Kelvin–Planck statement and entropy derivation

I'm only starting to "feel" what phenomenological entropy (not statistical) mean. Well, they mean the same thing, but I never "felt" classical Entropy definition as I could "feel" and understand Bolzmann definition.

So recently I had a thought:

Ok. Entropy $$S$$ is somehow related to energy dispersal and thus "hides" some part of the energy from being converted to work $$W$$. That is, the $$Q$$ you've received from hot sink $$T_h$$ will not be fully converted to $$W$$, because there should be "heat dump" onto cold sink at the temperature $$T_c$$.

Question 1: This(above) is a principle of "limit to a heat engine's efficeincy" in the action. Right?

If so,

Question 2: What if there were no such principle? I know the universe will disintegrade or something... but I care only about one thing Would this mean that there is no entropy in some sence? Since there is no limit to heat engine efficency, then could there be an engine, which takes $$Q$$ from $$T_h$$ and just completly converts $$Q$$ into $$W$$ without using any cold sinks $$T_c$$?

If so,

Question 3: then entropy should somehow be a consequence from Kelvin-Plank statment(see quote below) and it maybe possible to derive entropy existance from this statment, right?

If so,

Question 4: Can someone provide a link where entropy is derived as a consequense of Kelvin-Plank statment or as a consequence to "the limit to heat engine efficency".

WIKI: The Kelvin–Planck statement (or the heat engine statement) of the second law of thermodynamics states that it is impossible to devise a cyclically operating thermal engine, the sole effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work. This implies that it is impossible to build a heat engine that has 100% thermal efficiency.

• But there IS such a principle, the second law, so realistically you are breaking one "law" (probabilistic as it is), and asking for opinions in terms of other, presumably unbroken laws. This leads to a lot of what ifs, in my opinion. – user213900 Nov 25 '18 at 18:27
• Yes, I understand this. But, please, try to understand WHY I ask such a question: I'm trying to undersand the signinicance of entorpy. And all the way entropy function can be derived. Those derivations are best explanation of WHAT this $S$ actually means. So when breaking the given law, please, break it for a second and tell me only entropy related stuff. :) – coobit Nov 25 '18 at 18:31
• One example, (that I'm sure you know already), I can't easily give you a definition of the concept of time. But without entropy, I ABSOLUTELY can't give you a definition of time. I have upvoted your question because we are talking about a probabilistic statement. Best of luck with it, the derivation you ask about is in any textbook (as you know, also:). – user213900 Nov 25 '18 at 18:57
• mm, If it was in any book, then the Internets will be full of it, but I can't find any. Maybe I choose wrong words for search engines. Thanks. – coobit Nov 25 '18 at 19:00

Ok. Entropy $$S$$ is somehow related to energy dispersal and thus "hides" some part of the energy from being converted to work $$W$$, that is, the $$Q$$ you've received from hot sink $$T_h$$ will not be fully converted to $$W$$, because there should be "heat dump" onto cold sink at the temperature $$T_c$$.

Question 1: This (above) is a principle of "limit to a heat engine's efficiency" in the action. Right?

That is correct as it pertains to a heat engine operating in a cycle, because any such cycle requires rejection of heat to a lower temperature body (see answer to question 2, below). This precludes the possibility of a 100% efficient engine operating in a cycle.

Question 2: What if there were no such principle? I know the universe will disintegrate or something... but I care only about one thing. Would this mean that there is no entropy in some sense? Since there is no limit to heat engine efficiency, then could there be an engine, which takes $$Q$$ from $$T_h$$ and just completely converts $$Q$$ into $$W$$ without any cold sinks $$T_c$$?

The problem is, there IS such a principle. If there were no such (Kelvin-Plank) principle, then the heat engine you describe would constitute a perpetual motion machine of the second kind. It would be capable of continuously spontaneously transferring heat from a single thermal reservoir and convert it into work. No one has been able to create such a machine.

In order for a heat engine to operate in any cycle and perform net work, it will need to reject some heat in order to complete the cycle. For example, take again the case of the Carnot cycle. Heat is absorbed from the high temperature source during the reversible isothermal expansion and the gas does work. But for the device to perform this work in a cycle you need a path to get back to the initial state. You could reverse the process to isothermally compress the gas and get back using the same single temperature source, but the work done on the gas during the compression would equal the work done by the gas during the expansion and no net work would be done! Therefore, in the case of the Carnot cycle, you need to isentropically expand the gas to get to a lower temperature and then isothermally compress the gas rejecting heat to a second, lower temperature sink. The cycle then wraps up with an isentropic compression to the original state and net work is done.

Can you envision any thermodynamic cycle, reversible or not, in which net work is done without involving at least one process where heat is rejected? I can't.

Question 3: then entropy should somehow be a consequence from Kelvin-Plank statement (see quote below) and it maybe possible to derive entropy existence from this statement, right?

Yes, or the other way around, to the extent that entropy is associated with efficiency limits and the Kelvin-Plank statement prohibits a 100% efficient heat engine operating in a cycle. The example of a Carnot cycle can help to show this.

A Carnot cycle consists of two isothermal processes (expansion and compression) involving two thermal reservoirs (heat source and sink sufficiently massive that their temperatures remain constant for a given quantity of heat transfer). Let:

$$T_H$$ be the temperature of the high temperature thermal reservoir.

$$T_L$$ be the temperature of the low temperature thermal reservoir (heat sink).

$$Q_H$$ be the heat transferred from the high temperature reservoir to the system during the isothermal expansion (Work done by the system).

$$Q_L$$ be the heat transferred from the system to the low temperature reservoir (work done on the system).

Since the heat transfers occur isothermally (at constant temperature), the entropy changes of the two thermal reservoirs are as follows.

$$\frac {-Q_H}{T_H}$$ is the change in entropy of the high temperature reservoir (a reduction in entropy)

$$\frac {+Q_L}{T_L}$$ is the change in entropy of the low temperature reservoir (an increase in entropy). So the total entropy change is given by:

$$\Delta S_{total}=\frac {+Q_L}{T_L}+\frac {-Q_H}{T_H}$$

The second law tells us that the total change in entropy must be equal to or greater than zero. The equality applying to both transfers being reversible. Keep in mind that the change in entropy of the system has to be zero since it has undergone a complete cycle which restores all its properties, including entropy, to their original states.

Now, in order to maximize the net work, $$W$$, done in this cycle you would want to make the heat rejected, $$Q_L$$, be as low as possible relative to $$Q_H$$. In your case you are asking that $$Q_L=0$$ (no heat sink) so that you have 100% efficiency. But if $$Q_L=0$$, then from the previous equation $$\Delta S<0$$ in violation of the second law.

The best we can do is the case where $$\Delta S=0$$ (a reversible cycle), in which case we have:

$$\frac{Q_L}{T_L}=\frac{Q_H}{T_H}$$ or $$\frac{Q_L}{Q_H}=\frac{T_L}{T_H}$$

This last equation basically limits the maximum possible efficiency of all heat engines to convert heat into work in a thermodynamic cycle. Recalling that the definition of efficiency of a heat engine is given by

$$ζ=\frac{W}{Q_H}$$

And

$$W=Q_H-Q_L$$

We get

$$ζ=1-\frac{Q_L}{Q_H}=1-\frac{T_L}{T_H}$$

Which is the Carnot cycle efficiency.

Question 4: Can someone provide a link where entropy is derived as a consequence of Kelvin-Plank statement or as a consequence to "the limit to heat engine efficiency".

I have not been able to find a link that directly connects the two. But hopefully the example I gave above can help you connect entropy and the Kelvin-Plank statement.

Hope this helps.