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SCIENCE · INVESTIGATE AND EXPLAIN

Audit entropy on the full isolated thermal boundary

A component can lose entropy while the complete isolated process produces it.

Learning outcome: Vary the hot temperature and cold heat capacity. First close energy, then evaluate both entropy changes through their state-function expressions; reject trial final temperatures that violate isolation.

Starting recap

Refresh state-function boundaries, absolute temperature, dimensionless logarithms, written reaction/electron equivalents, conserved charge and formal orbital/identity bookkeeping.

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Try

Two100 J/K bodies at400 K and300 K reach350 K. Their entropy changes are100ln(350/400) and100ln(350/300), totaling about2.06193 J/K.

Open the experiment

Predict

Does the hot body losing entropy violate the second law?

Experiment

Vary the hot temperature and cold heat capacity. First close energy, then evaluate both entropy changes through their state-function expressions; reject trial final temperatures that violate isolation.

Conserve energy while computing total entropy of equilibration

Worked teaching model · synthetic values · Two isolated constant-heat-capacity bodies: hot C100 J/K, cold initial300 K and selected C. No work, phase change, chemical reaction or external loss. Tf is energy-weighted; entropy change of each body is C ln(Tf/Ti), evaluated via a hypothetical reversible path between its states.

Reset keeps your prediction, captures and practice history.

total entropy change (J/K)trial common temperature (K; only energy-balanced point is allowed)040028.76821-28.76821
  • Total entropy at trial common temperature: solid

Axes rescale. A multiplicity distribution, inverse-temperature relation and charge-budget graph have different meanings. Read each axis and do not treat a computed curve as additional measured evidence.

All calculated plot coordinates
Total entropy at trial common temperature
trial common temperature (K; only energy-balanced point is allowed)total entropy change (J/K)
300-28.76821
301.6667-27.66017
303.3333-26.55824
305-25.46235
306.6667-24.37243
308.3333-23.28841
310-22.21024
311.6667-21.13785
313.3333-20.07118
315-19.01017
316.6667-17.95476
318.3333-16.90489
320-15.8605
321.6667-14.82154
323.3333-13.78795
325-12.75967
326.6667-11.73665
328.3333-10.71883
330-9.706171
331.6667-8.698612
333.3333-7.696104
335-6.698596
336.6667-5.706038
338.3333-4.718382
340-3.735579
341.6667-2.757582
343.3333-1.784344
345-0.8158188
346.66670.1480385
348.33331.107273
3502.061929
351.66673.012049
353.33333.957678
3554.898856
356.66675.835626
358.33336.768028
3607.696104
361.66678.619893
363.33339.539435
36510.45477
366.666711.36593
368.333312.27296
37013.1759
371.666714.07478
373.333314.96963
37515.8605
376.666716.74742
378.333317.63043
38018.50955
381.666719.38482
383.333320.25628
38521.12396
386.666721.9879
388.333322.84811
39023.70465
391.666724.55753
393.333325.40678
39526.25245
396.666727.09456
398.333327.93313
40028.76821
Calculated under the declared model
QuantityValueUnit
Final common temperature350K
Hot-body energy change-5000J
Cold-body energy change5000J
Hot-body entropy change-13.35314J/K
Cold-body entropy change15.41507J/K
Total isolated entropy change2.061929J/K

The hot body loses entropy, but the complete isolated two-body system gains it.

Energy conservation fixes the final temperature; not every plotted trial temperature is an allowed isolated final state. Compute entropy from a reversible comparison path even though the actual finite-temperature heat transfer is irreversible. Do not divide a finite exchanged heat by one arbitrary endpoint temperature or call the hot body alone the universe.

These calculated packets are not performed chemical measurements or separation procedures. Keep feasible inventories, activity conventions, fitting observations and independent checks distinct.

The hot body loses entropy, but the complete isolated two-body system gains it.

Energy conservation fixes the final temperature; not every plotted trial temperature is an allowed isolated final state. Compute entropy from a reversible comparison path even though the actual finite-temperature heat transfer is irreversible. Do not divide a finite exchanged heat by one arbitrary endpoint temperature or call the hot body alone the universe.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Find the permitted final state

Two bodies exchange energy inside an isolated boundary, with no work or phase change. Constant heat capacities Ch and Cc give Ch(Tf−Th)+Cc(Tf−Tc)=0. Thus Tf is an energy-weighted temperature, not an arbitrary arithmetic mean unless capacities match.

For equal100 J/K capacities initially400 and300 K, Tf=350 K. The hot body loses5000 J and the cold gains5000 J. Conservation closes exactly, but it does not determine whether the entropy change is zero.

Use a reversible comparison path for a state function

Each entropy change is C ln(Tf/Ti). This expression can be evaluated by an imagined reversible path between the same endpoint states even though the actual finite-temperature heat exchange is irreversible. Hot entropy is about−13.3531 J/K; cold entropy is15.4151 J/K; their sum is about2.06193 J/K.

Dividing all finite heat by350 K or one starting temperature would silently replace a changing-temperature process with an inappropriate isothermal approximation. Nor should the hot body alone be called the universe while it exchanges heat with the cold body.

Distinguish an isothermal phase change

At phase coexistence, a reversible isothermal transition does permit ΔS=q_rev/T. For a separate fictional melting transition with latent enthalpy 4.00 kJ/mol at 400 K, the molar entropy change is +10.0 J/(mol K); freezing at that coexistence state reverses its sign. This does not justify dividing a changing-temperature body’s entire heat by one endpoint temperature.

Transfer and check

With Ch=100 J/K at450 K and Cc=50 J/K at300 K, energy gives Tf=400 K. Entropy is100ln(400/450)+50ln(400/300), about2.6058 J/K. Predict the signs before calculating. The graph’s other trial common temperatures are not alternate allowed isolated final states; only the energy-balanced point is feasible.

Source and evidence boundary

Conceptual reference checked 30 October 2026: OpenStax Chemistry 2e. All prose, cases, graphs and questions are original. These supplied thermodynamic, electrical, nuclear or orbital models are not actual measurements, apparatus instructions, synthesis protocols or practical certification. Retain conventions, independent constraints, changed-case reasoning and limits. No heating, wiring, electrolysis, source handling or chemical manipulation is requested.

Apply

In the400/300 K equal-capacity example, the isolated total entropy change is:

Public practice checks a specific claim. It does not establish independent mastery.

Why is finite exchanged heat divided by one endpoint temperature not the correct entropy change here?

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