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Greenhouse Gases Do NOT Violate The Stefan-Boltzmann Law



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15-08-2019 04:45
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:I think this is just a model of insulation/reduced heat right?

You get bonus points. You grasped the "insulation = reduced heat." Well done.

Heat is a *flow* of thermal energy. You "heat" something by flowing thermal energy into it. Temperature cannot increase without additional energy. If you reduce the flow then you reduce the heat.


.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
15-08-2019 05:02
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
tmiddles wrote:I think this is just a model of insulation/reduced heat right?

You get bonus points. You grasped the "insulation = reduced heat." Well done.

Heat is a *flow* of thermal energy. You "heat" something by flowing thermal energy into it. Temperature cannot increase without additional energy. If you reduce the flow then you reduce the heat.


.


OK right on I think I'm on a solid foundation then.

I was think about this too in terms of conduction and reduced heat.

(*) (*) (*)

If you think of three molecules and the first one is moving around violently and knocks into the second that's at rest, it will easily "push it" with no "push back" and then on to the 3rd. Like crocket or pool balls.

((((*)))) (*) (*)
(*) ((((*)))) (*)
(*) (*) ((((*))))


But if the other balls are moving too there will be "Push back" which will mean it doesn't get to send so much over all at once.

((((*)))) (((*))) (((*)))
(((*))) ((((*)))) (((*)))
(((*))) (((*))) ((((*))))

So the "push back" or "hand back" of some thermal energy, a bit of a two steps forward and one step back, is happening with either conduction or radiation of thermal energy.

The net movement is still in the same direction from hotter to colder though.

Sound about right?
15-08-2019 19:31
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:
(*) (*) (*)

If you think of three molecules and the first one is moving around violently and knocks into the second that's at rest, it will easily "push it" with no "push back" and then on to the 3rd. Like crocket or pool balls.

You get beaucoup credit for the illustrations and you clearly took time to be as clear as possible. Thank you.

I'm going to ask you to rephrase your question using the following proper terminology within the following context.

When molecules collide, they do not suffer dents or "fender benders." They don't get banged up, damaged or deformed. Their collisions are "elastic" Think of billiard balls or sliding one coin into another.

For the brief moment that they collide, some thermal energy flows from the molecule of higher temperature to the molecule of lower temperature. This particular flow of thermal energy is governed by conduction.



Conduction is the physical matter parallel to Radiance of Stefan Boltzmann

Conduction is per a cross-sectional area over a length, Radiance is just per a surface area.
Conduction is modified by a thermal conductivity coefficient, Radiance is modified by an emissivity coefficient.

... the main difference is that Conduction is based on temperature difference, Raidance is based on absolute temperature, but both are driven by a temperature value.

If two molecules rub together, friction causes kinetic energy to be converted to thermal energy based on the coefficient of friction and the contact force. That thermal energy will flow to other molecules of lower temperature in subsequent collisions.

Could I get you to frame your question under this context?


.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
15-08-2019 23:39
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
Could I get you to frame your question under this context?
.


Well that pretty much answers my question.

If I have three objects A and B and C. A has a 100C temperature and B has a 50C temperature and C has a 10C temperature.

If A is to the left of B is to the left of C. If all three are in physical contact then conduction will cause thermal energy to move from A to B to C. This is "heat". If A is in physical contact with C the "heat" would be increased. B reduces the heat if it's in the way.

However the thermal energy wouldn't really move "up stream". You wouldn't actually have thermal energy moving from B to A.

Now if the same three bodies are in a vacuum and only have radiance as a way for thermal energy to leave. A, gap, B, gap, C

A will radiate out and that radiance will create thermal energy in B. B will also radiate out but less than A did. B will have the thermal energy is started with, plus what it's getting from A, but it's not getting everything A is putting out. B's radiance does actually "move upstream" and reach A (the radiances would move past each other in the gap). B's radiance is actually converted to thermal energy by A, but it's less than what A lost.
Thermal energy still has the net movement from A to B to C but there is real movement upstream that could be put into the diagram.

Rally though it's the same deal as with conduction. B reduces the heat between A and C by being in the way in both cases.

Sound about right?
16-08-2019 03:36
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:
IBdaMann wrote:
Could I get you to frame your question under this context?
.


Well that pretty much answers my question.

If I have three objects A and B and C. A has a 100C temperature and B has a 50C temperature and C has a 10C temperature.

If A is to the left of B is to the left of C. If all three are in physical contact then conduction will cause thermal energy to move from A to B to C. This is "heat". If A is in physical contact with C the "heat" would be increased. B reduces the heat if it's in the way.

However the thermal energy wouldn't really move "up stream". You wouldn't actually have thermal energy moving from B to A.

Now if the same three bodies are in a vacuum and only have radiance as a way for thermal energy to leave. A, gap, B, gap, C

A will radiate out and that radiance will create thermal energy in B. B will also radiate out but less than A did. B will have the thermal energy is started with, plus what it's getting from A, but it's not getting everything A is putting out.

You were fine up to this point. Good job.

But now we get to the point that you violate thermodynamics and Planck's. Thermal energy cannot flow from a cooler body to a warmer body, either by conduction or by thermal radiation. Yet you state:

tmiddles wrote: B's radiance does actually "move upstream" and reach A (the radiances would move past each other in the gap). B's radiance is actually converted to thermal energy by A, but it's less than what A lost.

Nope. B is of a lower temperature than A. B cannot increase A's temperature, either by conduction or by thermal radiation.

If you pour cold water into warm coffee, the cold water cannot bring the warm coffee to a boil while the cold water freezes and becomes ice..


.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
16-08-2019 03:47
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
If you pour cold water into warm coffee, the cold water cannot bring the warm coffee to a boil while the cold water freezes and becomes ice..


I tend to think of people shoving into one another when I think of conduction.

So how's this!

Two sumo wrestlers, the large easy bet to win, and the much smaller sure to lose opponent. The big wrestler pushes the small one, the small one pushes back (flexing leg muscles, really trying). But there is a steady movement of the big guy pushing the little guy off the mat. It matters that the little guy is putting up a fight, it really slows the big guy down, but it never reverses the direction.

The NET flow of energy is in one direction, from hot to cold.

But my question is (going back to our ball shell): Doesn't a big flow of radiant energy move across the gap towards the shell as a small flow of radiant energy moves the other way? Don't they mover right through each other in opposite directions?

Separate question:
Is there anything that prevents a white hot object from getting hotter due to infra-red radiation?

What if the power of the infrared is much greater than that of the white hot object? Couldn't infrared actually get something white hot hotter?
Edited on 16-08-2019 03:50
16-08-2019 16:12
Into the NightProfile picture★★★★★
(9582)
IBdaMann wrote:
tmiddles wrote:
(*) (*) (*)

If you think of three molecules and the first one is moving around violently and knocks into the second that's at rest, it will easily "push it" with no "push back" and then on to the 3rd. Like crocket or pool balls.

You get beaucoup credit for the illustrations and you clearly took time to be as clear as possible. Thank you.

I'm going to ask you to rephrase your question using the following proper terminology within the following context.

When molecules collide, they do not suffer dents or "fender benders." They don't get banged up, damaged or deformed. Their collisions are "elastic" Think of billiard balls or sliding one coin into another.

For the brief moment that they collide, some thermal energy flows from the molecule of higher temperature to the molecule of lower temperature. This particular flow of thermal energy is governed by conduction.



Conduction is the physical matter parallel to Radiance of Stefan Boltzmann

Conduction is per a cross-sectional area over a length, Radiance is just per a surface area.
Conduction is modified by a thermal conductivity coefficient, Radiance is modified by an emissivity coefficient.

... the main difference is that Conduction is based on temperature difference, Raidance is based on absolute temperature, but both are driven by a temperature value.

If two molecules rub together, friction causes kinetic energy to be converted to thermal energy based on the coefficient of friction and the contact force. That thermal energy will flow to other molecules of lower temperature in subsequent collisions.

Could I get you to frame your question under this context?


.


It should be mentioned here that radiant heating ONLY occurs if the light emitted due to blackbody radiance is absorbed and results in thermal energy in some other mass.

Light in and of itself is not heat, although it can be a means of heating.


The Parrot Killer
16-08-2019 16:36
Into the NightProfile picture★★★★★
(9582)
tmiddles wrote:
IBdaMann wrote:
If you pour cold water into warm coffee, the cold water cannot bring the warm coffee to a boil while the cold water freezes and becomes ice..


I tend to think of people shoving into one another when I think of conduction.

So how's this!

Two sumo wrestlers, the large easy bet to win, and the much smaller sure to lose opponent.

Bad parallel. The smaller guy actually has a good chance in Sumo. Sorry, I've watched too much Sumo.
tmiddles wrote:
The big wrestler pushes the small one, the small one pushes back (flexing leg muscles, really trying).

Only if the small guy is an idiot. There is much more to Sumo than simply shoving straight into each other.
tmiddles wrote:
But there is a steady movement of the big guy pushing the little guy off the mat. It matters that the little guy is putting up a fight, it really slows the big guy down, but it never reverses the direction.

Nope. The ONLY winning conditions of Sumo are to get your opponent out of the ring. Shoving isn't the only way.
tmiddles wrote:
The NET flow of energy is in one direction, from hot to cold.

There is no net flow of energy. Thermal energy never flows from cold to hot. Not one bit of it. It ONLY flows from hot to cold. It doesn't matter if it's by conduction, convection, or radiance.
tmiddles wrote:
But my question is (going back to our ball shell): Doesn't a big flow of radiant energy move across the gap towards the shell as a small flow of radiant energy moves the other way?

No. This is where absorption comes in again. No molecule will absorb a photon that has less energy then the molecule already has. To that photon, the molecule is 'transparent'.
tmiddles wrote:
Don't they mover right through each other in opposite directions?

No. No molecule will ever absorb any photon that has less energy than the molecule already has.
tmiddles wrote:
Separate question:
Is there anything that prevents a white hot object from getting hotter due to infra-red radiation?

No.
tmiddles wrote:
What if the power of the infrared is much greater than that of the white hot object? Couldn't infrared actually get something white hot hotter?

Yes. If you examine the radiance curve of any white hot object on Earth, the peak is still in the infrared range. What you see as visible light is the edge of the curve intruding into the visible light bands. There are still plenty of infrared photons that could be absorbed and converted into thermal energy (a trait of infrared light) and make the white hot object even hotter.

Even in space, where a star is putting out a brilliant white light and can even have a Wien's law peak of radiance in the visible range, some molecules still have only a little energy (just less of them as a percentage). Those can still absorb infrared light, increasing the overall temperature of the star.

In all cases, a molecule will only absorb a photon that has greater energy than the molecule already has.


The Parrot Killer
16-08-2019 16:36
IBdaMannProfile picture★★★★★
(4920)
Into the Night wrote:It should be mentioned here that radiant heating ONLY occurs if the light emitted due to blackbody radiance is absorbed and results in thermal energy in some other mass.

Light in and of itself is not heat, although it can be a means of heating.

Great point. I'm posting this to show that it is worth repeating.



Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
16-08-2019 18:24
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:Two sumo wrestlers, the large easy bet to win, and the much smaller sure to lose opponent.

I really don't recommend a Sumo analogy.

tmiddles wrote: The big wrestler pushes the small one, the small one pushes back (flexing leg muscles, really trying). But there is a steady movement of the big guy pushing the little guy off the mat. It matters that the little guy is putting up a fight, it really slows the big guy down, but it never reverses the direction.

No. Just no. There is no thermal "push-back."

If you open the flood gates on a dam allowing the water to flow into the river below, roughly how much of the water in the river is fighting to flow up into the dam? Water does not struggle to flow upstream like a spawning salmon. Thermal energy does not wrestle to flow from lower temperature to higher temperature.

tmiddles wrote: The NET flow of energy is in one direction, from hot to cold.

Just "the flow" of energy is in one direction. Whenever you see someone squeezing the word "net" in front of "flow" then s/he is implying that some of the flow is in violation of the 2nd law of thermodynamics.

If you incorporate the term "net flow" within this context, the immediate assumption is that you are regurgitating warmizombie dogma from some source that you are allowing to manipulate you. If you want to discuss thermodynamics and black body science, then you discuss the "flow" of thermal energy. If you want the discussion to take a turn towards the warmizombies who are manipulating you, feel free to start talking about the "net flow" and I'm sure I'm not the only one who will oblige.

tmiddles wrote: But my question is (going back to our ball shell): Doesn't a big flow of radiant energy move across the gap towards the shell as a small flow of radiant energy moves the other way?

Is your question "How is it that physical bodies of matter crash into each other per the Pauli Exclusion Principle whereas non-matter photons do not?"?

tmiddles wrote: Don't they mover right through each other in opposite directions?

Yes. The Pauli Exclusion Principle does not apply to photons.

I'm just guessing but I believe Into the Night would probably like it pointed out that phtons can certainly interact with each other in a variety of ways, but not in the "crash into each other" sense, especially not in the manner of one "pushing" another "off the matt".

tmiddles wrote: Separate question: Is there anything that prevents a white hot object from getting hotter due to infra-red radiation?

Yes ... the infrared thermal radiation coming from a cooler object.

Otherwise, if the infrared is coming from a body of higher temperature, then no, there is nothing preventing absorption and an increase in temperature. The 2nd law of thermodynamics applies always and everywhere.

To your point, suppose you had a big chunk of steel packed in a magickal insulation that nullifies all but infrared radiation which it allows to pass through unabated. You then hurl the package towards the sun.

What do you believe will happen to the temperature of the steel? Would you agree that the sun, considering only infrared radiation, makes one hell of a heat lamp? Would the steel eventually become white hot? Would the steel's temperature exceed "white hot"? Would the steel eventually vaporize?

I would like to be clear. The reason this happens is due to the sun's Radiance, not simply due to the specific energy of the photons. I was looking over some of my previous posts and I noticed that I fell into the trap of implying an equivalence between Radiance (power) and individual photon energy and I need to correct that. I will also mention that this is not the first time I have been lax and allowed myself to slip on this matter. I apologize for any confusion I may have caused in doing so.

Photons are simply quanta quantities of energy.

Radiance is power over an area.

Bodies of higher temperature radiate particular wavelengths with greater Radiance (radiativity at that wavelength) than bodies of lower temperature (re: Wein's Displacement law) which causes the radiation of that wavelength to be absorbed, bringing the colder body up to a higher energy state (Re: Planck's)

tmiddles wrote: What if the power of the infrared is much greater than that of the white hot object? Couldn't infrared actually get something white hot hotter?

Absolutely. You got it. Apparently you were not confused.


.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
16-08-2019 23:21
Into the NightProfile picture★★★★★
(9582)
IBdaMann wrote:
tmiddles wrote:Two sumo wrestlers, the large easy bet to win, and the much smaller sure to lose opponent.

I really don't recommend a Sumo analogy.

tmiddles wrote: The big wrestler pushes the small one, the small one pushes back (flexing leg muscles, really trying). But there is a steady movement of the big guy pushing the little guy off the mat. It matters that the little guy is putting up a fight, it really slows the big guy down, but it never reverses the direction.

No. Just no. There is no thermal "push-back."

If you open the flood gates on a dam allowing the water to flow into the river below, roughly how much of the water in the river is fighting to flow up into the dam? Water does not struggle to flow upstream like a spawning salmon. Thermal energy does not wrestle to flow from lower temperature to higher temperature.

tmiddles wrote: The NET flow of energy is in one direction, from hot to cold.

Just "the flow" of energy is in one direction. Whenever you see someone squeezing the word "net" in front of "flow" then s/he is implying that some of the flow is in violation of the 2nd law of thermodynamics.

If you incorporate the term "net flow" within this context, the immediate assumption is that you are regurgitating warmizombie dogma from some source that you are allowing to manipulate you. If you want to discuss thermodynamics and black body science, then you discuss the "flow" of thermal energy. If you want the discussion to take a turn towards the warmizombies who are manipulating you, feel free to start talking about the "net flow" and I'm sure I'm not the only one who will oblige.

tmiddles wrote: But my question is (going back to our ball shell): Doesn't a big flow of radiant energy move across the gap towards the shell as a small flow of radiant energy moves the other way?

Is your question "How is it that physical bodies of matter crash into each other per the Pauli Exclusion Principle whereas non-matter photons do not?"?

tmiddles wrote: Don't they mover right through each other in opposite directions?

Yes. The Pauli Exclusion Principle does not apply to photons.

I'm just guessing but I believe Into the Night would probably like it pointed out that phtons can certainly interact with each other in a variety of ways, but not in the "crash into each other" sense, especially not in the manner of one "pushing" another "off the matt".

tmiddles wrote: Separate question: Is there anything that prevents a white hot object from getting hotter due to infra-red radiation?

Yes ... the infrared thermal radiation coming from a cooler object.

Otherwise, if the infrared is coming from a body of higher temperature, then no, there is nothing preventing absorption and an increase in temperature. The 2nd law of thermodynamics applies always and everywhere.

To your point, suppose you had a big chunk of steel packed in a magickal insulation that nullifies all but infrared radiation which it allows to pass through unabated. You then hurl the package towards the sun.

What do you believe will happen to the temperature of the steel? Would you agree that the sun, considering only infrared radiation, makes one hell of a heat lamp? Would the steel eventually become white hot? Would the steel's temperature exceed "white hot"? Would the steel eventually vaporize?

I would like to be clear. The reason this happens is due to the sun's Radiance, not simply due to the specific energy of the photons. I was looking over some of my previous posts and I noticed that I fell into the trap of implying an equivalence between Radiance (power) and individual photon energy and I need to correct that. I will also mention that this is not the first time I have been lax and allowed myself to slip on this matter. I apologize for any confusion I may have caused in doing so.

Photons are simply quanta quantities of energy.

Radiance is power over an area.

Bodies of higher temperature radiate particular wavelengths with greater Radiance (radiativity at that wavelength) than bodies of lower temperature (re: Wein's Displacement law) which causes the radiation of that wavelength to be absorbed, bringing the colder body up to a higher energy state (Re: Planck's)

tmiddles wrote: What if the power of the infrared is much greater than that of the white hot object? Couldn't infrared actually get something white hot hotter?

Absolutely. You got it. Apparently you were not confused.


.


You pointed it out. I don't have to.



The Parrot Killer
17-08-2019 04:00
tmiddlesProfile picture★★★★☆
(1322)
OK so the reason for the shoving/sumo visualization is I see molecules girateing around in every direction like a spiragraph(remember those!) and smacking into nearby molecules regardless of the direction they lie in.

What I'm not confused about, from the other post as well is the notion that radiance and conduction are not acting the same way. That an object will radiate and it's temperature change with no influence from objects nearby, or a void, unless the nearby object are hotter?

You might have missed this ITN:

tmiddles wrote:
tmiddles wrote:
IBdaMann wrote:...heats the inner ball's temperature to a 475degC equilibrium. It gets red hot and starts really radiating towards the outer shell, ...

So the shell arrangement would be the equivalent of a material that couldn't conduct from it's central mass to it's out mass.


I was thinking more about this. The heated ball with and without a shell and pulses of 100 joules of energy entering into it. Assuming that it will emit 100 joules of energy over a 4 second period.

Without a shell:
Ball emits 100 joules over a 4 second period and they go straight out into space unimpeded.
Ball
1 -25
2 -25
3 -25
4 -25
Total -100

With a shell:
Ball emits 100 joules over a 4 second period, they are transferred to the shell that takes one second to absorb and re-emit a joule. Due to the handing back of 12.5 joules each time it ends up taking 7 seconds for the 100 joules to make their exit.
Ball..................Shell
1 -25................+25
2 -25 (+12.5).....+25, -12.5in, -12.5out
3 -25 (+12.5).....+25, -12.5in, -12.5out
4 -25 (+12.5).....+25, -12.5in, -12.5out
5 -25 (+12.5).....+25, -12.5in, -12.5out
6 -25 (+12.5).....+25, -12.5in, -12.5out
7 -25 (+12.5).....+25, -12.5in, -12.5out
Total -100 (-175+75)

If I feed 100 joules per second into the ball without a shell and it take 4 seconds for that pulse of energy to leave then that's a thermal energy total of 250 joules at any instant.
Example with the pulses lettered and the energy loss shown
Seconds____1____2____3____4____5____6____7
A________ 100__75___50___25____0____100__75
B__________0__100___75___50___25____0___100
C_________25___0___100___75___50___25____0
D_________50___25___0___100___75___50___25
E_________75___50___25____0___100___75___50
Total--------250---250---250---250---250---250---250

If I feed 100 joules per second into the ball and it take 7 seconds for the energy to leave then that's a thermal energy total of 400 joules at any instant.
Seconds____1____2____3____4____5____6____7
A_________100__86___71___57____43___29___14
B__________0___100__86___71____57___43___29
C_________14___0____100__86____71___57___43
D_________29___14____0___100___86___71___57
E_________43___29___14___0____100___86___71
F_________57___43___29___14____0___100___86
G_________71___57___43___29____14____0___100
H_________86___71___57____43___29___14___0
Total--------400---400---400---400---400---400---400

So the ball has a higher temperature with the shell, 400 joules present vs. 250
17-08-2019 10:00
Into the NightProfile picture★★★★★
(9582)
tmiddles wrote:
OK so the reason for the shoving/sumo visualization is I see molecules girateing around in every direction like a spiragraph(remember those!) and smacking into nearby molecules regardless of the direction they lie in.

What I'm not confused about, from the other post as well is the notion that radiance and conduction are not acting the same way. That an object will radiate and it's temperature change with no influence from objects nearby, or a void, unless the nearby object are hotter?

You might have missed this ITN:

tmiddles wrote:
tmiddles wrote:
IBdaMann wrote:...heats the inner ball's temperature to a 475degC equilibrium. It gets red hot and starts really radiating towards the outer shell, ...

So the shell arrangement would be the equivalent of a material that couldn't conduct from it's central mass to it's out mass.


I was thinking more about this. The heated ball with and without a shell and pulses of 100 joules of energy entering into it. Assuming that it will emit 100 joules of energy over a 4 second period.

Without a shell:
Ball emits 100 joules over a 4 second period and they go straight out into space unimpeded.
Ball
1 -25
2 -25
3 -25
4 -25
Total -100

With a shell:
Ball emits 100 joules over a 4 second period, they are transferred to the shell that takes one second to absorb and re-emit a joule. Due to the handing back of 12.5 joules each time it ends up taking 7 seconds for the 100 joules to make their exit.
Ball..................Shell
1 -25................+25
2 -25 (+12.5).....+25, -12.5in, -12.5out
3 -25 (+12.5).....+25, -12.5in, -12.5out
4 -25 (+12.5).....+25, -12.5in, -12.5out
5 -25 (+12.5).....+25, -12.5in, -12.5out
6 -25 (+12.5).....+25, -12.5in, -12.5out
7 -25 (+12.5).....+25, -12.5in, -12.5out
Total -100 (-175+75)

If I feed 100 joules per second into the ball without a shell and it take 4 seconds for that pulse of energy to leave then that's a thermal energy total of 250 joules at any instant.
Example with the pulses lettered and the energy loss shown
Seconds____1____2____3____4____5____6____7
A________ 100__75___50___25____0____100__75
B__________0__100___75___50___25____0___100
C_________25___0___100___75___50___25____0
D_________50___25___0___100___75___50___25
E_________75___50___25____0___100___75___50
Total--------250---250---250---250---250---250---250

If I feed 100 joules per second into the ball and it take 7 seconds for the energy to leave then that's a thermal energy total of 400 joules at any instant.
Seconds____1____2____3____4____5____6____7
A_________100__86___71___57____43___29___14
B__________0___100__86___71____57___43___29
C_________14___0____100__86____71___57___43
D_________29___14____0___100___86___71___57
E_________43___29___14___0____100___86___71
F_________57___43___29___14____0___100___86
G_________71___57___43___29____14____0___100
H_________86___71___57____43___29___14___0
Total--------400---400---400---400---400---400---400

So the ball has a higher temperature with the shell, 400 joules present vs. 250

The ball heats the shell until they are the same temperature. Both will be cooler than the ball's original temperature. It's really very simple.


The Parrot Killer
17-08-2019 10:58
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
tmiddles wrote: Separate question: Is there anything that prevents a white hot object from getting hotter due to infra-red radiation?

Yes ... the infrared thermal radiation coming from a cooler object.

Otherwise, if the infrared is coming from a body of higher temperature, then no, there is nothing preventing absorption and an increase in temperature. The 2nd law of thermodynamics applies always and everywhere.


Alright I'm having a hard time with this so I'm asking for help. I'm thinking I've just misunderstood you guys.

So full disclosure I'm very good at math and have taught it but I went to art school so don't have a solid post HS education in math and science. So I'm trying to learn now.

What I thought you were both saying is that if we're in a vacuum, with convection and conduction out of the picture, and there are two objects, or easier still an object in a room or a shell, that the hotter object does it's thing and the cooler object just accepts the incoming radiation with no ability to radiate to the hotter object and to have that radiation absorbed.

Which I'm visualizing as the hotter object will cool just as fast in a vacuum/space with no shell in the way as it would with a shell there. But in trying to find lessons online I'm not finding that scenario.

"The Stefan-Boltzmann law predicts
the power of the radiative heat exchange between two objects
as σ((T2)^4-(T1)^4) where σ is the Stefan-Bolzmann constant and
T1,T2 are the temperatures of objects"


at 20:50 in this video (poor quality): https://youtu.be/LR5bYxC4syI
Hotter ball in a cooler room "Heat leaves the sphere but also enters the sphere"
151.4 W out, 106.4 W in, for a net of 44.5 W out


Does heat loss by radiation depend on the surrounding temperature?"net heat loss by radiation does depend on the surrounding temperature. Normally objects lose energy to their surroundings and gain energy at the same time. ","The net heat loss (overall heat loss) is equal to radiation emitted - the radiation absorbed. "

"some heat will also be absorbed from the
surroundings, and that will depend on the temperature of objects in the
line of sight of the surface. "


There was this:
No.
This is because the question asks about emission by radiation.... temperature of the surroundings is not a factor when considering emission.


Anyway thought I'd ask before more digging around.
17-08-2019 12:35
tmiddlesProfile picture★★★★☆
(1322)
tmiddles wrote:
IBdaMann wrote:
tmiddles wrote: Separate question: Is there anything that prevents a white hot object from getting hotter due to infra-red radiation?

Yes ... the infrared thermal radiation coming from a cooler object.

Otherwise, if the infrared is coming from a body of higher temperature, then no, there is nothing preventing absorption and an increase in temperature. The 2nd law of thermodynamics applies always and everywhere.


Alright I'm having a hard time with this so I'm asking for help. I'm thinking I've just misunderstood you guys.

So full disclosure I'm very good at math and have taught it but I went to art school so don't have a solid post HS education in math and science. So I'm trying to learn now.

What I thought you were both saying is that if we're in a vacuum, with convection and conduction out of the picture, and there are two objects, or easier still an object in a room or a shell, that the hotter object does it's thing and the cooler object just accepts the incoming radiation with no ability to radiate to the hotter object and to have that radiation absorbed.

Which I'm visualizing as the hotter object will cool just as fast in a vacuum/space with no shell in the way as it would with a shell there. But in trying to find lessons online I'm not finding that scenario.

"The Stefan-Boltzmann law predicts
the power of the radiative heat exchange between two objects
as σ((T2)^4-(T1)^4) where σ is the Stefan-Bolzmann constant and
T1,T2 are the temperatures of objects"


at 20:50 in this video (poor quality): https://youtu.be/LR5bYxC4syI
Hotter ball in a cooler room "Heat leaves the sphere but also enters the sphere"
151.4 W out, 106.4 W in, for a net of 44.5 W out


Does heat loss by radiation depend on the surrounding temperature?"net heat loss by radiation does depend on the surrounding temperature. Normally objects lose energy to their surroundings and gain energy at the same time. ","The net heat loss (overall heat loss) is equal to radiation emitted - the radiation absorbed. "

"some heat will also be absorbed from the
surroundings, and that will depend on the temperature of objects in the
line of sight of the surface. "


There was this:
No.
This is because the question asks about emission by radiation.... temperature of the surroundings is not a factor when considering emission.


Anyway thought I'd ask before more digging around.


Still digging but hook me up with anything good.
Found: http://www.climate-change-theory.com/evidence.html

Interesting stuff.
17-08-2019 16:29
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:
Still digging but hook me up with anything good.
Found: http://www.climate-change-theory.com/evidence.html

Interesting stuff.


I have not had a chance to read this so it is not vetted, but read it and we can discuss after I get a chance to look at it.

https://hockeyschtick.blogspot.com/2015/08/plancks-quantum-theory-explains-why-low.html


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
18-08-2019 04:42
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
https://hockeyschtick.blogspot.com/2015/08/plancks-quantum-theory-explains-why-low.html


Perfect! Thanks
18-08-2019 07:54
tmiddlesProfile picture★★★★☆
(1322)
tmiddles wrote:
IBdaMann wrote:
https://hockeyschtick.blogspot.com/2015/08/plancks-quantum-theory-explains-why-low.html


Perfect! Thanks


So this is an unclear presentation by hockeyshtick in my view. It give 4 pages from a text teaching Planck's constant and the all or nothing quantized energy states. That if EM radiation reaches an object and fails to achieve the necessary energy level it will not be thermalized.

But what happens to it?

If a hot plate A has two cooler plates B on either side: The cool plates thermalize the radiance from A, they are radiating into A and that radiance fails to be thermalized, but what happens to it?

Does it reflect back into B allowing B to re-thermalize it (so that B effectivly only has it's outer surface to shed energy through radiance).

Also would A shed energy through radiance just the same, just as quickly, without the plates B being there? Right up to the point when B has as hot as A?

This would happen too if B could not heat A:

A: 100 -10% on each side, B: 50 -10% on each side


-----[B(50)B]--------------[A(100)A]--------------[B(50)B]-----
<-5-[B(40)B]-5-><--10--[A(080)A]--10--><-5-[B(40)B]-5->
-----[B(50)B]<------5-----[A(080)A]------5----->[B(50)B]-----
-----[B(55)B]--------------[A(080)A]--------------[B(55)B]-----
<-5-[B(45)B]-5-><--10--[A(060)A]--10--><-5-[B(45)B]-5->
-----[B(55)B]<------5-----[A(060)A]------5----->[B(55)B]-----
-----[B(60)B]--------------[A(060)A]--------------[B(60)B]-----EQUAL

Now A and B would I guess thermalize each others EM since they're the same temperature.

Just doesn't seem right that A would behave the exact same way up until that point without B being there.
--------------[A(100)A]--------------
------<--10--[A(080)A]--10-->-----
--------------[A(080)A]--------------
-----<--10--[A(060)A]--10-->-----
--------------[A(060)A]--------------
Edited on 18-08-2019 07:54
18-08-2019 09:36
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:So this is an unclear presentation by hockeyshtick in my view.

I agree.

tmiddles wrote: That if EM radiation reaches an object and fails to achieve the necessary energy level it will not be thermalized.

That is correct; it is the 2nd law of thermodynamics hardcoded into Planck's

tmiddles wrote:But what happens to it?

You are going to have a difficult time finding a satisfying answer. This is all I can tell you without researching it further:
1) photons of the lower temperature object are not absorbed by the higher temperature object.
2) what the photons actually do is governed more by uncertainty than by any science that predicts what will happen. Like I said before, photons can deflect, do back-flips, take selfies and interact in any way other than being absorbed.

Disappointing, eh?

Now don't tell anyone I said this, and I'll deny it if you do ... but as much as I claim to know absolutely everything, there are actually a couple of things I don't know. This is one of them. There might be someone out there who has a model for the activity of thermal photons that are insufficient to be absorbed under Planck's, but I haven't seen it.

tmiddles wrote:Also would A shed energy through radiance just the same, just as quickly, without the plates B being there?

Yes. Radiance is driven by absolute temperature alone.


.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
18-08-2019 09:59
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
tmiddles wrote:But what happens to it?

You are going to have a difficult time finding a satisfying answer.
tmiddles wrote:Also would A shed energy through radiance just the same, just as quickly, without the plates B being there?

Yes. Radiance is driven by absolute temperature alone.
.


This seems easily proven in the lab in any case. Creating hot plates, cold plates and vacuums is very doable.

Confirming that a hot plate with a cold plate near it cools as quickly as a hot plate with nothing near it also seems easily done.

I'll keep hunting.
18-08-2019 12:28
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
tmiddles wrote:Also would A shed energy through radiance just the same, just as quickly, without the plates B being there?

Yes. Radiance is driven by absolute temperature alone.
.


OK I think that while radiance depends only on temperature that radiant heat loss is effected by the surrounding radiance. A hotter object may be absorbing it or not but it does effect the time it takes for the the energy to be shed by radiance.

I was thinking about how if you're in a vacuum and only the temperature of the object matters then it would be the same as being in deep space. I keep running into this equation:
Thermal Radiation

Qt=eσA(T^4skin−T^4ambient)

Q = heat loss in Joules
t = time in seconds
e = emissivity of skin (≈ 0.95 for human body)
σ = Stefan-Boltzmann constant
A = surface area of human body

And (here)
Net Radiation Loss Rate
If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as

q = ε σ (Th^4 - Tc^4) Ah

where
Th = hot body absolute temperature (K)
Tc = cold surroundings absolute temperature (K)
Ah = area of the hot object (m2)

And it's basically what I saw earlier here:
"The Stefan-Boltzmann law predicts
the power of the radiative heat exchange between two objects
as σ((T2)^4-(T1)^4) where σ is the Stefan-Bolzmann constant and
T1,T2 are the temperatures of objects"


So if these equations are correct then the time it takes for a hot object to radiate out is calculated along with the surrounding radiance.

The math come out so radically different that it doesn't seem possible the above equations could be incorrect and that they should disregard the ambient radiance. The temperatures are to the 4th power, if you had a 100 C ball in a 0 C shell you're dealing with 373 Kelvin and 273, so the equation would have (373^4-273^4)= 1.38^10 if incoming radiance matters and just 373^4 if it doesn't, 1.9^10. So the radiant heat loss is 38% greater/faster without the ambient radiance considered.

So I think the shell would slow the balls radiant heat loss.
Edited on 18-08-2019 12:39
18-08-2019 17:21
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:OK I think that while radiance depends only on temperature that radiant heat loss is effected by the surrounding radiance.

Great. Why do you think this? Are you confusing thermal radiation with conduction?

tmiddles wrote: A hotter object may be absorbing it or not ...

No. This is not the case. A hotter object may not be absorbing it.

tmiddles wrote: ... but it does effect the time it takes for the the energy to be shed by radiance.

Great. Why do you think this? You must be confusing thermal radiation with conduction.

tmiddles wrote: I keep running into this equation:
Thermal Radiation

Qt=eσA(T^4skin−T^4ambient)


This is where I walk to the board and circle "(T^4skin−T^4ambient)"

You have injected the temperature differential of conduction into Stefan-Boltzmann which is driven by absolute temperature alone:

Stefan-Boltzmann:

Radiance (Power/Area) = Temperature^4 * Emissivity * Boltzmann

Whereas:

Conduction (Power/Area) = Thermal_Coefficient * [Temperature(Higher)-Temperature(Lower)] / Length

Stefan-Boltzmann is not:

Radiance (Power/Area) = (Temperature(Higher)^4 - Temperature(Lower)^4) * Emissivity * Boltzmann

... which is what you have:

Qt=eσA(T^4skin−T^4ambient) ... which, as you can see, is a strange mix of the classic symbology for both conduction and Stefan-Boltzmann.

The shell does not "slow" the radiance emitted by the ball ....

... so no, "greenhouse gases" do not "slow" radiance emitted by the earth's surface.

(ask me how I knew you were "going there")


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
18-08-2019 21:20
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
tmiddles wrote:OK I think that while radiance depends only on temperature that radiant heat loss is effected by the surrounding radiance.

Great. Why do you think this? Are you confusing thermal radiation with conduction?


So all 3 examples, none of it my work, are in a vacuum.

First is a Quora question about freezing in space (granted pretty casual)

But this is not a casual reference:
tmiddles wrote:
And it's basically what I saw earlier here:
"The Stefan-Boltzmann law predicts
the power of the radiative heat exchange between two objects
as σ((T2)^4-(T1)^4) where σ is the Stefan-Bolzmann constant and
T1,T2 are the temperatures of objects"

.

It says "parallel plates of silicon carbide in vacuum"

Is σ((T2)^4-(T1)^4) not the right formula for radiative heat exchange?
18-08-2019 22:26
tmiddlesProfile picture★★★★☆
(1322)
tmiddles wrote:
Is σ((T2)^4-(T1)^4) not the right formula for radiative heat exchange?

That research was published in Physical Review an American peer-reviewed scientific journal established in 1893 by Edward Nichols. It publishes original research as well as scientific and literature reviews on all aspects of physics. It is published by the American Physical Society (APS).

Here are two videos:
LearnChemE
Properties of Radiative Heat Transfer 7 min
https://youtu.be/epioKYRRpPI
"G=irradiation which can be transmitted, reflected or absorbed...
Q=J-G"

AK Lecture, 9 min
fundamental explanation of radiative heat transfer:
https://youtu.be/93-_JhGNn1Y
" at the same time any object that radiates energy also absorbs energy from the surroundings"
Again: ((T2)^4-(T1)^4)
The instructor is very clear this is NOT conductive or convective but ONLY radiative
Edited on 18-08-2019 22:48
19-08-2019 02:00
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote: That research was published in Physical Review an American peer-reviewed scientific journal established in 1893 by Edward Nichols.

There is so much wrong with this statement above, only somebody being bent over furniture would say something this stupid.

(I apologize for the blunt nature of my comment)

1) I don't give a chit in what publication crap is published. Chit is chit.

2) It takes a special type of gullible to allow oneself to be conned into believing that "peer reviewed" has anything to do with science.

3) Edward Nichols is deceased. The manner in which other people manipulate the public in the name of his legacy with bogus gibber-babble does not overcome its violations of physics.

4) You referenced a magazine article, not science. Next time reference the Lord of the Rings. Maybe that will work out better for you.

5) I took the time to explain to you exactly what was wrong with what you posted and your only response is to claim that you found some gibber-babble on the internet. You were too busy allowing the author of that article bend you over furniture to pay attention to the specific science errors to which I was directing your attention. Go back, reread my post and address those points. I'm not interested in discussing the merits of your accpetance of everything you read on the internet as being true.

6) No publication owns science. Maybe I'm the first person to tell you this but your belief that this particular publication somehow does should be your huge red flag that you are waiste deep in a religion.

7) You are clearly frustrated that your attempt to get around physics to champion your faith in Global Warming did not succeed and you are now resorting to blaming me for not buying the gibber-babble crap in that article.

Did I miss anything?

tmiddles wrote:

[quote]tmiddles wrote:Here are two videos:

I don't like watching videos that I don't choose because they are normally a waste of my time, and I normally get frustrated because I can't ask the video questions, i.e. I can't cross-examine anything that appears to be wrong.

I nonetheless began watching the first video and immediately had a bunch of questions. The speaker didn't sound like she was an authority on the subject and I just gave up after a short while.

Sorry.

Is there any way you can bring any of these "experts" to this forum so I can have them field a few questions and support their assertions? Until then I have to go with science.

.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
19-08-2019 02:48
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
I don't like watching videos .... The speaker didn't sound like she was an authority on the subject
.


That was PHD Dr. Janet deGrazia: Teaching Professor At the University of Colorado.

I tried to quote why they were relevant. I figured you'd only want to watch them if you thought they made an error. Videos can make that easier.

But this isn't anything but the basics of physics she's teaching for college students.

I can't find any research, instruction or work of any kind that doesn't use σ((T2)^4-(T1)^4)

Also the answers to problems would be WAY off comparing the two versions (see example at the bottom).

Also if the hotter object isn't able to absorb from the cooler one wouldn't that make a graph of reaching thermal equilibrium look like two lines meeting at a severe angle? The rising/ falling temps would crash into the equilibrium temp and stop like they hit a wall. (like tron light cycles)

Here is another much cleaner and focused reference than a video. It's: University Physics Volume 2
1.6 Mechanisms of Heat Transfer
"The Stefan-Boltzmann equation needs only slight refinement to deal with a simple case of an object's absorption of radiation from its surroundings. Assuming that an object with a temperature T1 is surrounded by an environment with uniform temperature T2, the net rate of heat transfer by radiation is

Pnet=σeA(T2^4−T1^4)

where e is the emissivity of the object alone. In other words, it does not matter whether the surroundings are white, gray, or black: The balance of radiation into and out of the object depends on how well it emits and absorbs radiation. When T2>T1, the quantity Pnet is positive, that is, the net heat transfer is from hot to cold."


This equation works well whether the object or it's surroundings have a higher temperature.

It gives an example:
"EXAMPLE 1.13
Calculating the Net Heat Transfer of a Person
What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient temperature is 22.0°C? The person has a normal skin temperature of 33.0°C and a surface area of 1.50m2. The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes place.

Strategy
We can solve this by using the equation for the rate of radiative heat transfer.

Solution
Insert the temperature values T2=295K and T1=306K, so that

Qt=σeA(T2^4−T1^4)=(5.67×10−8J/s⋅m2⋅K4)(0.97)(1.50m2)[(295K)4−(306K)4]=−99J/s=−99W." It's actually: 98.5320

Now if you do that calculation without including T2, the surrounds, you get:
Qt=σeA(T2^4)=(5.67×10−8J/s⋅m2⋅K4)(0.97)(1.50m2)[(295K)4]=−723J/s=−723W , it's actually: 723.3221

Now of course losing thermal energy 7 times faster really would come up as an error. In fact it would be pretty noticeable in real life.

But this would only be until the person reached equilibrium at which point 723Watts would suddenly become 0? (or actually it would drop from 723 to just over 624.8Watts, the temp of the room before suddenly BAM! going to zero.

Come on IBdaMann!!! You either just got this one wrong or I've misunderstood your last posts.
Edited on 19-08-2019 03:43
19-08-2019 04:48
tmiddlesProfile picture★★★★☆
(1322)
I see it everywhere I look up radiative heat I must be misunderstanding what you mean.


engineersedge.com




myodesie.com/RadiantHeatTransfer
Edited on 19-08-2019 04:50
19-08-2019 06:49
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:
I see it everywhere I look up radiative heat I must be misunderstanding what you mean.


This T1^4 - T2^4 that you insist is everywhere, I'm having difficulty finding anywhere.

Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Help me find it here:


Frankly, I couldn't find the crap you posted anywhere, but I will readily admit up front that I didn't check any scientifically illiterate warmizombie websites that preach physics violations. Is that where I need to look?

Oh, I made an exception and looked at your Wiki-Bible to at least find it there so I could read a bit about what you are getting at ... but I'm going to need your help finding it there as well.





.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
19-08-2019 08:51
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
Help me find it here:

Which is an image at link:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/imgheat/stef1.png

1st one, Here you go (top of the page):
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html



Of course the stefan-boltzman law is where the formula for radiative heat transfer between bodies comes from. Which is why both are found at
hyperphysics.phy-astr.gsu.edu which I trust we can agree is not:
IBdaMann wrote:any scientifically illiterate warmizombie websites that preach physics violations.

You'll notice I haven't posted from wikipedia in a while! I'm considering these things.

Nothing I've talked about or linked even touches on anything related to "warmizombie" stuff. It's basic college Physics online to help students with their homework.

We were talking about a ball and a shell and that's how I got here. The thermal transfer between two bodies in a vacuum.

Now you haven't responded yet to most of my post:

tmiddles wrote:EXAMPLE 1.13...Calculating the Net Heat Transfer of a Person...
Qt=σeA(T2^4−T1^4)=(5.67×10−8J/s⋅m2⋅K4)(0.97)(1.50m2)[(295K)4−(306K)4]=−99J/s=−99W." It's actually: 98.5320

Now if you do that calculation without including T2, the surrounds, you get:
Qt=σeA(T2^4)=(5.67×10−8J/s⋅m2⋅K4)(0.97)(1.50m2)[(295K)4]=−723J/s=−723W , it's actually: 723.3221

Now of course losing thermal energy 7 times faster really would come up as an error. In fact it would be pretty noticeable in real life.

Do you think the answer should be 723 watts and not 98?

tmiddles wrote:... if the hotter object isn't able to absorb from the cooler one wouldn't that make a graph of reaching thermal equilibrium look like two lines meeting at a severe angle? [in the example problem] would [it] drop [gradually] from 723 to just over 624.8Watts, the temp of the room before suddenly BAM! going to zero.

Am I misunderstanding how this plays out if a hotter object is unaffected by the cooler one? (in a vacuum of course)
Edited on 19-08-2019 09:08
19-08-2019 10:19
tmiddlesProfile picture★★★★☆
(1322)
tmiddles wrote:
Which is an image at link:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/imgheat/stef1.png


Just to be as accurate as possible and to link to the same page you used:

The image you posted:


Came from this page:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

And just below it there is also:
"If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form"


Edited on 19-08-2019 10:19
19-08-2019 15:29
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:Do you think the answer should be 723 watts and not 98?

OK, let's say you are on to something. The issue before you is that you have been presented two equations, both as science, yet that are not internally consistent, i.e. they are not compatible. One tells you that a body's thermal radiation is based solely on the temperature of the body while the other tells you that the power of a body's emission is dependent upon the difference between the body's temperature and some other temperature. One equation is touted as the standard time-tested classical physics while the other, although not as common, appears more intuitive and follows the format of convection.

You want to know which one is correct but you also know that you shouldn't take any human/person solely on his/her word ... and that includes websites and published articles. What is one to do?

Please forget that I have been clinging to the classical RAD = T^4 * emiss * bolt. I am now thinking that I have gotten off track. The point to remember is that the version of Stefan-Boltzman that I have been referencing could very well be in error and you could be completely correct.

... so how do we resolve this? Do we go to some website, as if there is any human who gets to arbitrate what is science and what is not? Do we count the number of peer-reviewed articles and establish a "consensus"?

[Hint: Into the Night and I have discussed this very issue with you ad nauseum]

Let's presume you might very well be correct. What should we do?


.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
19-08-2019 21:27
tmiddlesProfile picture★★★★☆
(1322)
IBdaMann wrote:
The issue before you is that you have been presented two equations, both as science, yet that are not internally consistent, i.e.they are not compatible. One tells you that a body's thermal radiation is based solely on the temperature of the body while the other tells you that the power of a body's emission is dependent upon the difference between the body's temperature and some other temperature.
.


They are compatible.

I think the only conflict you have is the notion that radiance from a cooler body cannot be absorbed by a hotter one.

Assuming it can then σ((T2)^4-(T1)^4) doesn't contradict anything. The radiance out is still based on temperature but there is also a radiance in. It is a situation of net energy from hotter to colder and the 2nd law of thermodynamics isn't violated,

It makes the answer to the ball shell in a void consistent to what occurs with conduction,

And it means a naked person in a dark room from the example has a net heat loss of 100 watts and not 700.

I can feel the differences in radiance of two objects which are both colder than my hand so that's all consistent,
19-08-2019 22:24
Into the NightProfile picture★★★★★
(9582)
tmiddles wrote:
IBdaMann wrote:
The issue before you is that you have been presented two equations, both as science, yet that are not internally consistent, i.e.they are not compatible. One tells you that a body's thermal radiation is based solely on the temperature of the body while the other tells you that the power of a body's emission is dependent upon the difference between the body's temperature and some other temperature.
.


They are compatible.

I think the only conflict you have is the notion that radiance from a cooler body cannot be absorbed by a hotter one.

It can't. No molecule will absorb a photon of less energy than the molecule already has.
tmiddles wrote:
Assuming it can

It can't.
tmiddles wrote:
The radiance out is still based on temperature but there is also a radiance in. It is a situation of net energy from hotter to colder and the 2nd law of thermodynamics isn't violated,

There is no 'net'. Heat only flows one way. You cannot heat a warmer object with a cooler one. You are confusing light with heat.
tmiddles wrote:
It makes the answer to the ball shell in a void consistent to what occurs with conduction,

No. There is no coefficient of coupling.
tmiddles wrote:
And it means a naked person in a dark room from the example has a net heat loss of 100 watts and not 700.

A human body can generate about 100 watts of power, including the ability to maintain body temperature. If the room is cold enough, they will become hypothermic. Their body temperature drops. It doesn't matter if the room is light or dark.
tmiddles wrote:
I can feel the differences in radiance of two objects which are both colder than my hand so that's all consistent,

What you feel is ONLY what is absorbed by your hand. You do not absorb all photons either. Since your hand is mostly water, the absorption frequencies for your hand are very similar to that of liquid water.


The Parrot Killer
19-08-2019 22:35
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote: They are compatible.

I claim they are not, and that you have the math in front of you to this effect.

tmiddles wrote:I think the only conflict you have is the notion that radiance from a cooler body cannot be absorbed by a hotter one.

Correct. Max Planck enforced the 2nd law of thermodynamics by building the 2nd law of thermodynamics into his model of black body radiation.

If you are telling me that you have "science" that has radiation from a cooler body being absorbed by a body of higher temperature then your "science" is incompatible with Planck's model of black body radiation.

So, without EVADING the question, what should we do according to the scientific method? How do we establish which model is the best one? Based on their incompatibility we know that at least one of them is therefore incorrect. How do we falsify the model that is false?

tmiddles wrote: Assuming it can

... but I will not. We need to first resolve the fundamental issue of which model we should discard and which model we should keep, and then apply the model we should keep. Otherwise we are wasting time.

tmiddles wrote: σ((T2)^4-(T1)^4) doesn't contradict anything.

Yes it does. Nature cannot adhere to one unambiguous model while also adhereing to another unambiguous model. One must be discarded.

So, what do we do?


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
19-08-2019 23:06
tmiddlesProfile picture★★★★☆
(1322)
So guys just to be clear what are you saying about the source:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

ITN and IBdaMann what are you saying about that page?

It includes the text
"If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss..."
and
"The relationship governing the net radiation from hot objects is called the Stefan-Boltzmann law:
"

They are identifying the above as being governed by " the Stefan-Boltzmann law"
Is that not correct?

I'm just a student at this point and trying to learn.
Edited on 19-08-2019 23:20
20-08-2019 01:05
Into the NightProfile picture★★★★★
(9582)
tmiddles wrote:
So guys just to be clear what are you saying about the source:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

ITN and IBdaMann what are you saying about that page?

It includes the text
"If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss..."
and
"The relationship governing the net radiation from hot objects is called the Stefan-Boltzmann law:
"

They are identifying the above as being governed by " the Stefan-Boltzmann law"
Is that not correct?

I'm just a student at this point and trying to learn.

You are still confusing light with heat. There is no 'net' heat. There is just heat, or there is not. It only flows from hot to cold, never the reverse.


The Parrot Killer
20-08-2019 02:28
tmiddlesProfile picture★★★★☆
(1322)
Into the Night wrote:
tmiddles wrote:
So guys just to be clear what are you saying about the source:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

You are still confusing light with heat.


No hyperphysics.phy-astr.gsu.edu is confusing it according to what you're saying. You and IBdaMann have had a lot of advice on not trusting Wikipedia and biased sources. So are you saying hyperphysics.phy-astr.gsu.edu is biased?

I was also asking you above about a text book taught to college students to teach the basics of physics and thermodynamics.

So help me out here. How did you learn the Stefan-Boltzmann Law and why do you trust that it's not made up garbage?

I'm a student and not conducting my own research right now : )
Maybe one day but it's unlikely.
Edited on 20-08-2019 02:38
20-08-2019 02:51
tmiddlesProfile picture★★★★☆
(1322)
Here's another one:


This page includes a calculator:
https://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

So as you can see if the hot object is 100C there are several different transfers of energy to the colder object. I confirmed this using their calculator
Hot.....Cold....Loss watts
100....100.....0
100.....50......480
100......0.......783
100....-50......957
100....-100....1047

ε - emissivity coefficient used was 1.0
th - object hot temperature (oC) 100C
tc - surroundings cold temperature (oC) from -100 to 100C
Ac - object area (m2) 1.0

According to you it wouldn't change based on the temp of the colder object right?
Edited on 20-08-2019 03:03
20-08-2019 03:56
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote: So guys just to be clear what are you saying about the source:

That you shouldn't worship any person, website or article. I was not making any statement about any source nor was I making any claim other than I could not find your "science" until you pointed it out.


.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
20-08-2019 03:59
IBdaMannProfile picture★★★★★
(4920)
tmiddles wrote:So help me out here. How did you learn the Stefan-Boltzmann Law and why do you trust that it's not made up garbage?

Someone explained the science to me, as I have to you, and then I was able to go into a lab and apply the scientific method. I could not show it to be false so I can't call it garbage.

.


Global Warming: The preferred religion of the scientifically illiterate.

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

When the alt-physics birds sing about "indivisible bodies," we've got pure BS. - VernerHornung

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
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