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According to mainstream physics: Is heat "destroyed" in a heat engine?


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Posted
1 hour ago, Tom Booth said:

Maybe if the word "heat" can be avoided, or more strictly defined.

Heat is technically, I think, a temperature dependent transfer of energy.

Would that be an accurate statement?

If so, then somewhat euphemistically I guess, if there are two objects in close proximity, one at a high temperature and another at a low temperature then "heat" could be said to exist between them. There can be a temperature dependent transfer of energy between the two objects.

If, by some means, we were to cool down the hotter object (lower it's temperature by taking away some of its energy) so that there is no longer a temperature difference, then there can no longer be this temperature dependent transfer of energy. "Heat" between the two objects, then, no longer exists.

By "destroying" the ∆T we have also "destroyed" the potential for there to be any temperature dependent  transfer of energy. (as a statistical average at least, there are still energy transfers but they tend to balance out or nullify each other on the whole)

Are these fair statements so far?

Heat, as a temperature dependent transfer of energy can either exist, or not exist, it could be "destroyed" or "created" by either adding or subtracting energy from one side or the other (one object or the other).

This is a semantics argument, about phrasing.

One reason why we use equations to make things clearer. But since language us still used, we try and define things as clearly as possible.

Yes, heat is a transfer of energy between objects (or systems) owing to a temperature difference (sometimes called heat transfer). That distinguishes it from work, which is other energy transfer. There is also energy content. Sloppy use of terminology often leads to confusion 

 

Energy is conserved. Ei = Ef  There are different forms of energy, so each of those might be broken down, depending on the details of a problem (potential, kinetic, thermal, etc)

Energy can be transferred in our out of systems, so that’s another thing that must be accounted for, like balancing one’s checkbook.

Ei = Ef + W for example

It’s important to define the boundary of a system when applying these equations.

 

In the context of the first law, energy is not destroyed. Heat is not destroyed. The energy goes somewhere. Argument to the contrary is semantics, and trying to skirt the issue by stretching the narrow definitions we have in physics. When in doubt, refer to the equations.

 

Posted (edited)

I don't know what's confusing. I'm trying to settle the apparent concussion.

I'm simply saying, is it not true that with work output there is a lowering of the temperature of the working fluid as a result?

That is, there was no "heat" transfer, but there was energy (work) transfer.

In a gas, the energy is temperature dependent. So a work output results in a drop in temperature, but no "heat" went out, only "work", but the consequent drop in temperature is the same though there was not any heat transfer, only a work transfer.

If the quantity of heat that produced the ∆T is transformed into work, than the ∆T no longer exists.

Edited by Tom Booth
Posted
4 minutes ago, Tom Booth said:

I don't know what's confusing. I'm trying to settle the apparent concussion.

I'm simply saying, is it not true that with work output there is a lowering of the temperature of the working fluid as a result?

That is, there was no "heat" transfer, but there was energy (work) transfer.

In a gas, the energy is temperature dependent. So a work output results in a drop in temperature, but no "heat" went out, only "work", but the consequent drop in temperature is the same though there was not any heat transfer, only a work transfer.

So what? Heat is not conserved. Energy is conserved. The form of energy changes. There is no confusion there.

Posted (edited)
15 minutes ago, Tom Booth said:

I don't know what's confusing. I'm trying to settle the apparent concussion.

I'm simply saying, is it not true that with work output there is a lowering of the temperature of the working fluid as a result?

That is, there was no "heat" transfer, but there was energy (work) transfer.

In a gas, the energy is temperature dependent. So a work output results in a drop in temperature, but no "heat" went out, only "work", but the consequent drop in temperature is the same though there was not any heat transfer, only a work transfer.

If the quantity of heat that produced the ∆T is transformed into work, than the ∆T no longer exists.

Except that the quantity of heat that produced the ΔT cannot be all transformed into work, only part of it. 

Edited by exchemist
Posted
35 minutes ago, swansont said:

This is a semantics argument,

Exactly. It's like saying water is destroyed, when it evaporates, and vapour is created. Same goes for ice. 

Acutally, 72 years ago, my mother's pregnancy was destroyed. Luckily, by way of compensation, I was created as a baby at about the same time.

Posted
46 minutes ago, Tom Booth said:

If the quantity of heat that produced the ∆T is transformed into work, than the ∆T no longer exists.

Heat flow does not create a difference in temperature. If there is no temperature difference, there is no heat flow. Q=0, since Q depends on a temperature difference

 

Quote

In a gas, the energy is temperature dependent. So a work output results in a drop in temperature, but no "heat" went out, only "work", but the consequent drop in temperature is the same though there was not any heat transfer, only a work transfer.

Not true. If energy is transferred owing to a temperature difference between objects, it’s heat. Work is energy transfer that doesn’t depend on a temperature difference.

You don’t get to change these definitions.

Also, do not confuse a temperature change in time with a temperature difference between objects.

Posted (edited)
1 hour ago, Genady said:

So what? Heat is not conserved. Energy is conserved. The form of energy changes. There is no confusion there.

Thank you! My point exactly.

51 minutes ago, swansont said:
Quote

In a gas, the energy is temperature dependent. So a work output results in a drop in temperature, but no "heat" went out, only "work", but the consequent drop in temperature is the same though there was not any heat transfer, only a work transfer.

Not true.

What specifically in my above statement do you consider "not true".

Your remarks here bear no  correspondence to what I actually said:

51 minutes ago, swansont said:

If energy is transferred owing to a temperature difference between objects, it’s heat. Work is energy transfer that doesn’t depend on a temperature difference.

You don’t get to change these definitions.

Also, do not confuse a temperature change in time with a temperature difference between objects.

 

Edited by Tom Booth
Changed resemblance to correspondence
Posted

 

1 hour ago, Tom Booth said:

What specifically in my above statement do you consider "not true".

A drop in temperature might be the result of heat flow, and might be due to work being done. You did not provide enough detail to say further. But since heat flow could be present, concluding there is no heat flow is incorrect. 

 

1 hour ago, Tom Booth said:

Your remarks here bear no  correspondence to what I actually said:

I guess that’s the problem in a nutshell: You don’t see how my comments apply to what you said. But what you said wasn’t true.

2 hours ago, Genady said:

So what? Heat is not conserved. Energy is conserved. The form of energy changes. There is no confusion there.

To the point of the thread topic: Heat isn’t conserved, but it isn’t destroyed, either.

In mechanics, kinetic energy isn’t a generally conserved quantity, either, but it can change into other forms. It isn’t “destroyed” - such a description would be misleading. 

Ultimately you want to be able to quantify effects, and that means using equations. Equalities are generally the most useful. “heat is destroyed” doesn’t lend itself to solving an equation.

Posted (edited)
37 minutes ago, swansont said:

I guess that’s the problem in a nutshell: You don’t see how my comments apply to what you said. But what you said wasn’t true.

[Emphasis added]

If you don't mind, please quote the specific statement, words of mine, sentence, paragraph or whatever to which you are referring by the bolded words. "what you said wasn’t true."

What specifically, in your opinion, did I say that "wasn't true". Please quote the words in my statement you consider not true. Please 

You seem to be reading things into my statement that I did not actually say at all.

Edited by Tom Booth
Posted
20 minutes ago, Tom Booth said:

If you don't mind, please quote the specific statement, words of mine, sentence, paragraph or whatever to which you are referring by the bolded words.

Advice: Maybe you can ask about what established science agrees upon to be correct instead? That may allow for the discussion to progress.

Posted
5 minutes ago, Ghideon said:

Advice: Maybe you can ask about what established science agrees upon to be correct instead? That may allow for the discussion to progress.

Swansont is doubling down that I made an untrue, false statement. I'd like to know specifically what he has issue with. Others here don't seem to have a problem with it, "so what".

I agree, so what? I said nothing that is not accepted mainstream recognized, confirmed, experimentally verified science. AFAIK, so what exactly is Swansont's big issue with my statement.

He mistakenly, falsely, reads things into it that are not there then goes on the attack against these imagined errors in my statement. I'd very much like to know what he has an issue with, specifically.

Posted
3 minutes ago, Tom Booth said:

I said nothing that is not accepted mainstream recognized, confirmed, experimentally verified science.

Yes, you did. You said, that if temperature does not change, then the work is done without heat transfer. This is not so. For example, in isothermal process, a work is done, the heat is transferred from the environment to the system, and the temperature does not change.

Posted (edited)
6 minutes ago, Genady said:

Yes, you did. You said, that if temperature does not change, then the work is done without heat transfer. This is not so. For example, in isothermal process, a work is done, the heat is transferred from the environment to the system, and the temperature does not change.

That's not what I said at all.

Please quote my actual statements not these straw man, made up supposed twisted mischaracterizations, please.

Edited by Tom Booth
Lies changed to mischaracterizations
Posted
3 hours ago, Tom Booth said:

I'm simply saying, is it not true that with work output there is a lowering of the temperature of the working fluid as a result?

Not necessarily.

Some chemical reactions evolve or accept either work or heat or both sometimes with a temperature change, sometimes not.

You made no answer to my previous post so this is the last time I attempt to help you.

Posted
6 minutes ago, Tom Booth said:

That's not what I said at all.

Please quote my actual statements not these straw man, made up supposed twisted mischaracterizations, please.

Then, whatever you said, you didn't say it clear enough and it was open to misinterpretation. Too many words - easy to misinterpret. Try to say it concisely. Math helps a lot in this.

Frankly, I like to discuss science, not who said what.

Posted
9 minutes ago, studiot said:

Not necessarily.

Some chemical reactions evolve or accept either work or heat or both sometimes with a temperature change, sometimes not.

You made no answer to my previous post so this is the last time I attempt to help you.

The topic is clear "in a heat engine" not a chemical reaction, so irrelevant.

7 minutes ago, Genady said:

Then, whatever you said, you didn't say it clear enough and it was open to misinterpretation

Swansont quoted one paragraph. A few sentences. What words did you personally find unclear?

Posted
4 minutes ago, Tom Booth said:

What words did you personally find unclear?

Every word by itself is clear. Their combination is not.

Why do you want me to do the work? Just rephrase what you said.

Posted (edited)
3 hours ago, swansont said:

Quoting Tom Booth:

 

In a gas, the energy is temperature dependent. So a work output results in a drop in temperature, but no "heat" went out, only "work", but the consequent drop in temperature is the same though there was not any heat transfer, only a work transfer.

A bit repetitions but clear enough.

Work performed by the "working fluid" (air or gas) in a heat engine results in a temperature drop of the working fluid.

Work goes out, the temperature goes down. Period. A basic principle, widely accepted.

Perhaps the drop in temperature can sometimes be compensated for but that does not negate the principle: conversion of heat into work results in a temperature drop within the working fluid as a consequence of the energy output.

True or not?

Edited by Tom Booth
Posted

 

43 minutes ago, Tom Booth said:

A bit repetitions but clear enough.

Work performed by the "working fluid" (air or gas) in a heat engine results in a temperature drop of the working fluid.

Work goes out, the temperature goes down. Period. A basic principle, widely accepted.

Perhaps the drop in temperature can sometimes be compensated for but that does not negate the principle: conversion of heat into work results in a temperature drop within the working fluid as a consequence of the energy output.

True or not?

What does the bold part mean?

(Bold highlight by me)

Posted
1 hour ago, Tom Booth said:

Swansont is doubling down that I made an untrue, false statement. I'd like to know specifically what he has issue with. Others here don't seem to have a problem with it, "so what".

I agree, so what? I said nothing that is not accepted mainstream recognized, confirmed, experimentally verified science. AFAIK, so what exactly is Swansont's big issue with my statement.

He mistakenly, falsely, reads things into it that are not there then goes on the attack against these imagined errors in my statement. I'd very much like to know what he has an issue with, specifically.

 

57 minutes ago, Tom Booth said:

A bit repetitions but clear enough.

Work performed by the "working fluid" (air or gas) in a heat engine results in a temperature drop of the working fluid.

Work goes out, the temperature goes down. Period. A basic principle, widely accepted.

Perhaps the drop in temperature can sometimes be compensated for but that does not negate the principle: conversion of heat into work results in a temperature drop within the working fluid as a consequence of the energy output.

True or not?

It’s the “Period” which implies that there is nothing else. It means that no heat is transferred out, which is contrary to mainstream physics. The notion that the work is solely driven by the heat in and work out, and that these are equal, is not mainstream physics. “Work goes out, the temperature goes down. Period.” is not widely accepted.

There are examples of work without heat flow - adiabatic compression and adiabatic expansion against pressure or some other impediment, like a spring. But these are only individual steps - they are not the entirety of a heat engine. There’s no heat in or out. All there is is the work being done, which changes the internal energy.

https://en.m.wikipedia.org/wiki/Adiabatic_process#Adiabatic_free_expansion_of_a_gas

Posted
2 hours ago, Tom Booth said:

The topic is clear "in a heat engine" not a chemical reaction, so irrelevant.

Very clear thank you.

Clear that you have classed all combustion engines as not being heat engines.

Posted
1 hour ago, Ghideon said:

 

What does the bold part mean?

(Bold highlight by me)

Perhaps the drop in temperature can sometimes be compensated for .

If there is a simultaneous addition of heat. i.e. Isothermal expansion for example; the working fluid which would otherwise suffer a drop in temperature maintains the temperature due to additional simultaneous heat input alongside the work output. It may even increase in temperature, none of that however negates or contradicts the principle involved as far as heat being converted into work.

For every unit of work output a unit of the heat/energy input to the engine "disappears" or is subtracted from the account so to speak.

Posted (edited)
13 minutes ago, Tom Booth said:

Perhaps the drop in temperature can sometimes be compensated for .

"Prevented" or "avoided" would be a better word. The temperature of the system that's doing work won't drop if the outflow of energy is compensated for by inflow from some heat source (such as an exothermic chemical reaction, for instance).

Edited by Lorentz Jr
Posted
1 hour ago, swansont said:

 

It’s the “Period” which implies that there is nothing else. It means that no heat is transferred out, which is contrary to mainstream physics. The notion that the work is solely driven by the heat in and work out, and that these are equal, is not mainstream physics. “Work goes out, the temperature goes down. Period.” is not widely accepted.

There are examples of work without heat flow - adiabatic compression and adiabatic expansion against pressure or some other impediment, like a spring. But these are only individual steps - they are not the entirety of a heat engine. There’s no heat in or out. All there is is the work being done, which changes the internal energy.

https://en.m.wikipedia.org/wiki/Adiabatic_process#Adiabatic_free_expansion_of_a_gas

You contradict yourself saying "it's contrary to mainstream science" then proceed to give an example: "There are examples of work without heat flow - adiabatic compression and adiabatic expansion against pressure or some other impediment, like a spring."

I would simply add to that example the fact that there is a temperature drop within the working fluid.

 

 

Posted (edited)
15 minutes ago, Tom Booth said:

You contradict yourself saying "it's contrary to mainstream science" then proceed to give an example: "There are examples of work without heat flow - adiabatic compression and adiabatic expansion against pressure or some other impediment, like a spring."

I would simply add to that example the fact that there is a temperature drop within the working fluid.

Your phrasing ("Period.") suggested that there are no exceptions to your rule. @swansont was pointing out that there are plenty of exceptions, because your rule only describes the class of situations where there's no other heat source. All  re-usable heat engines have a source of heat, because that's the only way they can keep running. External combustion for steam engines, internal combustion for piston engines in automobiles.

Edited by Lorentz Jr
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