodynamic process:The lake receives heat $Q=C(T_i-T_0)=65675\mathrm{J}$, and thereforeAdding the two contributions givesThe positive result is consistent with the Second law of thermodynamics for heat transfer across a finite temperature difference. The calculation uses $c=710\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ exactly as printed in both the original PDF and the transcription; it does not replace the supplied value by a tabulated value for lead.

(b) The block starts and eventually finishes at the lake's temperature, so its net entropy change is approximately zero. Neglecting buoyancy, air resistance and the final immersion depth, the lost gravitational potential energy $mgh$ becomes internal energy in the lake after the motion and waves have dissipated. The lake remains at $T_0$, givingTemporary heating of the block or temporary storage in kinetic energy does not change this final energy and entropy balance once thermal equilibrium is restored.

Entropy is a thermodynamic state function. For a reversible heat transfer, $dS=\delta Q_{\mathrm{rev}}/T$; the Clausius inequality generalizes this to irreversible processes. The combined entropy of a system and its surroundings does not decrease. A subsystem can lose entropy by exporting heat or matter, so the second-law statement must specify the whole isolated combination. In a simple compressible system,which identifies temperature, pressure and chemical potential as derivatives of the energy representation of the state function. Thermodynamic equilibrium maximizes entropy under fixed total energy, volume and particle number. A heat reservoir instead leads to minimizing the Helmholtz free energy at fixed $T,V,N$, or the Gibbs free energy at fixed $T,P,N$.

Statistical mechanics explains the state count. In a microcanonical ensemble, equally probable compatible microstates give Boltzmann's entropy formula, $S=k_B\ln \Omega$. With general discrete probabilities, the Gibbs entropy isThe microcanonical formula follows by setting every nonzero probability to $1/\Omega$. For a canonical ensemble, $p_j=e^{-\beta E_j}/Z$, soThis turns microscopic energy levels and their populations into a measurable thermodynamic quantity. Classical phase-space counting needs a consistent microscopic normalization and indistinguishability convention; quantum state counting supplies that normalization. Independent systems have multiplicative state counts and therefore additive entropies.

For an ideal gas freely expanding from volume $V$ to $2V$ in an insulated enclosure, energy remains fixed while each particle gains twice the accessible position volume. The state count increases by $2^N$, giving $\boxed{\Delta S=N{k}_B\ln 2}$. The actual irreversible process transfers no heat. A reversible isothermal comparison path transfers heat $N{k}_{B}T\ln 2$ and computes the same entropy change because the initial and final thermodynamic equilibrium states are the same.

Calorimetry supplies another connection to experiment: at fixed pressure and within one phase, $S(T)-S(0)={\int }_0^T{C}_P(T^{\prime })d{T}^{\prime }/T^{\prime }$, with latent-heat contributions $\mathcal{L}/T$ at first-order transitions. The third law of thermodynamics fixes $S(0)=0$ for an ideal perfect crystal with a unique ground state. Degeneracy or frozen disorder can produce residual entropy. Entropy thus links heat-capacity measurements, phase thermodynamic equilibrium and the number of accessible microscopic configurations.

The Second law of thermodynamics states that total entropy of an isolated composite system cannot decrease. For a cyclic heat engine exchanging heat only with reservoirs at $T_h>T_c>0$, it gives $-Q_h/T_h+Q_c/T_c\ge 0$. Since $W=Q_h-Q_c$,The bound is attained by a reversible heat engine: no friction, no finite-temperature-difference heat exchange and no other entropy production. A Carnot cycle provides an implementation with two reversible isotherms and two reversible adiabats. All temperatures in these ratios are absolute.

Let $n_{\mathrm{mol}}$ be the fixed gas amount, $r=V_2/V_1>1$, $\Delta T=T_h-T_c$, and let $C_V$ denote the heat capacity of the whole sample. First take the unregenerated Stirling cycle, so that heat rejected during isochoric cooling is not stored and returned during isochoric heating. The reversible quasistatic gas work iswhile the heat inputs are $Q_{12}=n_{\mathrm{mol}}R{T}_h\ln r$ and $Q_{41}=C_V\Delta T$. HenceThe strict inequality follows because $C_V\Delta T>0$ adds heat input without additional work. Quasistatic motion alone does not make heat exchange with a single hot or cold reservoir reversible while the gas temperature changes on an isochore. Indeed, using those two reservoirs gives total external entropy production $C_V\Delta T(1/T_c-1/T_h)>0$ per cycle.

The strict comparison requires the no-regeneration assumption. A perfect reversible regenerator stores $C_V\Delta T$ during $2\to 3$ and returns it during $4\to 1$; only the isothermal heat transfers then enter the external heat budget, giving $\eta =1-T_c/T_h$ exactly. For example, perfect regeneration with $T_h=600\mathrm{K}$, $T_c=300\mathrm{K}$ and $r=2$ gives $\eta =1/2={\eta }_{\max }$. Thus the four gas paths by themselves do not enforce an efficiency strictly below the Carnot bound.

For the gas itself, the four entropy changes areExpansion increases the available positional states; compression reverses that change. Isochoric cooling reduces thermal populations and isochoric heating restores them. The sum is zero because entropy is a state function. This does not imply zero total entropy production in the gas plus external reservoirs.

Taking $S_1$ as reference, the isotherms are horizontal on a $T$–$S$ diagram. The isochore through state 1 obeys $S-S_1=C_V\ln (T/T_h)$, and the isochore through state 2 is shifted right by $n_{\mathrm{mol}}R\ln r$. For the internally reversible quasistatic gas paths, $\delta Q=TdS$, soThe enclosed area is the net work per engine cycle, with a clockwise orientation. It is not the total heat input.

For the refrigerator, the stated rates give the cooling coefficient of performance $\mathcal{C}={\dot{Q}}_c/P=1.9/0.1=19$. But the absolute reservoir temperatures are $T_c=277.15\mathrm{K}$ and $T_h=293.15\mathrm{K}$, so Carnot's theorem requiresAt $0.1\mathrm{kW}$ input the largest allowed cooling rate is only $1.7322\mathrm{kW}$. Equivalently the stated $2.0\mathrm{kW}$ heat rejection gives the reservoir entropy ratea direct violation of the Second law of thermodynamics for this cyclic two-reservoir device.

The unregenerated neon model makes the algebraic obstruction explicit. With $C_V/(n_{\mathrm{mol}}R)=3/2$, reversing the cycle givesSetting this to 19 produces $\ln r=-0.89385$, or $r=0.4091$, contradicting the required expansion ratio $r>1$. Perfect regeneration can raise performance only to the Carnot bound and does not fix the problem. There is no physically admissible $V_2/V_1$ for the stated data.

With a thermally insulating boundary, $\delta Q=0$. If the slow piston motion is frictionless and reversible, it is an isentropic process:The first law of thermodynamics gives $dU=-pdV$, so expansion lowers the internal energy and temperature. For constant heat capacity ratio $\gamma$, $p{V}^{\gamma }$ and $T{V}^{\gamma -1}$ are constant combinations of state functions. Insulation alone does not guarantee constant entropy: an irreversible adiabatic process can produce entropy. The conclusion uses the intended reversible, quasistatic piston idealization.

A state function is determined by the equilibrium state, rather than the path used to reach it. Its differential is exact and its change around a closed thermodynamic cycle is zero. Internal energy, entropy and enthalpy are examples; transferred heat and work are not state functions.

For slow expansion in contact with a large thermal reservoir, the fixed temperature of that reservoir maintains an isothermal process. For a fixed amount of ideal gas, internal energy and enthalpy depend only on temperature, soThe ideal gas law also gives $pV=\mathrm{constant}$. For a reversible expansion the absorbed heat equals the work done by the gas, while its entropy increases by $n_{\mathrm{mol}}R\ln (V_2/V_1)$; constant internal energy does not mean zero heat transfer.

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.

A state function assigns a value to each equilibrium thermodynamic state of a specified thermodynamic system. Its change depends only on the initial and final states. Internal energy $U$, entropy $S$ and enthalpy $H=U+pV$ are state functions; here $p$ is pressure and $V$ is volume. Transferred heat $Q$ and work $W$ belong to a thermodynamic process, rather than to a state. A system stores internal energy; it does not store separate inventories of “heat received” and “work received”.^[ref]

Use the explicit convention $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work done by the thermodynamic system on its surroundings. For a closed system with unchanged endpoint bulk kinetic energy and external potential energy, the first law of thermodynamics readsThe notation $dU$ denotes an exact differential of a state function; $\delta Q$ and $\delta W_{\mathrm{by}}$ denote incremental transfers. Along a reversible thermodynamic process of a simple compressible system,where $T$ is absolute temperature. These are rules for particular thermodynamic processes, not definitions of stored heat or stored work. In particular, a quasistatic process with friction can remain near thermodynamic equilibrium while producing entropy.^[ref]

For any state function $F$ and two equilibrium states $A,B$, an integral of its exact differential givesThe proof is telescoping: add the changes $F(x_{i+1})-F(x_i)$ along a subdivided path, and all intermediate values cancel. The smooth limit gives the integral. Conversely, if an integrable differential form has zero integral around every accessible closed path in a connected region, define $F(B)$ by integrating from a reference state. Two routes differ by a closed path, so the definition is unambiguous. This argument needs every accessible loop, rather than a local test that might miss a hole in the domain; see path independence.

The same cancellation does not hold separately for heat and work. A thermodynamic cycle has $\Delta U=0$ but can have nonzero net transfers, with $\oint \delta Q=\oint \delta W_{\mathrm{by}}$. Nor does the fact that entropy is a state function make $Q/T$ its general change: the formula $dS=\delta Q_{\mathrm{rev}}/T$ uses a reversible thermodynamic process and its varying temperature.

This distinction makes endpoint calculations powerful. We can compute an internal energy, entropy or enthalpy difference along a convenient reversible reference path even when the actual thermodynamic process is irreversible. Its actual heat and work must still be obtained from its own boundary interactions. The appealing cancellation is visible in two reversible ideal-gas paths between the same states: a different mechanical output requires a matching difference in thermal input, while stored internal energy arrives at the same value.

A free energy is a thermodynamic potential that compares possible states under specified environmental constraints. For a closed system of fixed composition, treated as a simple compressible system with only pressure-volume work, stable thermodynamic equilibrium minimizes Helmholtz free energy at fixed temperature and volume, or Gibbs free energy at fixed temperature and pressure. The choice follows the experiment: a rigid sealed vessel and a sealed cylinder with a freely moving loaded piston impose different constraints. Both may exchange heat with a thermal reservoir. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19.

The useful insight is that internal energy alone cannot rank these alternatives: exchanging heat also changes the surroundings' entropy. Subtracting $TS$ packages that accounting into a property of the chosen system. Adding $pV$ additionally accounts for displacing the surroundings at fixed pressure. A free energy is a state function; supplied heat describes a transfer during a process. Under the stated constraints, supplied heat is $Q=\Delta U$ at fixed volume or $Q=\Delta H$ at fixed pressure, whereas neither $\Delta F$ nor $\Delta G$ generally equals $Q$.

A lower free energy identifies a thermodynamic preference, without specifying a timescale. A barrier can trap a system in a local minimum, producing metastability although a lower global minimum exists. The enzyme example illustrates the corresponding distinction for chemical reactions: changing the route can change the reaction rate while preserving the endpoint Gibbs free energy difference and chemical equilibrium.

A state function used to describe a thermodynamic system with a chosen set of natural variables. Internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy use different variables and serve different environmental constraints.

A thermodynamic process changes a thermodynamic system between states. A quasistatic process supplies equilibrium intermediate states and therefore a path on equilibrium coordinates. An irreversible thermodynamic process can have equilibrium endpoints while lacking a uniform intermediate temperature or pressure. Endpoint values still determine changes in a state function; computing heat and work requires the actual interaction history.

Thermodynamics studies energy transfers and the constraints on macroscopic changes. Its vocabulary connects internal energy, entropy, temperature, pressure and volume to heat and work. The first law of thermodynamics balances energy; the Second law of thermodynamics decides which changes can occur spontaneously and limits how much work a heat engine can deliver. These principles let us bound performance before choosing a material or a detailed machine design. Applications include power generation, propulsion and refrigerators. ^[ref]

Start by choosing a system boundary. A closed system can exchange energy but no matter; an isolated system exchanges neither. A state function, such as internal energy or entropy, depends on the equilibrium state. A process specifies how that state changes. Thus a body's temperature is a property it has, whereas heat records energy crossing its boundary during a process. A body contains internal energy, rather than a stored quantity of heat.

Two insulated bodies initially at different positive temperatures approach thermal equilibrium when brought into thermal contact. Their combined energy is conserved, but their combined entropy increases. They do not spontaneously separate again into a hotter body and a colder one. This familiar asymmetry makes thermodynamics interesting: conserving energy leaves many imaginable changes available, and entropy supplies a further constraint. The same distinction underlies both entropy maximization under thermal contact and the thermal efficiency limit in Carnot's theorem.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.