How to reverse the process of rubbing two stones in water?

• Chemistry
• zenterix
zenterix
Homework Statement
I am reading the book "Heat and Thermodynamics" by Zemansky and Dittman. Chapter 6 entitled "The 2nd Law of Thermodynamics" starts with multiple sections on heat engines and then on the refrigerator.
Relevant Equations
My question is about dissipative effects. Before I get to my question, which has to do with the 2nd law of thermodynamics, I want to go through some of the topics presented in the chapter before the section on dissipative effects.
There is a statement of the 2nd law by Kelvin-Planck

It is impossible to construct an engine that, operating in a cycle, will produce no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.

and then a statement of the 2nd law by Clausius

It is impossible to construct a refrigerator that, operating in a cycle, will produce no effect other than the transfer of heat from a lower-temperature reservoir to a higher-temperature reservoir.

Then, it is shown that these statements are equivalent. This is done by showing the following (this is how I understood it, anyways)

a) If the Clausius statement is false, then we can make a refrigerator that extracts ##|Q_L|## from a cold reservoir without requiring any work. We can use this heat in the cycle of a heat engine: ##|Q_L|## goes into its cold reservoir and ##|Q_H|## is extracted from its hot reservoir such that ##|Q_H|-|Q_L|## of work is done on its surroundings. Thus, such a refrigerator together with the heat engine constitutes a form of perpetual motion machine (of the second kind): in a cycle, ##|Q_H|## is extracted from the hot reservoir, work is done on the surroundings, but there is no net rejection of heat to the cold reservoir. This violates the Kelvin-Planck statement.

b) If the Kelvin-Planck statement is false, then we can make a heat engine that extracts ##|Q_H|## from its hot reservoir without rejecting any heat to the cold reservoir. The associated work can be used in the cycle of a refrigerator: compression of the refrigerant in the isothermal compression part of its cycle, with ##|Q_H|## going into its hot reservoir. ##|Q_L|## is removed from its cold reservoir. Taken together, the refrigerator and the heat engine constitute a form of perpetual motion machine (of the second kind): in a cycle, ##|Q_L|## of heat is removed from the cold reservoir but no net heat into the hot reservoir. This violates the Clausius statement.

Then there is a discussion on reversibility and irreversibility.

A reversible process is one that is performed in such a way that, at the conclusion of the process, both the system and the local surroundings may be restored to their initial states without producing any changes in the rest of the universe.

The question immediately arises as to whether natural processes, namely, the familiar processes of nature, are reversible or not. Since dissipation is present in all real processes, it follows that all natural processes are irreversible.

Finally, we reach the part that I have some major questions about.

There is a large class of processes involving the isothermal transformation of work through a system (which remains unchanged) into internal energy of a reservoir. This type of process is depicted schematically in the following figure

Five examples of this type of process are

1) Friction from rubbing two solids in contact with a reservoir.

2) Irregular stirring of a viscous liquid in contact with the reservoir.

3) Inelastic deformation of a solid in contact with the reservoir.

4) Transfer of charge through a resistor in contact with the reservoir.

5) Magnetic hysteresis of a material in contact with the reservoir.

In order to restore the system and its local surroundings to their initial states without producing changes elsewhere, ##|Q|## units of heat would have to be extracted from the reservoir and converted completely into work. Since this would involve a violation of the second law (Kelvin statement), all processes of the above type are irreversible.

Let's consider just the first example.

Take two stones, one in each hand, submerge the stones in water (a cold reservoir) and start rubbing them together. We are performing mechanical work which is somehow being converted into heat that is passing into the water.

To restore the stones and local surroundings to their initial states without producing changes elsewhere, we would have to extract heat from the water and convert it completely into work.

I don't quite understand what it would mean to convert the heat to work in this example with the stones.

In terms of violating the second law, it seems we would have to be able to devise a machine that would extract heat from the reservoir (now a hot reservoir) and completely turn this into work without rejecting any heat, thus violating the 2nd law (in particular as it is stated by Kelvin-Planck.

The essence here seems to be that we can dissipate heat when we perform work but we can't recover the heat completely.

In other words, we can dissipate work freely, but we can't extract heat freely (as this would violate the 2nd law and imply the possibility of perpetual motion machine).

Is this a correct interpretation of the topics above?

There is yet another related topic.

If we could produce a machine with no dissipation of work then we would have a "perpetual motion machine of the third kind". The machine could run indefinitely without violating either of the two laws of thermodynamics; that is, it would run but produce no work.

I'd like to understand this a bit better.

What would be a simple example of such a machine?

zenterix said:
What would be a simple example of such a machine?
How about a flywheel, disconnected from any load, that can store it's rotational energy forever.

zenterix
zenterix said:
Homework Statement: I am reading the book "Heat and Thermodynamics" by Zemansky and Dittman. Chapter 6 entitled "The 2nd Law of Thermodynamics" starts with multiple sections on heat engines and then on the refrigerator.
Relevant Equations: My question is about dissipative effects. Before I get to my question, which has to do with the 2nd law of thermodynamics, I want to go through some of the topics presented in the chapter before the section on dissipative effects.

There is a statement of the 2nd law by Kelvin-Planck

and then a statement of the 2nd law by Clausius

Then, it is shown that these statements are equivalent. This is done by showing the following (this is how I understood it, anyways)

a) If the Clausius statement is false, then we can make a refrigerator that extracts ##|Q_L|## from a cold reservoir without requiring any work. We can use this heat in the cycle of a heat engine: ##|Q_L|## goes into its cold reservoir and ##|Q_H|## is extracted from its hot reservoir such that ##|Q_H|-|Q_L|## of work is done on its surroundings. Thus, such a refrigerator together with the heat engine constitutes a form of perpetual motion machine (of the second kind): in a cycle, ##|Q_H|## is extracted from the hot reservoir, work is done on the surroundings, but there is no net rejection of heat to the cold reservoir. This violates the Kelvin-Planck statement.

b) If the Kelvin-Planck statement is false, then we can make a heat engine that extracts ##|Q_H|## from its hot reservoir without rejecting any heat to the cold reservoir. The associated work can be used in the cycle of a refrigerator: compression of the refrigerant in the isothermal compression part of its cycle, with ##|Q_H|## going into its hot reservoir. ##|Q_L|## is removed from its cold reservoir. Taken together, the refrigerator and the heat engine constitute a form of perpetual motion machine (of the second kind): in a cycle, ##|Q_L|## of heat is removed from the cold reservoir but no net heat into the hot reservoir. This violates the Clausius statement.

Then there is a discussion on reversibility and irreversibility.

Finally, we reach the part that I have some major questions about.

There is a large class of processes involving the isothermal transformation of work through a system (which remains unchanged) into internal energy of a reservoir. This type of process is depicted schematically in the following figure

View attachment 347770

Five examples of this type of process are

1) Friction from rubbing two solids in contact with a reservoir.

2) Irregular stirring of a viscous liquid in contact with the reservoir.

3) Inelastic deformation of a solid in contact with the reservoir.

4) Transfer of charge through a resistor in contact with the reservoir.

5) Magnetic hysteresis of a material in contact with the reservoir.

In order to restore the system and its local surroundings to their initial states without producing changes elsewhere, ##|Q|## units of heat would have to be extracted from the reservoir and converted completely into work. Since this would involve a violation of the second law (Kelvin statement), all processes of the above type are irreversible.

Let's consider just the first example.

Take two stones, one in each hand, submerge the stones in water (a cold reservoir) and start rubbing them together. We are performing mechanical work which is somehow being converted into heat that is passing into the water.

To restore the stones and local surroundings to their initial states without producing changes elsewhere, we would have to extract heat from the water and convert it completely into work.

I don't quite understand what it would mean to convert the heat to work in this example with the stones.

In terms of violating the second law, it seems we would have to be able to devise a machine that would extract heat from the reservoir (now a hot reservoir) and completely turn this into work without rejecting any heat, thus violating the 2nd law (in particular as it is stated by Kelvin-Planck.
No. The stones and water can be returned reversibly to their initial states by putting them in contact with a sequence of ideal infinite reservoirs at temperatures ranging from the final temperature of the stones and water to the initial temperature (no work involved). Of course the states of these auxiliary reservoir would change. Reversing the process does not require recovery of the work.
zenterix said:
Is this a correct interpretation of the topics above?

Chestermiller said:
The stones and water can be returned reversibly to their initial states
The book seems to say the opposite. I am now confused.

Is this incorrect, then?

It depends. I can help you work this out, but we would have to be more precise in analyzing this than they do.

If you are interested in working with me to analyze this, then please specify precisely the system and immediate surroundings that is being analyzed, and the initial state of this system and surroundings.

berkeman

• Biology and Chemistry Homework Help
Replies
1
Views
178
• Biology and Chemistry Homework Help
Replies
2
Views
287
• Thermodynamics
Replies
1
Views
948
• Biology and Chemistry Homework Help
Replies
13
Views
410
• Thermodynamics
Replies
20
Views
1K
• Other Physics Topics
Replies
8
Views
2K
• Thermodynamics
Replies
12
Views
1K
• Thermodynamics
Replies
26
Views
3K