Cosmology · Conditional futures of the Universe

Entropy, usable energy and heat-death scenarios

Compare reversible work extraction with direct thermal contact in a finite system, then identify the additional assumptions needed for cosmic heat-death scenarios.

Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.

Back to the chapter

After reading, you can

Distinguish energy, temperature, entropy and available work.
Compute two different outcomes for the same initial thermal bodies.
State why a finite thermal example does not establish a unique cosmic endpoint.

A resource depends on a usable difference

A system can contain substantial energy while offering little opportunity to extract work through a particular process. A heat engine needs a suitable temperature difference; energy content alone does not specify that opportunity.

Temperature describes thermal state, while entropy tracks how reversible heat exchange relates to temperature and helps constrain possible processes. These quantities are not interchangeable with energy.

Begin with two ordinary nongravitating thermal bodies, each with constant heat capacity C = 10 J/K. Their initial temperatures are 400 K and 100 K. They are isolated from other thermal resources.

Supply the thermal-energy convention U_i = CT_i relative to a chosen zero. Their initial combined thermal energy is 5000 J. Only energy changes, stored work and entropy differences will enter the comparison; this is not their total rest-mass energy.

Let an ideal engine use the temperature difference

Place an ideal reversible cyclic engine between the bodies, storing its extracted work separately. As energy is transferred, the hot body cools and the cold body warms.

For constant heat capacities, dS = C dT/T. Reversible operation keeps the bodies’ total entropy unchanged, so T_hotT_cold stays equal to its initial value, 40000 K².

When the two temperatures meet, each is 200 K, the geometric mean of the initial temperatures. The bodies then retain 4000 J in the specified thermal-energy convention, and 1000 J has been stored as work.

No heat engine can extract further work from those two equal temperatures alone. This conclusion concerns the stated resource and process; another external resource would define a different problem.

Reversible extraction; T_* = 100 K, energy unit CT_* = 1000 J, and y is entropy transferred divided by C
y	T_hot/T_*	T_cold/T_*	Remaining thermal energy/(CT_*)	Extracted work/(CT_*)
0	4	1	5	0
(ln 2)/2	2.828427	1.414214	4.242641	0.757359
ln 2	2	2	4	1

The same starting energy can yield no stored work

Now consider a separate process starting from the original 400 K and 100 K bodies. Put them in direct thermal contact and extract no work.

Energy conservation for this isolated pair gives a common final temperature of 250 K, the arithmetic mean. All 5000 J remains in the bodies under the supplied energy convention.

The entropy change is ΔS = C ln(250/400)+C ln(250/100) = C ln(25/16), about 4.462871 J/K. It is positive, unlike the reversible extraction process.

Both processes conserve energy when the work store is included. The difference is how the initial temperature gradient is used: reversible operation stores its maximum obtainable work, while direct contact dissipates that opportunity.

Optional mathematical path: an entropy-transfer coordinate

Define y as the entropy transferred from the hot body to the cold body, divided by C. Along the reversible path, T_hot = 4T_*exp(−y) and T_cold = T_*exp(y).

The extracted work is W/(CT_*) = 5−4exp(−y)−exp(y). The finite path ends at y = ln 2, where the temperatures coincide.

Its derivative is d[W/(CT_*)]/dy = 4exp(−y)−exp(y), positive before equality and zero at equality. Thus work increases to the displayed value CT_* = 1000 J.

The initial and final body energies plus the stored work balance exactly. Entropy conservation describes this ideal reversible path, rather than claiming that every possible thermal process is reversible.

Two ordinary nongravitating bodies have equal constant heat capacities C = 10 J/K and initial temperatures 400 K and 100 K. Set T_* = 100 K and U_i = CT_i relative to a supplied zero, not their rest-mass energy. An ideal reversible cyclic engine stores work separately. Panel A has entropy-transfer coordinate y from 0 to ln 2 and a zero-based temperature/T_* ordinate from 0 to 4.2. The hot body is solid orange with circles, 4exp(−y); the cold body is dashed blue with squares, exp(y). Samples at y = 0, (ln 2)/2 and ln 2 are (4,1), approximately (2.828427,1.414214) and (2,2). Panel B has its own zero-based work/(CT_*) axis from 0 to 1.1, with CT_* = 1000 J. Dotted green diamonds show 5−4exp(−y)−exp(y), giving 0, approximately 0.757359 and 1 at those samples. Total entropy change is zero on this reversible path. The native table compares separate processes from the original state: reversible extraction ends at 200 K, with 4000 J in the bodies and 1000 J stored as work; direct contact ends at 250 K, with 5000 J in the bodies and zero work. Direct contact is not another point of the reversible path. Its entropy increase is 10ln(25/16) ≈ 4.462871 J/K. Equal final temperatures cannot drive further heat-engine work from these bodies alone, although energy remains. No gravity, horizon entropy or global cosmic equilibrium is modeled. Original Ark teaching model.

A cosmic scenario needs gravity and horizon assumptions

Heat-death scenarios concern the depletion of accessible work resources under specified future conditions. They do not mean that every form of energy literally vanishes.

The two-body example isolates one thermal gradient. It does not model gravitational clustering, systems with negative heat capacity, black-hole entropy, horizon entropy or a total entropy of the Universe.

Cosmic expansion alone does not establish one final global equilibrium or one uniquely known final temperature. The future stress-energy, horizons, matter stability and accessible physical processes affect the proposed scenario.

A useful comparison names the resources, which observer or system can access them, and what dynamics are assumed to continue. The finite calculation explains why energy and useful work differ, while the cosmic conclusion remains conditional.

Worked reasoning

Account for energy after reversible extraction

How much energy remains in the two bodies when their temperature difference is exhausted?

Reversibility gives a common final temperature √(400×100) = 200 K.
The bodies retain 2×10×200 = 4000 J in the supplied thermal-energy convention.
The initial 5000 J minus 4000 J is 1000 J stored as work.

4000 J remains in the bodies, even though the specified temperature-gradient resource for further heat-engine work is exhausted.

Try the reasoning

What is the final temperature under direct contact with no work extraction?
Show the answer and reasoning
250 K.
Equal constant heat capacities and conserved body energy give the arithmetic mean.
What is the entropy change for direct contact?
Show the answer and reasoning
10 ln(25/16) J/K ≈ 4.462871 J/K.
Add C ln(T_final/T_initial) for each body; the result is positive.
Does this two-body result prove that the Universe approaches a unique global equilibrium?
Show the answer and reasoning
No.
Gravity, horizons, expansion and other work resources lie outside the finite nongravitating model.

What to carry forward

Energy content and energy available for a specified kind of work are different.
Reversible and irreversible paths can use the same initial temperature gradient differently.
Cosmic heat-death claims require additional assumptions about dynamics and accessible resources.

References

Sources supporting the explanations, physical distinctions and examples in this lesson.