Heat is never hot or cold. Internal energy can be hot or cold.

[Fool]

Heat is never hot or cold. Internal energy can be hot or cold.

Speaking only thermodynamically, of course.

Heat is a description of the transfer of energy from hot to cold.

Internal energy is a condition, a state, of how hot something is compared to absolute zero Kelvin. Internal energy is calculated by measuring temperature, mass and material properties, using equations such as the following:

U=M•Cv•T

Please take note that it is a state function.

U= internal energy, in Joules
M= Mass, in kilograms
Cv= Coefficient of heat, in Joules per kilogram per Kelvin
T= Temperature, in Kelvin

Heat is in Joules AKA energy, or Joules per seconds AKA power in Watts. Heat is calculated from measurements of a temperature difference, two temperatures are needed. Heat can only result from a difference in temperature between two masses. Zero delta T, equals, zero heat transfer.

Heat can be calculated with equations such as the following:

Q=(T1-T2)A/R

Q= Heat from source to sink positive heat leaving, in Joules.
T1= Temperature of the source, in Kelvin.
T2= Temperature of the sink, in Kelvin.
A= Area of thermal path, in meters squared.
R= R-value of material between source and sink, in Joules per Kelvin per meters squared.

Please note that it is a path function going from state T1 to state T2.

Systems can change temperature without heat flow. It should be called adiabatic temperature change. It can drop or rise. This is done by expansion or compression.

Most compressions and expansions are not purely adiabatic. There is always some, at times very little, heat exchange.

For expansions and compressions with large amounts of temperature differences, large amounts of heat transfer, the temperature can rise or fall for either process. Temperature rise or fall is independent of compression or expansion when two or more masses with large temperature differences are present.

For expansions it should be called adiabatic energy loss. For compression it should be called adiabatic energy gain. This would be clearer as it is true regardless of temperature change. It is in defiance of, "heat of compression", a misleading statement. No heat involved.

The temperature of a system dictates the internal energy. A single temperature will have no correlation to heat, it requires two.

This language is in defiance of common, colloquial, terminology. A red hot bolt doesn't contain heat. It contains internal energy. A red hot bolt radiates heat to a colder room-temperature mass. Heat radiation can be felt as it warms the cooler, than red hot, skin and sensory cells, becoming internal energy of the skin. Radiant heat can be blocked by any room-temperature shield mass. You don't technically feel the heat, you feel the temperature of the internal energy of the skin as it increases.

This has nothing to do with the ideal gas law. It also doesn't forbid the use of ideal gas law.

[Tom Booth]

...
"heat of compression", a misleading statement. No heat involved.
...

Your dissertation does not address Work.

Seems to deny a work-heat relationship.

"Heat of compression" is a transfer of energy. Work input results in an increase in temperature.

Calling that increase in temperature "heat" may be technically incorrect, but what would you propose as an alternative?

Denying the phenomenon altogether is just silly IMO.

The work of compression is converted to thermal energy that manifests as an increase in temperature.

In common language work converted to heat and vice versa.

If nothing else, a lot easier to say and I for one thing the meaning is perfectly clear, or clear enough.

If you want to suggest better terminology, that's fine, trying to deny the phenomenon altogether, as you seem to be doing, isn't.

May as well just say there's no such thing as a heat engine.

[Jack]

My 2 cents.

Ideal compression is a bunching up of internal energy.
The fluid particles have less space to move around with their internal energy and have to give it off to their surroundings in the form of heat.
In the real world some heat will be added during compression due to friction and all that though.

So while compression is done by work and it'll get things hot, it might not fall in the heat category like fool describes it. Only the part that's added due to friction.

[matt brown]

Q=(T1-T2)A/R

Please note that it is a path function going from state T1 to state T2.

Kudos Fool - the proper way to 'mix' state and path functions.

Thermo notation is almost an art form that ranges from simple statements to complex expressions. The problem is how to 'say' something technically correct amidst a plethora of values. The typical calculus expressions are the most accurate, but least accessible due to heavy math. Meanwhile, the simple statements are more accessible, but limited in scope and ripe for error. I'm not much for the calculus stuff, but love the statements since these tend to convey bite size info vs a plate full of distractions. Over the years, various conventions have arisen as with any other 'language' and the big one in thermo revolves around mixing state functions with path functions.

As Fool points out, Q is a path function while T is a state function...and there's no direct way to equate them. However, Q=(T1-T2) resolves this apparent 'dilemma' by Q equal the path between two state functions, sweet !!! It takes practice to read notation, but a lot more to write it (properly).

[Fool]

Matt,
Thanks for the kudos. I've procrastinated responding to this thread because I've been busy and the second and third posts have questions and comments that require careful wording. The easiest to explain is Jack's point. He is very close to understanding the difference between heat, work, compression, and internal energy.

Rubbing a mass on a mass, friction, is putting work into a system, not heat. The temperature change/rise is an indication of increased internal energy, U. It is considered irreversible energy transfer. It's irreversible because a heat engine operating on the temperature difference from friction will output less work than was put in. The total work is irrecoverable. Some work can be reclaimed if good engineering design is used. Zero if the masses are very large and input work energy is very small, miniscule temperature difference.

Temperature change is not the same as two temperatures. Heat only flows between temperature differences of two masses with different temperatures. Work input just changes one temperature to one higher temperature, still only one.

Friction puts energy into the system in the process/path of work, that energy manifests itself as a rise it temperature. Since the temperature has increased, heat may begin flowing to a colder surrounding or ambient mass that didn't experience work input. It requires two temperatures, hotter and colder.

Work input in the form of compression is considered a reversible temperature rise, because it will be higher than for friction. (COP greater than one). Ideally all the work put in can be gotten back out. 100% ideally. (COP=1/n).

Energy put in as heat will never be converted all to work even ideally. Efficient max n=(Th-Tc)/Th=1/COP , ideal.

COP•n= 1 proving ideality.

4 • 0.25 = 1.0 the same as 100%

[Jack]

You're right. Friction results in a higher internal energy, but because of work, not heat. I overlooked that.

[Fool]

...
"heat of compression", a misleading statement. No heat involved.
...

Your dissertation does not address Work.

Seems to deny a work-heat relationship.

"Heat of compression" is a transfer of energy. Work input results in an increase in temperature.

Calling that increase in temperature "heat" may be technically incorrect, but what would you propose as an alternative?

Denying the phenomenon altogether is just silly IMO.

The work of compression is converted to thermal energy that manifests as an increase in temperature.

In common language work converted to heat and vice versa.

If nothing else, a lot easier to say and I for one thing the meaning is perfectly clear, or clear enough.

If you want to suggest better terminology, that's fine, trying to deny the phenomenon altogether, as you seem to be doing, isn't.

May as well just say there's no such thing as a heat engine.

"Your dissertation does not address Work.

Seems to deny a work-heat relationship."

"Denying the phenomenon altogether is just silly IMO."

Wow! My "dissertation"/post does address Work, it is disguised as compression and expansion. All compression and expansion, inside a cylinder from a pistons motion, is expansion or compression with work. With or without any external load. The temperature drop or rise will always be at a maximum, unless the piston is so light that it's speed approaches supersonic. This speed isn't even close in the engines we work with here. The work manifests itself as momentum/kinetic energy of the piston/system.

There is no heat-work relationship. There is only a work heat equivalence, meaning they both are energy and can be described as Joules. Heat going into an engine Q becomes the internal energy U. They are used differently in equations. The internal energy has a temperature. The heat is only the energy that was added to that internal energy, and then ceased to be heat. In fact it never existed, U1 lost energy directly to U2, never in any other form. The internal energy and assisted temperature and pressure does the work of expansion/volume change.

Why would you continue such a malformed rebuttal with the ad homonym "silly"? No reason. Please stop.

"The work of compression is converted to thermal energy that manifests as an increase in temperature."

Yes. In engineering this is called internal energy, U. And used that way.

"Calling that increase in temperature "heat" may be technically incorrect, but what would you propose as an alternative?"

Already explained that. Increase in internal energy. Hotter. Hotter energy. Higher energy. Heat is misleading. Misleading leads a person to incorrect conclusions and makes it more difficult to understand. I'm not proposing any change other than the explanation as to how it is used in engineering equations and diagrams.

"In common language work converted to heat and vice versa.

If nothing else, a lot easier to say and I for one thing the meaning is perfectly clear, or clear enough."

It is very common. I'm hoping for an improvement far and above "common", because it isn't used that way. It just leads to the wrong conclusions. Heat of compression leads to adiabatic heat transfer an oxymoron, or even worse thinking that heat is both hot and internal energy.

It's your choice as to wether you want to stay common or rise up far and above.

"If you want to suggest better terminology, that's fine, trying to deny the phenomenon altogether, as you seem to be doing, isn't.

May as well just say there's no such thing as a heat engine."

Again. Wow! Left field. Have no idea why that is relevant to engineering. Perhaps some method of clinging to colloquialisms? Sad.

Does a heat engine take in heat? Yes. Does it output all the heat as work? No. Does it run on a temperature difference? Yes. I have no problems calling it a heat engine, or more accurately a temperature difference engine.

That leads correctly to the question: Why does it need a temperature difference? Because it needs to reject.heat to get back to the beginning of the cycle.

Does heat go into an IC engine? No. Chemicals go into an IC engine. I have no problem calling it a gasoline engine.

[Tom Booth]

What I object to mostly is how (added) "heat" and (existing) "internal energy" are treated as equivalent for the purposes of the "Carnot limit" equation

η=(Th -Tc)/Th ⋅100%

Tc is the internal energy before adding heat
Th is the internal energy after adding heat

Th - Tc would then represent the heat added.

Let's say 100 joules are added
Tc = 300

η=(400 - 300)/400 ⋅100%

η=100/400 ⋅100%

η=1/4 ⋅100%

η=25%

All that really says is that the temperature was raised 25% or the internal energy was increased by 25%

To illustrate: suppose I have a gas tank that contains 300 ounces of gasoline.

I add 100 ounces making 400

Suppose my gas tank was empty and I added 100 ounces and drove 50 miles.

My mileage (efficiency) is 0.5 miles per ounce

Now suppose my tank has 300 ounces and I add 100 to make a total of 400.

I again drive 50 miles and burn 100 ounces.

My car efficiency is still 0.5 miles per ounce.

The "reserve" 300 gallons that never gets touched has zero influence on efficiency. It is how much is added vs.how much is burned or actually used that determines efficiency.

The 300 "internal energy" is balanced by an equivalent 300 external ambient energy These simply cancel out, may as well not exist. It's just the given environmental heat.

So "Carnot efficiency" is actually the percentage of heat that can be utilized of the environmental "internal energy" plus the added heat rather than just the added heat.

Why should the pre-existing "internal energy" be included along with the increase in "internal energy" that results exclusively from the added heat?

Actual efficiency is, or should be, heat added in joules vs internal energy (resulting from the added heat exclusively) converted to work in joules.

I didn't add 100+300=400 joules. I added 100.

Efficiency is (or IMO should be) a .measure of Joules added vs joules converted to work NOT joules added PLUS the pre-existing 300 joules of internal energy vs joules converted to work.

100 joules added, 100 converted = 100%

Not

100 joules added 100 converted = 25% because 100 is 25% of 400 which is joules added PLUS ambient "internal energy".

I see no justification for such a convoluted measure of efficiency.

[Fool]

I'm assuming that you see, and perhaps accept the current engineering use of the difference of heat and internal energy.

I'm moving and addressing the off topic post into the, "The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)", thread.

The only way you could have an objection is if you accept the engineering differentiation.

You said "What I object to mostly is how (added) "heat" and (existing) "internal energy" are treated as equivalent for the purposes of the "Carnot limit" equation."

I will address that in the other thread. They aren't treated as equivalent. The differences just cancel out in the ratio.