Entropy and the Second Law

Entropy change and thermal reservoirs

The first law balances energy transfers without a criterion for their spontaneous direction. Comparison of initial and final energies alone permits heat transfer in either direction. Experience and the second law add the direction condition: heat flows spontaneously from a higher-temperature body to a lower-temperature body, not from cold to hot without other changes. Entropy is the state quantity used to express this condition quantitatively.

Reversible transfer of heat $Q$ to a system held at absolute temperature $T$ changes entropy by $Δ S = Q / T$ . The unit is joules per kelvin. A reversible transfer is an ideal limiting process in which the temperature difference between the transferring bodies is infinitesimal, so reversing that difference reverses the transfer without leaving a net change elsewhere. The formula defines entropy change through a reversible reference path even when the actual process is irreversible. Entropy itself is a state function; heat transfer is not.

A large thermal reservoir remains effectively constant in temperature during heat exchange with a system. If the system receives $Q$ , the reservoir loses the same energy and has entropy change $- Q / T_{res}$ . The entropy change of the combined system and reservoir determines the second-law result. In a reversible exchange at the same temperature, the positive and negative changes sum to zero. In an irreversible exchange across a finite temperature difference, the total is positive. A negative total entropy change for an isolated combination is not permitted by the second law.

Heat capacity connects entropy change to a measurable temperature path. For a sample of approximately constant heat capacity $C$ heated reversibly from $T_{i}$ to $T_{f}$ , integration gives $Δ S = C ln (T_{f} / T_{i})$ . The logarithm requires absolute temperatures. A body with positive heat capacity gains entropy when heated and loses entropy when cooled. The result depends only on the endpoint temperatures for the specified heat-capacity model, even though the actual heating apparatus can transfer energy at different rates.

Entropy accounting includes both bodies exchanging heat. Each divides the transferred

Q

by its own temperature, so transfer from a hotter to a colder body gives a positive total change

Q (1/ T_{C} - 1/ T_{H})

Heat engines and the second law

A heat engine operates cyclically between a hot reservoir and a cold reservoir. During one cycle it absorbs heat $Q_{H}$ from the hot reservoir, delivers work $W$ to the surroundings, and rejects heat $Q_{C}$ to the cold reservoir. Because the engine returns to its starting state, its internal-energy change over the cycle is zero. The first law then gives $W = Q_{H} - Q_{C}$ . A heat engine cannot convert all of the absorbed heat into work while operating in a cycle with only one thermal reservoir; some energy must be rejected to a lower-temperature reservoir.

Thermal efficiency is the useful work divided by heat absorbed from the hot reservoir: $e = W / Q_{H} = 1 - Q_{C} / Q_{H}$ . It is a ratio, not an energy-conservation law. The first law permits $Q_{C} = 0$ algebraically, but the second law excludes such a cyclic engine. The limit is established by comparing the engine with a reversible Carnot engine operating between the same reservoir temperatures. No engine can exceed the Carnot efficiency $e_{C} = 1 - T_{C} / T_{H}$ , with both temperatures measured in kelvins.

The Carnot expression depends only on reservoir temperatures, not on working-gas identity, cylinder size, or the amount of heat processed per cycle. A hot reservoir at $600 K$ and a cold reservoir at $300 K$ set an upper efficiency bound of $0.50$ . An actual engine has lower efficiency because friction, finite-temperature heat transfer, turbulence, and incomplete combustion or other losses generate additional entropy. The Carnot value is a benchmark for a specified temperature pair, not a performance prediction for an ordinary machine.

Refrigerators use work to move heat from a cold region to a hot region. Their operation does not violate the second law because the required work and the heat delivered to the hot reservoir are included in the total account. A claim that heat has moved from cold to hot must therefore state what work source or other external change accompanied the transfer. The system boundary determines whether that work is visible in the equation.

A heat engine absorbs energy from a hot reservoir, rejects part to a colder reservoir, and delivers the difference as work

W = Q_{H} - Q_{C}

. The mandatory rejection is the second-law limit on cyclic heat-to-work conversion.

Reversible and irreversible processes

A reversible process is an ideal sequence of equilibrium states that can be reversed by an infinitesimal change in external conditions. It has no friction, no finite temperature jump during heat transfer, and no unrestrained expansion or mixing. Reversibility is a limiting model, not a claim that ordinary processes can be performed with zero elapsed time or zero engineering complexity. Its value is that it sets the maximum-work or maximum-efficiency reference for processes between specified equilibrium states.

Irreversible processes include heat flow across a finite temperature difference, frictional sliding, viscous flow, diffusion, electrical resistance, and free expansion. These processes produce positive entropy change for an isolated system. Direct heat transfer of $Q$ from a reservoir at $T_{H}$ to one at $T_{C}$ gives combined entropy change $Q (1/ T_{C} - 1/ T_{H})$ , which is positive when $T_{H} > T_{C}$ . The energy transfer is the same amount entering and leaving, but the entropy changes do not cancel because the temperatures differ.

The phrase entropy increases must specify the system. A cooling hot object has negative entropy change; a warming cold object has positive entropy change. The second law applies to the total of all participating systems and surroundings. An open device can reduce its own entropy while exporting a larger entropy increase by heat, matter, or work-related dissipation. A correct balance therefore names the objects included and all transfers crossing the boundary.

An irreversible path cannot generally be replaced by a reversible path for work calculation without accounting for the difference in transfers. It can, however, use a reversible path to calculate the entropy change between the same equilibrium end states because entropy is a state function. This distinction is central: use the actual path for work and heat bookkeeping; use a convenient reversible path only to evaluate the entropy difference. Confusing the two uses turns a state calculation into an incorrect claim about the real apparatus.

Clausius inequality for thermal cycles

For any cyclic process, the Clausius inequality states that $\oint δ Q / T \leq 0$ , where each heat-transfer element is divided by the absolute temperature of the boundary at which that transfer occurs. Equality holds for a reversible cycle. A strictly negative value identifies irreversible effects within the cycle or at the interfaces through which heat is transferred. The inequality does not replace the first law: it adds a restriction on the possible combinations of heat transfers and temperatures after energy conservation has been satisfied.

The sign of $δ Q$ follows the selected working system. Heat entering the system is positive, and heat leaving is negative. A simple engine absorbs positive heat from the hot reservoir and rejects negative heat to the cold reservoir. Because the cold-reservoir temperature is smaller, the rejected term has a larger magnitude after division by temperature unless the process is the reversible limiting case. A reversible engine has two entropy-transfer terms whose sum is zero over a cycle because the working substance returns to its initial state.

An irreversible engine has additional positive entropy production. Friction, viscous flow, finite temperature differences in heat exchangers, electrical resistance, and unrestrained expansion all contribute. The working substance can still return to its original state after a cycle, so its own net entropy change is zero, but the reservoir entropy changes no longer cancel. The total entropy change of reservoirs and apparatus is positive. This is the physical content hidden by a strict Clausius inequality rather than an equality.

The inequality is also a diagnostic for proposed machines. If measured heat transfers in a complete cycle give a positive value for $\oint δ Q / T$ , the signs, temperature assignments, or measurements are inconsistent. If the result is zero within uncertainty, the data are compatible with a reversible limit but do not prove that every internal process is reversible. A negative value is expected for ordinary machines and quantifies the departure from the ideal limit.

Entropy change of an ideal gas

Entropy change for an ideal gas can be evaluated from any convenient reversible path between the same equilibrium endpoints. With constant heat capacity at constant volume, the result is $Δ S = n C_{V, m} ln (T_{f} / T_{i}) + n R ln (V_{f} / V_{i})$ . The first term represents the entropy change associated with changing temperature at fixed volume; the second represents the change associated with volume at fixed temperature. The actual process need not follow those two artificial stages. They are used because entropy is a state function and the stages make the integral tractable.

For one mole of a monatomic ideal gas changing from $300 K$ and $10.0 L$ to $360 K$ and $15.0 L$ , $C_{V, m} = (3/2) R$ . The temperature term is $(3/2) R ln (360/300) = 2.27 J K^{- 1}$ , and the volume term is $R ln (15.0/10.0) = 3.37 J K^{- 1}$ . The total entropy change is $5.64 J K^{- 1}$ . Both increased temperature and increased volume raise the number of accessible molecular arrangements in this example.

The signs need not always agree. Isothermal compression has negative entropy change because volume decreases. Cooling at fixed volume also has negative entropy change. An expansion accompanied by enough cooling can have either sign depending on the relative sizes of the two logarithmic terms. The endpoint formula replaces informal descriptions such as expansion increases disorder with a quantitative comparison of temperature and volume changes.

The result requires an ideal gas and equilibrium endpoint states. It does not require the actual expansion to be reversible. A free expansion into vacuum is irreversible and has no simple boundary-work integral, yet the entropy change can still be obtained from a hypothetical reversible isothermal expansion connecting the same initial and final volumes. The entropy increase belongs to the state change; the irreversibility belongs to the actual path and appears in the total entropy balance with the surroundings.

An ideal-gas entropy change splits into a temperature term and a volume term. A reversible reference path may heat at fixed volume and then expand at fixed temperature, even when the real process follows the dashed direct route.

Real engines have efficiency below the Carnot value because their heat transfers occur across finite temperature differences and their moving parts dissipate energy. The Carnot relation should not be used with a working-gas temperature guessed from one point in a cycle; the relevant temperatures are those of the thermal reservoirs supplying and receiving heat. It is also not a formula for power. Power depends on how much heat is processed per cycle and how many cycles occur per unit time, whereas Carnot efficiency is a limit on energy conversion per cycle.

Reversible operation sets a benchmark; first-law energy balance still applies. The work output equals hot heat input minus cold heat output. The second law relates the two heat transfers through reservoir temperature, while the first law relates them through energy. Both statements are needed to derive the efficiency limit. Omitting either one can produce an algebraic answer that conserves energy but permits an impossible heat engine.

An entropy sign audit begins with the stated system. Heat entering a body at temperature $T$ contributes positively to that body's entropy by the reversible reference value $Q / T$ ; heat leaving contributes negatively. The same transfer must appear with the opposite energy sign in the other body. Adding those two changes before inserting reservoir temperatures prevents a local entropy decrease from being mistaken for a violation of the second law.

Entropy generation and the total balance

Entropy generation is the nonnegative contribution produced by irreversible processes. A closed-system balance sets entropy change equal to entropy transferred with heat plus entropy generated internally. The generated term is zero for a reversible limiting process and positive for friction, mixing, finite-temperature heat transfer, viscous flow, electrical resistance, and other dissipative effects. It is not an additional form of energy. Energy remains conserved through the first law; entropy generation records the loss of available ways to direct that energy into useful work.

Heat transfer between two reservoirs gives a direct calculation. Let $500 J$ flow from a $500 K$ reservoir to a $300 K$ reservoir. The hot reservoir entropy change is $- 500/500 = - 1.00 J K^{- 1}$ . The cold reservoir entropy change is $+ 500/300 = + 1.67 J K^{- 1}$ . The combined entropy generation is therefore $+ 0.667 J K^{- 1}$ . No energy has disappeared: the same $500 J$ leaves one reservoir and enters the other. The positive total entropy identifies the transfer as irreversible.

Entropy generation depends on the actual mechanism, whereas the entropy change of an equilibrium system depends only on its endpoints. A gas can be brought between two states by a slow sequence of small temperature differences or by direct contact with a much hotter reservoir. The endpoint entropy change of the gas can be the same, but the surroundings entropy change and generated entropy differ. An entropy balance therefore includes the sample, reservoirs, and every other interacting body whose entropy changes.

The generated term sets a performance bound. In a device receiving heat from a finite temperature source, any positive entropy generation reduces the work that could have been obtained in a reversible limit. Reducing friction or using smaller temperature differences can reduce generation, but cannot make a real finite-rate device perfectly reversible. The sign is a practical data check: a calculated negative generation for an isolated combination indicates missing heat transfer, an incorrect temperature, or a sign error.

Refrigerators, heat pumps, and coefficient of performance

A refrigerator uses work input to remove heat $Q_{C}$ from a cold region and reject $Q_{H}$ to a warmer region. The first-law cycle balance is $Q_{H} = Q_{C} + W_{in}$ . The desired output of a refrigerator is cooling, so its coefficient of performance is $COP_{R} = Q_{C} / W_{in}$ . A coefficient of performance can exceed one because it compares heat moved to work supplied; it is not a heat-engine efficiency and is not limited by unity. The work input enables energy already present in the cold region to be pumped uphill in temperature.

A heat pump uses the same physical cycle but has a different useful output: heat delivered to the warm region. Its coefficient is $COP_{H P} = Q_{H} / W_{in} = COP_{R} + 1$ . The same appliance can be described as a refrigerator or a heat pump depending on whether cooling the cold space or warming the hot space is the intended service. The reservoir labels and the chosen useful output must be written before selecting a performance ratio.

A reversible refrigerator has the maximum coefficient for reservoir temperatures $T_{C}$ and $T_{H}$ : $COP_{R, C} = T_{C} / (T_{H} - T_{C})$ . The reversible limit for a cold space at $270 K$ with surroundings at $300 K$ is 9.0, while the heat-pump limit is 10.0. Actual equipment has lower values because compression losses, pressure drops, finite temperature differences in heat exchangers, and electrical losses generate entropy. A small temperature lift gives a high theoretical COP; this does not remove the need for work input.

The second law excludes a refrigerator whose only effect is to move heat from cold to hot. Adding work makes the process possible because the work source and the extra heat delivered to the hot reservoir are included in the total change. A complete refrigerator calculation checks both the energy balance and the entropy balance. The first law alone would allow arbitrary heat flows if work were adjusted; the second law restricts the minimum work required for a given temperature pair.

A refrigerator uses work input to move heat

Q_{C}

out of a cold region and reject a larger

Q_{H} = Q_{C} + W

to the warm surroundings. Its coefficient of performance compares cold-side heat removal with the required work.

Kelvin--Planck and Clausius statements.

The Kelvin--Planck form of the second law states that no cyclic engine can have as its sole effect the absorption of heat from a single reservoir and the conversion of all of that heat into work. A cyclic engine must reject some heat or produce another change outside itself. This statement rules out a perpetual-motion machine of the second kind. It does not prohibit conversion of stored chemical, electrical, or mechanical energy entirely into work; the restriction concerns a cyclic device whose only energy source is one thermal reservoir.

The Clausius form states that no process can have as its sole effect the transfer of heat from a colder body to a hotter body. Refrigerators appear to contradict this only when their work input is omitted. Their sole effect is not heat transfer from cold to hot: they also consume work and reject a larger amount of heat to the warm side. The statement makes the required external change explicit.

The two forms are equivalent. If a device violated the Clausius statement by moving heat from cold to hot without work, it could be coupled to an ordinary heat engine so that the cold-reservoir heat rejection is returned freely to the hot reservoir. The combined apparatus would then convert heat from one reservoir completely into work, violating Kelvin--Planck. Conversely, a Kelvin--Planck violation could power a refrigerator without external work and violate Clausius. The equivalence shows that both statements express the same direction restriction in different language.

These statements are more specific than a vague instruction that entropy must increase. They identify an impossible claimed device by listing its sole effect. When another change is present, such as work input, fuel consumption, or a second reservoir, the first and second laws must be applied to the full arrangement. The wording prevents an incomplete system boundary from being mistaken for a physical violation.

The Kelvin–Planck statement forbids a cyclic engine drawing on one reservoir and converting all of that heat to work; the Clausius statement forbids heat passing from cold to hot with no work. Both name a device by its sole effect.

Entropy balance for a closed system.

A closed system separates entropy change into transfer and generation: $Δ S_{sys} = \int δ Q / T_{b} + S_{gen}$ . The boundary temperature $T_{b}$ is the temperature at the location where the heat crosses the system boundary. The generation term satisfies $S_{gen} \geq 0$ . It is zero only in a reversible limit. The balance is valid for an actual irreversible process; unlike the reversible reference formula $d S = δ Q_{rev} / T$ , it does not pretend that the real heat transfer occurred through an infinitesimal temperature difference.

An insulated closed system has no entropy transfer with heat, so its entropy change equals its internally generated entropy. A gas freely expanding into an evacuated region is a standard example. No heat crosses the outer boundary and no boundary work is done against an external pressure, yet the gas entropy increases. The increase is not created from energy. It records the irreversible spreading of the molecular distribution into a larger accessible volume. The first law gives zero change in internal energy for an ideal-gas free expansion, while the second law gives positive entropy generation.

A system receiving heat $Q$ across a boundary held at temperature $T_{b}$ receives entropy $Q / T_{b}$ . If the system entropy change is larger than this transfer term, the difference is the generated entropy. If it is smaller, the calculation has omitted an entropy transfer out or used the wrong boundary temperature. The balance applies directly to real heat exchangers, where a finite temperature difference generates entropy even when the energy transferred from one stream to the other is equal and opposite.

The boundary must remain fixed during the accounting interval. Moving a piston can perform work but does not itself carry entropy across the boundary in the way heat does. Matter crossing an open-system boundary carries entropy with it and requires additional terms; that case is outside the closed-system form used here. Declaring the system as closed, insulated, or in thermal contact is therefore a mathematical choice with physical consequences, not a label added after the calculation.

A Carnot-cycle calculation

A reversible Carnot engine gives a complete numerical use of both thermodynamic laws. Let the hot and cold reservoirs have temperatures $T_{H} = 500 K$ and $T_{C} = 300 K$ . During one cycle the engine absorbs $Q_{H} = 1200 J$ from the hot reservoir. The Carnot efficiency is $e_{C} = 1 - T_{C} / T_{H} = 0.40$ , so the maximum work output is $W = e_{C} Q_{H} = 480 J$ . The first-law cycle balance then requires cold-side heat rejection $Q_{C} = Q_{H} - W = 720 J$ .

The entropy calculation checks the result independently. The hot reservoir loses $Q_{H} / T_{H} = 1200/500 = 2.40 J K^{- 1}$ . The cold reservoir gains $Q_{C} / T_{C} = 720/300 = 2.40 J K^{- 1}$ . Their sum is zero, as required for a reversible cycle. The working substance also returns to its starting state, so its net entropy change over the cycle is zero. A nonzero total would signal either an irreversible process or inconsistent heat and temperature data.

The calculation shows why work is not the same as hot-side heat input. Forty percent of the absorbed energy appears as work only because the reservoir temperatures permit that reversible limit. The remaining sixty percent must be rejected to the cold reservoir. Replacing the cold reservoir by one at a lower temperature would raise the Carnot limit; increasing the hot reservoir temperature has the same effect. In either case, the engine still requires two reservoirs and a cyclic working substance.

Actual engine data can be compared with the same structure. Measured work divided by measured hot heat input gives an actual efficiency. It must not exceed the Carnot value computed from the reservoir temperatures. If it appears to exceed the limit, the usual causes are an unmeasured heat input, use of Celsius rather than kelvin temperatures, or a mismatch between the temperatures measured and the boundary temperatures at which heat transfer occurred.

Mixing and free-expansion entropy

Ideal-gas free expansion makes the difference between energy and entropy especially clear. One mole of an ideal gas expands from volume $V$ to $2 V$ into vacuum inside an insulated container. No heat enters, and no work is done against an external pressure. The first law therefore gives $Δ E_{int} = 0$ . An ideal gas has internal energy depending only on temperature, so the initial and final temperatures are equal. Nevertheless, the entropy change is $Δ S = n R ln (2 V / V) = R ln 2$ , about $5.76 J K^{- 1}$ for one mole.

The entropy increase can be computed with a hypothetical reversible isothermal expansion between the same endpoints. That reference path transfers heat and does work, unlike the actual free expansion, but it gives the correct state change. The actual process is irreversible because the gas does not spontaneously reassemble into the original half-volume region. The entropy generation in the insulated container equals the positive gas entropy change.

Mixing two different ideal gases follows the same logic. Removing a partition lets each gas occupy a larger accessible volume. For equal amounts initially occupying equal volumes, each species gains $n R ln 2$ of entropy, so the total mixing entropy is the sum of the two positive terms. There is no heat or work requirement in the idealized insulated mixing process. The increased entropy reflects the larger set of molecular arrangements compatible with the macroscopic mixture.

Mixing identical gases is different. Removing a partition between equal samples of the same gas produces no observable composition change and no entropy of mixing in the thermodynamic description. The distinction requires molecules of the same species to be treated as indistinguishable. This result prevents counting a macroscopically unchanged arrangement as a new accessible state on the basis of an imaginary molecular label attached before partition removal.

Free expansion and mixing raise ideal-gas entropy with no heat input and no boundary work. A reversible reference expansion gives the endpoint entropy change

n R ln 2

, while the actual partition removal generates that entropy internally.

Entropy calculations should state whether a quoted quantity is a system change, a reservoir change, or generated entropy. The same irreversible process can have a positive system entropy change and zero heat transfer, as in free expansion, while another process can have entropy transfer through heat with no internal generation in the reversible limit. Adding unlike terms without their system labels obscures the second-law balance rather than simplifying it.

Entropy of a phase change.

At a phase-change temperature and pressure, heat can be transferred reversibly while the temperature remains constant. The entropy change of the material is then $Δ S = Q_{rev} / T = m L / T$ , where $L$ is the appropriate latent heat. Melting and vaporization have positive entropy changes because heat enters the material; freezing and condensation have negative changes because heat leaves. The entropy formula does not treat the phase change as an exception to energy conservation. The latent heat is the energy transfer required to change molecular arrangement and interaction energy without changing average translational kinetic energy.

Melting $0.200 kg$ of ice at $273.15 K$ requires $Q = (0.200) (333.5 kJ k g^{- 1}) = 66.7 kJ$ . If the heat enters reversibly from a reservoir at the same temperature, the ice--water system entropy change is $66.7 kJ /273.15 K = 244 J K^{- 1}$ . The reservoir loses the same entropy, so the total change is zero. The two phases can coexist at the transition temperature because their entropy and energy changes are balanced by the latent-heat transfer under those equilibrium conditions.

Actual melting often occurs with a finite temperature difference between heat source and ice. A reservoir warmer than the melting temperature supplies the same latent heat, but its entropy loss has smaller magnitude because the heat is divided by a larger temperature. The ice entropy gain remains determined by its reversible reference path at the phase-change temperature. The combined change is positive; the difference is entropy generation caused by transfer across the finite temperature difference. The phase-change entropy is a state change, whereas the generation depends on the actual heat-transfer arrangement.

Phase-change entropy also explains why a temperature equation alone is incomplete. During melting, $m c Δ T$ is zero while $m L$ is not. A calorimetry problem must first determine whether heat is changing temperature, changing phase, or doing both in sequence. The associated entropy account follows the same staging: use $C ln (T_{f} / T_{i})$ for a temperature interval and $m L / T$ for a reversible phase change. Adding only one term undercounts the state change whenever both occur.

The sign should be attached to the selected material. Water freezing releases heat and has negative entropy change; the colder surroundings receiving that heat have positive entropy change. Spontaneous freezing requires the combined total to be nonnegative. A statement that a substance becomes more ordered is not a complete second-law calculation until the entropy of the surroundings that receive the latent heat is included.

The ledger separates energy from entropy quantities. Work has units of joules and does not carry an entropy-transfer term in this simple cycle account. Heat transfers have units of joules, while their associated reservoir entropy changes have units of joules per kelvin. Adding $Q_{H}$ and $Q_{C}$ produces an energy balance; adding $Q_{H} / T_{H}$ and $Q_{C} / T_{C}$ produces an entropy balance. Mixing the two equations is a dimensional error even when the numerical values happen to look comparable.

Loss mechanisms can be located qualitatively. Finite-temperature heat exchangers generate entropy at the hot and cold interfaces. Friction converts organized shaft work into internal energy. Flow resistance and incomplete combustion or electrical losses can add further generation. The ledger does not require resolving every microscopic mechanism to establish the total generation, but component measurements can identify which parts of a real engine most limit efficiency.

A reversible engine with the same $Q_{H}$ and reservoir temperatures would have cold rejection $Q_{C} = Q_{H} T_{C} / T_{H} = 750 J$ and work $750 J$ . Comparing this value with the real engine's $500 J$ shows the work lost to irreversibility for the stated heat input. The comparison is valid only for the same hot heat input and reservoir temperatures; changing either changes the available-work reference.

The table uses positive energy for energy entering the listed item.

part of the account	energy entry per cycle	entropy entry per cycle	value for this engine
hot reservoir	$- Q_{H}$	$- Q_{H} / T_{H}$	$- 1500 J$ , $- 2.50 J K^{- 1}$
cyclic working substance	$Q_{H} - Q_{C} - W = 0$	$Δ S_{ws} = 0$	returns to its initial state
cold reservoir	$+ Q_{C}$	$+ Q_{C} / T_{C}$	$+ 1000 J$ , $+ 3.33 J K^{- 1}$
work receiver	$+ W$	no entropy-transfer term for ideal work	$+ 500 J$
reservoirs and working substance	energy is balanced with the work receiver	$S_{gen}$	$+ 0.833 J K^{- 1}$

A real-engine ledger tracks energy and entropy separately. The cold-side entropy gain exceeds the hot-side loss, so the positive difference is entropy generation and the efficiency lies below the Carnot limit.

Worked example. A freezer at $255 K$ runs in a room at $295 K$ . In one hour the unit removes $Q_{C} = 1.20 MJ$ from the freezer and consumes $W_{in} = 0.480 MJ$ of electrical work, so it rejects $Q_{H} = Q_{C} + W_{in} = 1.68 MJ$ to the room. The coefficient of performance is $COP_{R} = Q_{C} / W_{in} = 2.50$ : cooling delivered per unit work, not a fraction of energy converted to work.

The reservoir entropy changes over the hour are $- 1.20 MJ /255 K = - 4.71 kJ K^{- 1}$ for the freezer and $+ 1.68 MJ /295 K = + 5.69 kJ K^{- 1}$ for the room, a generation of about $0.99 kJ K^{- 1}$ from dissipation and finite-temperature heat exchange.

The reversible limit for these temperatures is $COP_{R, C} = T_{C} / (T_{H} - T_{C}) = 255/ (295 - 255) = 6.38$ , so the real value of $2.50$ uses more than twice the minimum reversible work for this cooling load. Better insulation lowers the load $Q_{C}$ ; a better compressor and heat exchangers raise the COP at fixed load — distinct changes that should not be conflated.

An energy account must include the boundary selected for the freezer. Heat leaking through its insulation enters the cold compartment and becomes part of the cooling load. Electrical work entering the refrigerator crosses a different boundary and eventually appears as heat delivered to the room, together with the extracted freezer heat. A measurement of only the freezer temperature cannot determine COP: the heat load and electrical work must both be measured over a stated interval.

A freezer illustrates the Clausius statement. Heat moves from the colder freezer to the warmer room while electrical work enters the refrigerator. The room receives the sum of extracted heat and work input. The complete ledger obeys energy conservation and produces positive total entropy, as required by the second law.

Thermal ledgers are most reliable when all quantities are reported over one common time interval. Heat and work may be listed as energies per cycle, per hour, or as rates, but the entries must not be mixed. Dividing every entry by the same interval converts an energy ledger into a power ledger without changing the first-law relations. Entropy transfer and generation must similarly be reported per cycle, per hour, or as rates consistently. A refrigerator data sheet can otherwise combine a cooling rate with an electrical energy measurement and produce a meaningless COP.

Phase-change and reservoir calculations require the temperature at the actual heat boundary. The melting material may be at its phase-change temperature while the heating element is much warmer. Using the heater temperature for the material entropy change understates the increase; using the material temperature for the heater entropy loss overstates its magnitude. Their difference equals the entropy generated by finite-temperature transfer. The same boundary temperature distinction applies to real engine and freezer heat exchangers.

An audit of a complete device therefore proceeds in two passes. First, sum heat and work with signs to verify energy conservation for the chosen system. Second, divide each reservoir heat transfer by that reservoir temperature and add any internal generation to verify a nonnegative total entropy change. A device can satisfy the energy balance while failing the entropy audit. The second pass rules out an apparently efficient but physically impossible thermal machine.

Measurement uncertainty should be carried through both ledgers. A small uncertainty in a reservoir temperature affects every entropy-transfer term because temperature appears in the denominator. Heat-flow calibration affects the energy balance and the entropy balance simultaneously. A reported entropy generation that is smaller than its propagated uncertainty is compatible with a reversible limit but does not establish zero irreversibility. Consistent units, common intervals, and uncertainty estimates turn a schematic entropy ledger into a testable experimental account.