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I quote the coming lines from Igneous & Metamorphic Petrology by Myron G. Best (2003):

One statement of the second law of thermodynamics is that spontaneous natural processes tend to even out the concentration of some form of energy, smoothing the energy gradient. A hot lava flow extruded from a lofty volcano cools to atmospheric T as it descends down slope, thereby reducing differences in thermal and gravitational potential energy between initial and final states in accordance with the second law.

Eventually, billions of years from now, all of the thermal energy in the Earth will be consumed in tectonism, volcanism, and other processes and dispersed into outer space. No mountains or volcanoes will be erected and erosion in the solar powered hydrologic system will wear everything down to some common level (assuming the Sun does not run out of nuclear energy!).

Without differences in the concentration of thermal and gravitational potential energy no geologic work can be accomplished and the planet will be geologically dead! The measure of the uniformity in concentration of energy in a system is called the entropy, S. The more uniform the concentration of some form of energy, the greater the entropy. The geologically dead planet will have maximal entropy.

Is the last statement is right?... because it seems that dead planet will have the minimal entropy not the maximum, if we consider that entropy is the measure of disorder? just like ice vs water vs vapor.. ice will eventually have the minimal entropy as temperature decreases (thermal stability increases).

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You always have to be careful at understanding entropy as a "measure of disorder". For example, let's say we have a cup of coffee that is well mixed, so it looks very uniform and ordered. Then let's say we dump in some creamer. Very soon after we dump the creamer in everything looks very disordered: swirls of coffee and creamer intertwined as things mix up. But if we wait long enough (and maybe even help the process by stirring the coffee) we will see that our coffee becomes uniform again (albeit a bit lighter in color than when we started). But we know that mixing the coffee and creamer results in maximum entropy. So what happened? We went from order, to disorder, and then back to order, but our entropy was increasing the entire time.

The problem is that disorder is a subjective term. Entropy is better understood as how your text says it how "spread out" energy is, or more objectively as most text books have it is a counting of the number of microstates that the system can be in.

Some videos that talk about this:

https://www.youtube.com/watch?v=vSgPRj207uE

https://www.youtube.com/watch?v=w2iTCm0xpDc

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    $\begingroup$ Thank you for this. The malignant connection between entropy and disorder must be tamed. $\endgroup$ – SuchDoge Jul 12 '18 at 15:21
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    $\begingroup$ @SuchDoge It is one of my largest pet peeves. I can't stand it haha. $\endgroup$ – BioPhysicist Jul 12 '18 at 15:56
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I think it’s kind of right, but poorly worded from a thermodynamics standpoint. Any natural process involving heat or work transfers results in an increase in overall entropy (entropy of system + surroundings). Moreover, all natural processes are irreversible and the result of disequilibrium (e.g., differences in temperature, pressure, potential energy and so on). The increase in entropy diminishes when the differentials shrink and equilibrium is approached. The overall entropy change becomes zero when equilibrium is reached at which point the processes cease. In that sense the overall entropy is maximized. I like to use the following heat transfer example to show the connection between entropy and disequilibrium (in this case, thermal disequilibrium).

Consider a system H (a hot body) and its surroundings C (a cold body). Further consider both H and C to be thermal reservoirs, that is, they are so massive that a heat transfer between them doesn’t change their temperatures. The temperature of H is TH and the temperature of C is TC. We bring the bodies together and desire to transfer heat Q from H to C. Since the temperature of either does not change, the heat transfer occurs isothermally. Let’s look at the entropy changes:

For Body A (System): ΔSA = -Q/TH (a drop in entropy)

For Body B (Surroundings): ΔSB = +Q/TC (a rise in entropy)

The total entropy change: ΔSTot = ΔSA + ΔSB

Then for any TH > TC: Q/T C - Q/TH > 0

In order for the total entropy change to approach zero, the temperature difference must approach zero. This results in the heat transfer rate approaching zero and the time it takes to transfer Q infinitely long. In order for the total entropy change to actually equal zero, the temperatures would have to be the same- but if that were the case we would have no heat transfer at all!

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  • $\begingroup$ Your calculation of the entropy change of A and B is valid only for a tiny incremental transfer of heat. If the two bodies are allowed to transfer enough heat such that they equilibrate with one another, then your equation is incorrect. $\endgroup$ – Chet Miller Jul 12 '18 at 14:23
  • $\begingroup$ Not so. The example explicitly states that the two bodies are considered to be thermal reservoirs. Specifically it states "the heat transfer between them doesn't change their temperatures". That is the assumption. "Tiny" is a relative term. $\endgroup$ – Bob D Jul 13 '18 at 12:22
  • $\begingroup$ Would 1,000,000 BTUs be a tiny amount of heat? It would be if it was transferred to or from the Atlantic Ocean in terms of the change in temperature of the ocean. $\endgroup$ – Bob D Jul 13 '18 at 12:31
  • $\begingroup$ Oops. I missed that. Sorry. $\endgroup$ – Chet Miller Jul 13 '18 at 13:31
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It is correct that a maximum uniformity is a synonymous to a maximum of disorder, thus a maximum of entropy.

Another way to look at this is to consider entropy as a complete lack of information. If a medium is completely uniform, it will be no different than the same medium with particles shuffled to random positions. In this case, the lack of information is maximum, the entropy is maximum too.

The case of ice you quote is misleading. Indeed, a mass of ice has a smaller entropy than the same mass of liquid water: it is colder, more organized. But for the liquid water to become ice, heat had to be transferred out of the system. And if you consider the mass of water together with whatever received the thermal energy (another body, radiation,...), then the total entropy has increased, "because of" irreversible processes (second principle of thermodynamics).

So, the entropy of the earth, as it becomes more uniform (in density, in temperature), will increase because it will be less organized. But at the same time, as thermal energy is radiated away (simply from black body radiation), the entropy of the earth (alone) will tend to decrease.

So I guess there are two competing processes here. Still, in an infinite time from now (assuming earth still exists which is extremely unlikely), when the universe reaches thermodynamic equilibrium, the entropy of the earth will tend to a value which depends on its final temperature.

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    $\begingroup$ Thermodynamic entropy, in the classical sense, is not a measure of disorder or information. Using these analogies only promotes more unnecessary confusion about entropy. $\endgroup$ – SuchDoge Jul 12 '18 at 15:19

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