Physics · Thermodynamics

Entropy change in reversible isothermal heat exchange

The entropy change of a system is the integral of reversible heat divided by absolute temperature along a reversible comparison path; at constant temperature this reduces to $\frac{Q_{\mathrm{rev}}}{T}$ .

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Formula

Calculate

Δ S = \int_{i}^{f} \frac{δ Q _{re v}}{T}, T = const. : Δ S = \frac{Q _{re v}}{T}

ΔS in J/K; δQrev and Qrev in J; T in K.

Variables & units

Variables & units
Symbol	Meaning	Unit
$ΔS$	Change in system entropy	J/K
$Q_{\mathrm{rev}}$	Reversible heat transfer, with sign	J
$T$	Absolute thermodynamic temperature	K

LaTeX

\Delta S=\int_{i}^{f}\frac{\delta Q_{rev}}{T},\qquad T=\mathrm{const.}:\ \Delta S=\frac{Q_{rev}}{T}

QUANTA STUDYquanta-study.de

Two paths from 2 to 5 litres at equal final temperature. Top: free expansion, initially gas occupying two fifths with vacuum beside it. Bottom: reversible comparison. Entropy changes are equal but transferred heats differ. — The same ideal-gas endpoints at 340 K: free expansion in an isolated rigid vessel has Q = W = 0. The reversible isothermal comparison gives Qrev = Wrev ≈ 1942.6 J. Both paths have ΔS ≈ 5.7135 J/K.

Entropy change calculator

Find the entropy change for reversible heat exchange at constant temperature, plus for an ideal gas and for heating a body.

Calculation model

Find

Reversibly exchanged heat Q

Heat taken in (positive) or given off (negative) by the system, e.g. 33.4 kJ to melt 100 g of ice.

Absolute constant temperature T

Constant temperature during the heat exchange, e.g. 0 °C while ice melts; the calculation uses kelvin.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for ΔS, substitute, evaluate

1. Rearrange for ΔS
$\Delta S=\frac{Q_{\rm rev}}T$
2. Substitute values (in SI units)
- Qrev = 33.4 kJ = 33 400 J
- T = 0 °C = 273.15 K
$\frac{33400}{273.15}=122.277$
3. Result
ΔS = 122.3 J/K

Result: System entropy change ΔS

ΔS = 122.3 J/K

With Qrev = 33.4 kJ and T = 0 °C, ΔS = 122.3 J/K.

Change Reversibly exchanged heat Q

If Qrev doubles, ΔS doubles too: they are proportional.

Quick tips

ΔS = Q/T needs absolute temperature: convert °C to kelvin (0 °C = 273.15 K), otherwise melting ice would have you divide by zero.
The sign follows the heat: if the system takes in heat its entropy rises, if it gives heat away it falls. J divided by K gives J/K.

Assumptions and limits

Closed system and reversible heat exchange at one constant absolute temperature T>0. Qrev>0 means heat absorbed by the system; ΔS may be negative. This shortcut does not apply to nonisothermal paths or actual irreversible heat.

Derivation & background

Entropy is a state property. Change between equilibrium states follows from $\Delta S=\int\frac{\delta Q_{\mathrm{rev}}}{T}$ on a reversible comparison path. Only constant $T$ permits $\frac{Q_{\mathrm{rev}}}{T}$ . Actual irreversible heat and comparison heat may differ. Closed-system balance is $\Delta S=\int\frac{\delta Q_{\mathrm{actual}}}{T_b}+S_{\mathrm{gen}}$ , using local boundary temperature and $S_{\mathrm{gen}}\ge0$ . Individual system entropy can decrease; an isolated total cannot. Equal-temperature ideal-gas endpoints give $\Delta S=nR\ln\frac{V_2}{V_1}$ . Reversible phase transitions give signed contributions $\pm\frac{mL}{T}$ . Reversible closed comparison integrals vanish; an actual cycle satisfies $\oint\frac{\delta Q_{\mathrm{actual}}}{T_b}=-S_{\mathrm{gen}}\le0$ .

Exam blueprint

Validity range

Calculate ΔS between equilibrium states along a reversible comparison path, with T in kelvin. $\frac{Q_{\mathrm{rev}}}{T}$ requires constant T on that path. For actual irreversible processes, $\frac{Q_{\mathrm{actual}}}{T}$ is not generally ΔS; boundary temperatures and generation require separate accounting. A closed system may have negative ΔS while generation remains nonnegative.

Derivation steps

Calculate a state change on a reversible comparison path and only then factor out constant T.

1Specify two equilibrium states. State difference $\Delta S=S_2-S_1$ is independent of the actual path.
2Construct a reversible comparison path between exactly those states.
3On this path $dS=\frac{\delta Q_{\mathrm{rev}}}{T}$ , using absolute temperature and signed heat entering the system.
4Integration gives $\Delta S=\int_1^2\frac{\delta Q_{\mathrm{rev}}}{T}$ . Comparison heat may differ from actual heat.
5Only constant $T$ along this path allows factoring it out: $\Delta S=\frac{Q_{\mathrm{rev}}}{T}$ .
6For varying temperature integrate the appropriate material model, such as $\int\frac{C(T)}{T}\,dT$ for reversible heating.
7Check the actual process separately using boundary entropy transfer and nonnegative generation.

Rearrangements

Reversible heat

Q_{\mathrm{rev}}=T\Delta S

Only at constant absolute temperature along the reversible comparison path. Retain the heat sign; actual irreversible heat need not equal this comparison value.

Task variant

A system absorbs 900 J reversibly at 300 K. Find ΔS and state the general relation.

In general, $\Delta S=\int\frac{\delta Q_{\mathrm{rev}}}{T}$ .
At constant 300 K, $\Delta S=\frac{900\,\mathrm J}{300\,\mathrm K}=3.0\,\mathrm{J\,K^{-1}}$ .

Common mistakes

Applying the short form directly to actual heat in an irreversible process.

Entropy is a state change; calculate it using a reversible comparison path.

Exam context

Second law, heat engines and heat transfer between reservoirs.

Formula cluster

Entropy and heat engines

Entropy supplements energy conservation with the direction of possible processes.

η = 1 − T_k/T_w

Carnot Efficiency

ΔU = Q − W

First Law of Thermodynamics

Q = m·c·ΔT

Heat Capacity: Q = m·c·ΔT

Q = mL

Latent heat during a phase change

Worked example

A system absorbs 600 J reversibly at constant 300 K. Approach at constant temperature: $\Delta S=\frac{Q_{\mathrm{rev}}}{T}$ . Substitution: $\Delta S=\frac{600\,\mathrm J}{300\,\mathrm K}=2.0\,\mathrm{J\,K^{-1}}$ . If temperature changes, this quotient is not sufficient; reversible heat must be integrated with $\frac1T$ . Heat released has a negative sign for the system.

Applications

Heat engines, phase changes, reversible heat transfer and the second law of thermodynamics.

How do you calculate with Entropy change along a reversible path?

Here is how to work through a typical Entropy change along a reversible path (ΔS = ∫δQrev/T; bei konstantem T: ΔS = Qrev/T) task step by step:

1
Task
A system absorbs 900 J reversibly at 300 K. Find ΔS and state the general relation.
Solution path
1. In general, $\Delta S=\int\frac{\delta Q_{\mathrm{rev}}}{T}$ .
2. At constant 300 K, $\Delta S=\frac{900\,\mathrm J}{300\,\mathrm K}=3.0\,\mathrm{J\,K^{-1}}$ .

Understand and apply the equation

A gas can expand without receiving heat and still gain entropy. This seems contradictory only if actual heat is confused with a reversible comparison path. Entropy change depends on the initial and final states. To calculate it, choose a reversible path connecting precisely those states. The short form $\Delta S=\frac{Q_{\mathrm{rev}}}{T}$ applies only when temperature remains constant along that path.

Separate state change from the actual path

First identify the two equilibrium states being compared. Their difference $\Delta S=S_2-S_1$ is independent of the actual path. During a rapid or irreversible process, one temperature for the whole system may not even exist. A reversible comparison path therefore serves as a calculation route between the same endpoints. Its heat $Q_{\mathrm{rev}}$ may differ from actual heat, but its temperature is well defined for the small equilibrium steps.[1]

\begin{aligned}dS&=\frac{\delta Q_{\mathrm{rev}}}{T}\\\Delta S&=\int_1^2\frac{\delta Q_{\mathrm{rev}}}{T}\end{aligned}

T is absolute temperature along the reversible comparison path.

Entropy is measured in J/K and, as an extensive quantity, increases with system size. Specific entropy is per mass; molar entropy is per amount of substance. Entropy of an individual body may decrease, for instance during cooling. Nonnegativity applies to generation or to the total change of an isolated assembly. Do not confuse one system’s change with entropy generated by the overall process.[4]

When does the reversible-heat quotient apply at constant temperature?

T=\mathrm{const.}\quad\Longrightarrow\quad\Delta S=\frac1T\int_1^2\delta Q_{\mathrm{rev}}=\frac{Q_{\mathrm{rev}}}{T}

Constant T is a path condition, not merely an initial-temperature specification.

If a system reversibly absorbs 600 J at constant 300 K, its entropy increases by 2 J/K. Reversibly rejecting the same amount at that temperature changes entropy by −2 J/K. Constant $T$ allows it to be taken outside the integral. However, isothermal alone does not establish reversibility: an irreversible isothermal process can generate entropy. Likewise, adiabatic only means no actual heat flows. Only additionally reversible adiabats have $\Delta S=0$ .[3]

Does reversibility mean zero entropy change?

\Delta S_{\mathrm{system}}=\int\frac{\delta Q_{\mathrm{actual}}}{T_b}+S_{\mathrm{gen}},\qquad S_{\mathrm{gen}}\ge0

Closed system; Tb is local absolute temperature at the heat-transfer boundary. Add contributions for multiple boundary locations.[3]

For the actual process, separate entropy transferred by heat from entropy generation. $T_b$ is local absolute temperature at the chosen heat-transfer boundary, not an arbitrary mean body temperature. An ideal large reservoir’s own entropy change is $\frac{Q_{\mathrm{reservoir}}}{T_{\mathrm{reservoir}}}$ because its state retains constant temperature. Which sub-balance includes a transfer layer and its generation depends on the system boundary. The complete process total remains unambiguous.

In this closed macroscopic balance, work transports energy but not entropy through a $\frac WT$ term. Dissipation of work by friction can nevertheless generate entropy within the system. In open systems, incoming and outgoing matter also carries entropy. These streams require additional terms; the closed-system balance shown here is then insufficient.

Evaluate a reversible gas path explicitly

\begin{aligned}dU&=T\,dS-p\,dV\\dU&=nC_{V,m}\,dT\\p&=\frac{nRT}{V}\end{aligned}

For a simple compressible system of fixed composition, the first is a state relation; the next two assume an ideal gas.[2]

\begin{aligned}dS&=nC_{V,m}\frac{dT}{T}+nR\frac{dV}{V}\\\Delta S&=nC_{V,m}\ln\frac{T_2}{T_1}+nR\ln\frac{V_2}{V_1}\end{aligned}

The integrated form uses constant molar CV; otherwise integrate its temperature dependence.

The first law and ideal-gas equation make the comparison calculable. Solve the state relation for $dS$ , then insert ideal-gas internal energy and pressure. The integral separates into temperature and volume contributions. Equal endpoint temperatures leave $nR\ln\frac{V_2}{V_1}$ . A reversible isothermal comparison also has $Q_{\mathrm{rev}}=W_{\mathrm{rev}}$ . These heat and work values are not automatically those of the actual irreversible path. Both logarithms require dimensionless ratios. The isothermal work of an ideal gas here belongs to the selected reversible comparison path.[2]

Example: free expansion with Q = 0

There are two paths with the same endpoints: actual free expansion and an imagined reversible isothermal expansion. Only the second is used in the entropy integral. The actual path’s energy balance remains unchanged.[6][5]

Separate the reversible comparison from the experiment

Given: $n=0.750\,\mathrm{mol}$ of ideal gas initially at 340 K in 2.00 L. Removing a partition allows free expansion into vacuum to 5.00 L in an isolated rigid vessel. Equilibrium endpoints; no other energy changes. $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ .
Find: Find T₂, ΔS and comparison heat Qrev.

Approach: First balance the actual isolated expansion into vacuum. Then find the entropy difference through a hypothetical reversible isothermal path between the same equilibrium states.
$\begin{aligned}Q_{\mathrm{actual}}&=W_{\mathrm{actual}}=0\\\Delta U&=0\ \Longrightarrow\ T_2=T_1\\\Delta S&=nR\ln\frac{V_2}{V_1}\\Q_{\mathrm{rev}}&=T_1\Delta S\end{aligned}$
First balance energy for the actual process.
$\begin{aligned}Q_{\mathrm{actual}}&=0\\{} W_{\mathrm{actual}}&=0\\{} \Delta U&=0\quad\Longrightarrow\quad T_2=340\,\mathrm K\end{aligned}$
Use a reversible isothermal gas path between endpoints.
$\Delta S=0.750\cdot8.314\ln(\frac{5}{2})=5.71353086\,\mathrm{J/K}$
Calculate heat on the comparison path only.
$Q_{\mathrm{rev}}=340\Delta S=1942.60049\,\mathrm J=W_{\mathrm{rev}}$

Result: In the ideal model the gas remains at 340 K, while entropy increases by about 5.714 J/K. No heat was actually transferred.

Check: The isolated whole vessel generates $S_{\mathrm{gen}}=\Delta S>0$ . The reversible comparison expansion would receive heat and deliver work; its 1942.6 J must not be assigned to the actual free expansion. For real gases, $\Delta U=0$ does not generally imply unchanged T.

Example: finite bodies equalise temperatures

Direct temperature equalisation conserves energy but generates entropy. First use the two heat capacities to determine their common final temperature. Then calculate each body’s state change through an imagined reversible heating or cooling path. Precisely because the bodies have different temperatures, the total transferred heat cannot be divided by one temperature.[5]

Combine conservation with two entropy integrals

Given: Two rigid bodies without phase changes have constant total heat capacities $C_A=180\,\mathrm{J\,K^{-1}}$ and $C_B=300\,\mathrm{J\,K^{-1}}$ . Initial temperatures are 600 K and 300 K. They contact in an isolated assembly with no work and negligible contact-layer energy.
Find: Find final temperature and total entropy change.

Approach: The isolated assembly conserves energy. For each body use its constant heat capacity and absolute initial and final temperatures to determine entropy change.
$\begin{aligned}C_A(T_f-T_A)+C_B(T_f-T_B)&=0\\\Delta S_{\mathrm{total}}&=C_A\ln\frac{T_f}{T_A}+C_B\ln\frac{T_f}{T_B}\end{aligned}$
Energy lost by A equals energy gained by B.
$T_f=\frac{180\cdot600+300\cdot300}{480}=412.5\,\mathrm K$
Integrate reversible heating or cooling comparisons for each body.
$\Delta S_A=180\ln(\frac{412.5}{600})=-67.44482\,\mathrm{J/K}$
Calculate the second contribution and sum.
$\begin{aligned}\Delta S_B&=300\ln(\frac{412.5}{300})=95.53612\,\mathrm{J/K}\\{} \Delta S_{\mathrm{total}}&=28.09130\,\mathrm{J/K}\end{aligned}$

Result: Final temperature is 412.5 K, transferred energy 33,750 J and total entropy increases by 28.09 J/K.

Check: $180(600-412.5)=300(412.5-300)=33750\,\mathrm J$ . Assembly energy stays constant although its entropy grows. Dividing heat by only the initial or final temperature would be wrong for each finite body.

View diagram: entropy during temperature equalisation

Energy equalisation generates entropy

QUANTA STUDYquanta-study.de

Original state curves for the two bodies in the second example, plotted against energy transferred from hot to cold. Only the interval up to 33.75 kJ, where temperatures equalise, is shown. The curves do not determine transfer time.[1][5]

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Treat temperature change and phase change separately

\begin{aligned}\Delta S&=\int_{T_1}^{T_2}\frac{C(T)}{T}\,dT\\\Delta S&=C\ln\frac{T_2}{T_1}\qquad(C=\mathrm{const.})\end{aligned}

C is total heat capacity appropriate to the chosen reversible heating path.[1]

A reversible phase transition maintains constant coexistence temperature. With positive appropriate $L$ , melting or vaporisation gives $Q_{\mathrm{rev}}=+mL$ and $\Delta S=+\frac{mL}{T}$ ; reversal gives negative values. Substance, transition and pressure must match the tabulated value. For example, a hypothetical transition involving 1500 J at 280 K gives +5.357 J/K and −5.357 J/K on reversible reversal. Heating before or after transition occurs at varying temperature and needs its own integral.[1]

Count melting and warming separately

Two temperature conditions in one calculation

Given: 60.0 g pure ice at 0 °C is completely melted near normal atmospheric pressure and then warmed to 20 °C. Specified constant model values: latent heat 333000 J/kg and water heat capacity 4182 J/(kg·K). Calculate the water’s state change, not entropy of the whole heating apparatus.
Find: Find heat requirement and entropy change of the substance.

Approach: a reversible comparison melts at constant $T_m$ and then warms at varying $T$ . Add the two state contributions.
$\begin{aligned}Q&=mL_f+mc(T_f-T_m)\\\Delta S&=\frac{mL_f}{T_m}+mc\ln\frac{T_f}{T_m}\end{aligned}$
Insert absolute temperatures and calculate both heat stages.
$\begin{aligned}T_m&=273.15\,\mathrm K,&T_f&=293.15\,\mathrm K\\Q&=0.0600\cdot333000+0.0600\cdot4182\cdot20\,\mathrm J\\&=24998.4\,\mathrm J\end{aligned}$
Evaluate the melting contribution and temperature integral separately.
$\begin{aligned}\Delta S_{\mathrm{melt}}&=\frac{19980}{273.15}\,\mathrm{J\,K^{-1}}\simeq73.1466\,\mathrm{J\,K^{-1}}\\\Delta S_{\mathrm{warm}}&=250.92\ln\frac{293.15}{273.15}\,\mathrm{J\,K^{-1}}\simeq17.7308\,\mathrm{J\,K^{-1}}\end{aligned}$

Result: Heat is about 25.00 kJ. Entropy increases by about 90.88 J/K; both contributions are positive.

Check: The melting contribution satisfies $T_m\Delta S_{\mathrm{melt}}=19980\,\mathrm J$ . Warming is not isothermal, so its contribution must lie between $\frac{5018.4}{293.15}$ and $\frac{5018.4}{273.15}$ , in J/K. The substance’s entropy change alone does not establish whether the actual heating apparatus is reversible.

Explore further: entropy in a cycle

Every complete cycle gives zero entropy change of the working medium: its final state is its initial state. A reversible closed path therefore has $\oint\frac{\delta Q_{\mathrm{rev}}}{T}=0$ . The actual closed cycle instead has $\oint\frac{\delta Q_{\mathrm{actual}}}{T_b}=-S_{\mathrm{gen}}\le0$ . An irreversible cycle can export generated entropy through heat rejection and still restore its initial state. Equality in a sub-balance means no generation inside its boundary; fully reversible operation additionally requires reversible coupling to the surroundings.[4]

Entropy is not energy in joules or a direct measure of disorder in an arbitrary picture. The thermodynamic calculation used here requires equilibrium endpoints, absolute temperatures and a suitable material model. Reactions, mixing and open mass transport need additional contributions. The gas-volume formula cannot be applied to those cases without checking.

Five solved entropy exercises

Calculate · Foundation

A system rejects 900 J reversibly at constant 360 K. Find its ΔS and that of an equal-temperature reservoir.

Show hint

Q is negative for the system.

Learning goal: Balance system and reservoir entropy during reversible rejection.

Show solution and calculation

Approach: reversible isothermal rejection has $\Delta S=\frac{Q_{\mathrm{rev}}}{T}$ , using negative system heat.
$\Delta S_{\mathrm{system}}=-\frac{900}{360}\,\mathrm{J\,K^{-1}}=-2.50\,\mathrm{J\,K^{-1}}$ ; reservoir $+2.50\,\mathrm{J\,K^{-1}}$ ; total zero.
System entropy may decrease.

\Delta S_{\mathrm{system}}=-2.50\,\mathrm{J/K}

Transfer · Application

A body with constant $C=120\,\mathrm{J\,K^{-1}}$ heats from 300 to 450 K. Find ΔS.

Show hint

Temperature varies; integrate $\frac CT$ .

Learning goal: Integrate entropy change for varying temperature.

Show solution and calculation

Approach: constant $C$ and changing $T$ give $\Delta S=\int_{T_1}^{T_2}\frac C T\,dT=C\ln\frac{T_2}{T_1}$ .
$\Delta S=120\ln(1.5)\,\mathrm{J\,K^{-1}}=48.6558\,\mathrm{J\,K^{-1}}$ .
$Q=18000\,\mathrm J$ ; $\frac{Q}{300\,\mathrm K}=60\,\mathrm{J\,K^{-1}}$ would not be the correct state change.

\Delta S=48.6558\,\mathrm{J/K}

Calculate · Application

600 J flows directly from a 500 K reservoir to a 300 K reservoir. Find ΔStotal.

Show hint

Use each reservoir’s own constant temperature.

Learning goal: Find entropy production across a finite temperature difference.

Show solution and calculation

Approach: each reservoir retains its own temperature.
Divide each signed heat amount by that temperature and add.
The hot reservoir loses 1.20 J/K and the cold one gains 2.00 J/K.
Their sum is +0.80 J/K.
Direct transfer across the finite temperature difference generates entropy and is irreversible.
Energy is nevertheless conserved.

\begin{aligned}\Delta S_H&=-\frac{600}{500}=-1.20\,\mathrm{J\,K^{-1}}\\\Delta S_C&=\frac{600}{300}=2.00\,\mathrm{J\,K^{-1}}\\\Delta S_{\mathrm{total}}&=0.80\,\mathrm{J\,K^{-1}}\end{aligned}

Interpret · Challenge

An ideal gas freely expands into vacuum inside an isolated rigid overall vessel, doubling its gas volume. Initial and final equilibrium temperatures match. For the gas $Q=0$ , $W=0$ . Give ΔS in units of nR and explain why $\frac QT=0$ is not the entropy change.

Show hint

Entropy is a state function; Qrev belongs to a comparison path.

Learning goal: Use a reversible comparison path despite irreversible actual motion.

Show solution and calculation

Approach: a reversible isothermal comparison path between the same states gives $\Delta S=nR\ln2>0$ .
The actual path transfers no heat but generates entropy.
In this isolated model $S_{\mathrm{gen}}=\Delta S$ .

Check: the actual first-law balance gives $\Delta U=0$ , consistent with equal T but not requiring $\Delta S=0$ . The shortcut $\frac{Q_{\mathrm{rev}}}{T}$ cannot use actual irreversible heat instead.

\begin{aligned}\Delta S&=nR\ln\frac{V_2}{V_1}=nR\ln2\\S_{\mathrm{gen}}&=\Delta S>0\qquad(Q_{\mathrm{actual}}=0)\end{aligned}

[6][2][3]

Error analysis · Foundation

A system absorbs 1000 J reversibly and isothermally at 27 °C. A solution gives $\Delta S=\frac{1000}{27}\,\mathrm{J\, K^{-1}}\approx37.04\,\mathrm{J\, K^{-1}}$ . Correct it and explain why the Celsius number cannot be substituted.

Show hint

Convert temperature to kelvin first.

Learning goal: Use absolute temperature in the entropy formula.

Show solution and calculation

Approach: $T=300.15\,\mathrm K$ , giving $\Delta S=\frac{1000\,\mathrm J}{300.15\,\mathrm K}\approx3.33167\,\mathrm{J\, K^{-1}}$ .
Celsius has an offset zero; entropy uses absolute temperature, not a temperature difference.

Check: ΔS×T reconstructs 1000 J. Moving a temperature-scale zero cannot change a thermodynamic state change, but the incorrect calculation would do exactly that.

\Delta S=\frac{Q_{\mathrm{rev}}}{T}=\frac{1000\,\mathrm J}{(27+273.15)\,\mathrm K}\approx3.33167\,\mathrm{J\,K^{-1}}

[1]

In brief

Identify equilibrium endpoints and choose a reversible comparison connecting them. Only constant comparison temperature turns the integral into a simple quotient. Actual heat instead belongs in a balance with the appropriate boundary temperature and entropy generation. One system may lose entropy; total change of an isolated assembly is nonnegative.

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Highlighted constant comparison temperature and the distinction between entropy change and reversibility in question headings; added relevant gas-work and phase-transition links.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Reworked the introduction and explanations; added a third complete example, an original quantitative diagram and a short summary. Retained existing exercises, models and sources; clarified mathematical typography and model conditions.

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Quanta. Entropy change along a reversible path. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/entropie-reversibel-isotherm

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Frequently asked questions about Entropy change along a reversible path

What does entropy change along a reversible path describe physically?+

The relation describes a state-property change between equilibrium states. Integrate $\frac{\delta Q_{\mathrm{rev}}}{T}$ along a reversible comparison path, whose heat is not automatically the actual heat transferred. Only constant T reduces the result to $\frac{Q_{\mathrm{rev}}}{T}$ . Gas can increase its entropy during free expansion in an isolated vessel even though actual $Q=0$ . A reversible comparison between identical endpoints nevertheless yields the correct ΔS. Units are J/K rather than J. Individual system entropy may decrease by rejecting heat; generation or total change of an appropriately isolated assembly must be nonnegative. Work has no entropy-transfer term $\frac WT$ in this closed-system description.

How can I check a calculation involving entropy change along a reversible path?+

First check that heat and temperature belong to a reversible path and whether its temperature is constant. Absorbing 600 J reversibly at constant 300 K gives $\Delta S=2\,\mathrm{J\,K^{-1}}$ ; surroundings lose the same amount during reversible transfer. For an ideal gas with equal endpoint temperatures, independently use $nR\ln\frac{V_2}{V_1}$ . With 0.750 mol and volume ratio 2.5 this gives 5.7135 J/K. Actual free expansion can have $Q=0$ ; $Q_{\mathrm{rev}}=T\Delta S$ belongs only to the comparison path. Alongside units and signs, carefully check which energy transfer belongs to which path rather than assigning comparison heat to the actual process.

What assumptions and limits apply to entropy change along a reversible path?+

The integral gives ΔS between equilibrium states using a reversible comparison path. The shortcut $\frac{Q_{\mathrm{rev}}}{T}$ additionally requires constant absolute temperature along that path. An actual irreversible process need not have a single system temperature throughout, so actual heat must not blindly be divided by an initial reading. Closed-system entropy accounting includes heat transfer at boundary temperature and nonnegative generation. Mass exchange additionally brings entropy carried by streams. Gas logarithm expressions require an ideal gas, and the simple integrated temperature term assumes constant heat capacity. Adiabatic alone means $Q=0$ , not necessarily $\Delta S=0$ ; reversibility is an additional condition.

How do I check units and scale for entropy change along a reversible path?+

Heat in joules divided by absolute temperature in kelvin gives J/K. Specific or molar entropy additionally requires division by the relevant mass or amount of substance. Logarithms contain dimensionless ratios such as $\frac{T_2}{T_1}$ or $\frac{V_2}{V_1}$ ; a single temperature with units is not a valid logarithm argument. Reversible heat absorption at fixed temperature gives positive ΔS and rejection gives negative ΔS. A body with $C=120\,\mathrm{J\,K^{-1}}$ heating from 300 to 450 K has $\Delta S=120\ln(1.5)\,\mathrm{J\,K^{-1}}\simeq48.656\,\mathrm{J\,K^{-1}}$ . $\frac{Q}{300\,\mathrm K}=60\,\mathrm{J\,K^{-1}}$ would have correct units but the wrong value because temperature varies.

When should I combine entropy change along a reversible path with another relation?+

Combine entropy with energy conservation when final temperature or heat transfer is unknown. Two isolated connected bodies with constant heat capacities reach a common weighted final temperature from the energy balance; calculate each entropy change afterwards using $C\ln\frac{T_f}{T_i}$ . A heat engine’s working medium has $\Delta U=0$ and $\Delta S=0$ over a complete cycle while its reservoirs can generate entropy overall. The reversible comparison gives $\frac{Q_C}{T_C}=\frac{Q_H}{T_H}$ and hence the Carnot bound. For reversible phase changes, signed heat $Q_{\mathrm{rev}}=+mL$ for melting/vaporisation or $-mL$ for freezing/condensation supplies the constant-temperature quotient; L denotes a positive magnitude. Each added expression requires its own material, pressure and process conditions; entropy accounting does not replace them.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §4.6: Entropy

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Entropy change in reversible isothermal heat exchange

Variables & units

Entropy change calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Entropy and heat engines

Worked example

Applications

How do you calculate with Entropy change along a reversible path?

Separate state change from the actual path

When does the reversible-heat quotient apply at constant temperature?

Does reversibility mean zero entropy change?

Evaluate a reversible gas path explicitly

Example: free expansion with Q = 0

Separate the reversible comparison from the experiment

Example: finite bodies equalise temperatures

Combine conservation with two entropy integrals

Energy equalisation generates entropy

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Five solved entropy exercises

A system rejects 900 J reversibly at constant 360 K. Find its ΔS and that of an equal-temperature reservoir.

A body with constant C=120 J K−1C=120\,\mathrm{J\,K^{-1}}C=120JK−1 heats from 300 to 450 K. Find ΔS.

600 J flows directly from a 500 K reservoir to a 300 K reservoir. Find ΔStotal.

An ideal gas freely expands into vacuum inside an isolated rigid overall vessel, doubling its gas volume. Initial and final equilibrium temperatures match. For the gas Q=0Q=0Q=0, W=0W=0W=0. Give ΔS in units of nR and explain why QT=0\frac QT=0TQ​=0 is not the entropy change.

A system absorbs 1000 J reversibly and isothermally at 27 °C. A solution gives ΔS=100027 J K−1≈37.04 J K−1\Delta S=\frac{1000}{27}\,\mathrm{J\, K^{-1}}\approx37.04\,\mathrm{J\, K^{-1}}ΔS=271000​JK−1≈37.04JK−1. Correct it and explain why the Celsius number cannot be substituted.

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Retain Entropy change along a reversible path for exams

A body with constant $C=120\,\mathrm{J\,K^{-1}}$ heats from 300 to 450 K. Find ΔS.

An ideal gas freely expands into vacuum inside an isolated rigid overall vessel, doubling its gas volume. Initial and final equilibrium temperatures match. For the gas $Q=0$ , $W=0$ . Give ΔS in units of nR and explain why $\frac QT=0$ is not the entropy change.

A system absorbs 1000 J reversibly and isothermally at 27 °C. A solution gives $\Delta S=\frac{1000}{27}\,\mathrm{J\, K^{-1}}\approx37.04\,\mathrm{J\, K^{-1}}$ . Correct it and explain why the Celsius number cannot be substituted.