The Heart of the AGW Premise Fails Empirical Review.

[SSDD]

If stating what is, rather than what you believe or wish makes me a troll then I guess we can chalk up one more word you don't know the meaning of.

Nope, lying about what science is about - that makes you a troll.

Modelled experiments...not observed measured experiments...it is models all the way down with you guys. And why can't you bring yourself to simply admit that neither you nor all of science has the first observed, measured instance of energy moving spontaneously from cool to warm?

Yep, physics is based on mathematical models. Can't do much science these days without it. 372 thousand scientist know that, and know the CMB penetrated the earth. You have no proof that says otherwise. Troll.

I have provided you with the article that stated explicitly that they received a resonant radio frequency that equated to CMB...it isn't my fault that you think that means that the resonant radio frequency was actually cmb.

and models without physical verification mean exactly jack.

 

[Toddsterpatriot]

If stating what is, rather than what you believe or wish makes me a troll then I guess we can chalk up one more word you don't know the meaning of.

Nope, lying about what science is about - that makes you a troll.

Modelled experiments...not observed measured experiments...it is models all the way down with you guys. And why can't you bring yourself to simply admit that neither you nor all of science has the first observed, measured instance of energy moving spontaneously from cool to warm?

Yep, physics is based on mathematical models. Can't do much science these days without it. 372 thousand scientist know that, and know the CMB penetrated the earth. You have no proof that says otherwise. Troll.

I have provided you with the article that stated explicitly that they received a resonant radio frequency that equated to CMB...it isn't my fault that you think that means that the resonant radio frequency was actually cmb.

and models without physical verification mean exactly jack.

I love the idea that a signal is detected at the Earth's surface without that signal entering the atmosphere.

Magic waves!

 

[Wuwei]

I have provided you with the article that stated explicitly that they received a resonant radio frequency that equated to CMB...it isn't my fault that you think that means that the resonant radio frequency was actually cmb.

You never provided an article that said nor implied the CMB did not penetrate the air to a warm dish.

models without physical verification mean exactly jack.

All models of EM radiation have been verified by experiments to parts per billion accuracy.

 

[Wuwei]

Magic waves!

First smart photons. Now magic waves! What wonders are next?

 

[SSDD]

I have provided you with the article that stated explicitly that they received a resonant radio frequency that equated to CMB...it isn't my fault that you think that means that the resonant radio frequency was actually cmb.

You never provided an article that said nor implied the CMB did not penetrate the air to a warm dish.

models without physical verification mean exactly jack.

All models of EM radiation have been verified by experiments to parts per billion accuracy.

models and models and models and not the first piece of physical evidence. Is EM radiation an observable, measurable, testable physical entity? Wouldn't you think that you might have physical evidence supporting claims regarding a physical entity? You are laughable.

 

[SSDD]

Magic waves!

First smart photons. Now magic waves! What wonders are next?

Now you are just being a pissy little girl out of frustration over the fact that I won't join you in your faith.

 

[CrusaderFrank]

I looked at the equation for heat loss and it's a gross number T1-T for greater temperature minus lower temperature. Where do you add back the imaginary heat transferred from the cooler object back to the warmer one

The Stefan-Boltzmann constant, symbolized by the lowercase Greek letter sigma (

), is a physical constant involving black body radiation. A black body, also called an ideal radiator, is an object that radiates or absorbs energy with perfect efficiency at all electromagnetic wavelength s. The constant defines the power per unit area emitted by a black body as a function of its thermodynamic temperature .

What is Stefan-Boltzmann constant? - Definition from WhatIs.com

Power emitted. Not heat.

So does heat radiates from cooler to warm?

Plug in the temperature of the cool object.
That tells you how much it radiates.

Now plug in the temperature of the warm object.
That tells you how much it radiates.

You see, they both radiate at the same time.

But the formula states, e.g. 200F- 100F = 100F, it's all one way. There's no heat flowing upstream.

But the formula states, e.g. 200F- 100F = 100F, it's all one way.

The formula does not state one way.

There's no heat flowing upstream.

Who said anything about heat? We're discussing radiation. Power emitted.

Heat Transfer Through Conduction: Equation & Examples - Video & Lesson Transcript | Study.com

Q over t is the rate of heat transfer - the amount of heat transferred per second, measured in Joules per second, or Watts. kis the thermal conductivity of the material - for example, copper has a thermal conductivity of 390, but wool has a thermal conductivity of just 0.04. T1 is the temperature of one object, and T2 is the temperature of the other. Since it's a temperature difference, you can actually use Celsius or Kelvin, whichever is most convenient. And d is the thickness of the material we're interested in. So the rate of heat transfer to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

Calculation Example
Okay, let's go through an example. Let's say you're going to a water park, and you're going to take a Styrofoam cooler with you. The cooler has a total surface area of 1.2 meters squared, and the walls are 0.03 meters thick. The temperature inside the cooler is 0 Celsius, and at the hottest part of the day it's 38 degrees Celsius. During this time of day, how much heat energy per second is lost by the cooler? And how much heat energy is lost in three hours at the water park assuming the temperature stays at 38 degrees? (Note: The thermal conductivity of Styrofoam is 0.01.)"

So you're saying their equation is incorrect, that the computation is not T2-T1, but T2-T1 + imaginary heat radiated from the cooler to the warmer

 

[ChesBayJJ]

I have provided you with the article that stated explicitly that they received a resonant radio frequency that equated to CMB...it isn't my fault that you think that means that the resonant radio frequency was actually cmb.

You never provided an article that said nor implied the CMB did not penetrate the air to a warm dish.

models without physical verification mean exactly jack.

All models of EM radiation have been verified by experiments to parts per billion accuracy.

models and models and models and not the first piece of physical evidence. Is EM radiation an observable, measurable, testable physical entity? Wouldn't you think that you might have physical evidence supporting claims regarding a physical entity? You are laughable.

What kind of evidence do you need to confirm the physical presence of EM radiation?

 

[SSDD]

I have provided you with the article that stated explicitly that they received a resonant radio frequency that equated to CMB...it isn't my fault that you think that means that the resonant radio frequency was actually cmb.

You never provided an article that said nor implied the CMB did not penetrate the air to a warm dish.

models without physical verification mean exactly jack.

All models of EM radiation have been verified by experiments to parts per billion accuracy.

models and models and models and not the first piece of physical evidence. Is EM radiation an observable, measurable, testable physical entity? Wouldn't you think that you might have physical evidence supporting claims regarding a physical entity? You are laughable.

What kind of evidence do you need to confirm the physical presence of EM radiation?

The evidence would be of energy moving spontaneously from a cool object to a warm object in spite of the second law of thermodynamics which says: It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.

 

[ChesBayJJ]

I have provided you with the article that stated explicitly that they received a resonant radio frequency that equated to CMB...it isn't my fault that you think that means that the resonant radio frequency was actually cmb.

You never provided an article that said nor implied the CMB did not penetrate the air to a warm dish.

models without physical verification mean exactly jack.

All models of EM radiation have been verified by experiments to parts per billion accuracy.

models and models and models and not the first piece of physical evidence. Is EM radiation an observable, measurable, testable physical entity? Wouldn't you think that you might have physical evidence supporting claims regarding a physical entity? You are laughable.

What kind of evidence do you need to confirm the physical presence of EM radiation?

The evidence would be of energy moving spontaneously from a cool object to a warm object in spite of the second law of thermodynamics which says: It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.

I think we are talking apples and oranges. Do I think there will be a net flow of energy from a colder body to a warmer one, with no work done? No. But let me propose this idea - put two steel balls, suspended on thin wires, in an enclosed space with the air removed. One ball is at 0C, the other at 100C, and the walls of the enclosure are at 25C. All will radiate - both steel balls and the walls of the enclosure. Eventually, assuming the temperature of the walls are held constant, everything will move to the same temp, 25C. So the colder ball will gain energy, the warmer will lose energy, and everything will continue to radiate.

 

[SSDD]

I think we are talking apples and oranges. Do I think there will be a net flow of energy from a colder body to a warmer one, with no work done? No. But let me propose this idea - put two steel balls, suspended on thin wires, in an enclosed space with the air removed. One ball is at 0C, the other at 100C, and the walls of the enclosure are at 25C. All will radiate - both steel balls and the walls of the enclosure. Eventually, assuming the temperature of the walls are held constant, everything will move to the same temp, 25C. So the colder ball will gain energy, the warmer will lose energy, and everything will continue to radiate.

Radiate to where that wouldn't result in spontaneous two way energy flow?

 

[ChesBayJJ]

I think we are talking apples and oranges. Do I think there will be a net flow of energy from a colder body to a warmer one, with no work done? No. But let me propose this idea - put two steel balls, suspended on thin wires, in an enclosed space with the air removed. One ball is at 0C, the other at 100C, and the walls of the enclosure are at 25C. All will radiate - both steel balls and the walls of the enclosure. Eventually, assuming the temperature of the walls are held constant, everything will move to the same temp, 25C. So the colder ball will gain energy, the warmer will lose energy, and everything will continue to radiate.

Radiate to where that wouldn't result in spontaneous two way energy flow?

There will be two way energy flow. Everything radiates that is above absolute zero. But the NET energy flow will be from the warmer to the cooler objects.

And why do we need evidence of unforced energy flow from cooler to warmer to confirm EM radiation?

 

[SSDD]

I think we are talking apples and oranges. Do I think there will be a net flow of energy from a colder body to a warmer one, with no work done? No. But let me propose this idea - put two steel balls, suspended on thin wires, in an enclosed space with the air removed. One ball is at 0C, the other at 100C, and the walls of the enclosure are at 25C. All will radiate - both steel balls and the walls of the enclosure. Eventually, assuming the temperature of the walls are held constant, everything will move to the same temp, 25C. So the colder ball will gain energy, the warmer will lose energy, and everything will continue to radiate.

Radiate to where that wouldn't result in spontaneous two way energy flow?

There will be two way energy flow. Everything radiates that is above absolute zero. But the NET energy flow will be from the warmer to the cooler objects.

And why do we need evidence of unforced energy flow from cooler to warmer to confirm EM radiation?

Back to net...there is no net energy flow. But if you can provide observed, measured evidence of spontaneous two way energy flow, wuwei and toddster would be so happy. They have been stymied on that bit of evidence for years. I know what the models say...models are not, however observed, measured evidence.

And it wasn't evidence of EM radiation I was asking for..it was once again, observed measured evidence of spontaneous two way energy flow.

 

[ChesBayJJ]

I think we are talking apples and oranges. Do I think there will be a net flow of energy from a colder body to a warmer one, with no work done? No. But let me propose this idea - put two steel balls, suspended on thin wires, in an enclosed space with the air removed. One ball is at 0C, the other at 100C, and the walls of the enclosure are at 25C. All will radiate - both steel balls and the walls of the enclosure. Eventually, assuming the temperature of the walls are held constant, everything will move to the same temp, 25C. So the colder ball will gain energy, the warmer will lose energy, and everything will continue to radiate.

Radiate to where that wouldn't result in spontaneous two way energy flow?

There will be two way energy flow. Everything radiates that is above absolute zero. But the NET energy flow will be from the warmer to the cooler objects.

And why do we need evidence of unforced energy flow from cooler to warmer to confirm EM radiation?

Back to net...there is no net energy flow. But if you can provide observed, measured evidence of spontaneous two way energy flow, wuwei and toddster would be so happy. They have been stymied on that bit of evidence for years. I know what the models say...models are not, however observed, measured evidence.

And it wasn't evidence of EM radiation I was asking for..it was once again, observed measured evidence of spontaneous two way energy flow.

Going to my example, an IR sensor would detect energy flowing in all directions from both steel balls. That would be direct, measured evidence of two way energy flow. And what do you mean there is no net energy flow? Are you saying the cold steel ball wouldn't get warmer and the warm steel ball wouldn't get colder?

 

[Toddsterpatriot]

The Stefan-Boltzmann constant, symbolized by the lowercase Greek letter sigma (

), is a physical constant involving black body radiation. A black body, also called an ideal radiator, is an object that radiates or absorbs energy with perfect efficiency at all electromagnetic wavelength s. The constant defines the power per unit area emitted by a black body as a function of its thermodynamic temperature .

What is Stefan-Boltzmann constant? - Definition from WhatIs.com

Power emitted. Not heat.

So does heat radiates from cooler to warm?

Plug in the temperature of the cool object.
That tells you how much it radiates.

Now plug in the temperature of the warm object.
That tells you how much it radiates.

You see, they both radiate at the same time.

But the formula states, e.g. 200F- 100F = 100F, it's all one way. There's no heat flowing upstream.

But the formula states, e.g. 200F- 100F = 100F, it's all one way.

The formula does not state one way.

There's no heat flowing upstream.

Who said anything about heat? We're discussing radiation. Power emitted.

Heat Transfer Through Conduction: Equation & Examples - Video & Lesson Transcript | Study.com

Q over t is the rate of heat transfer - the amount of heat transferred per second, measured in Joules per second, or Watts. kis the thermal conductivity of the material - for example, copper has a thermal conductivity of 390, but wool has a thermal conductivity of just 0.04. T1 is the temperature of one object, and T2 is the temperature of the other. Since it's a temperature difference, you can actually use Celsius or Kelvin, whichever is most convenient. And d is the thickness of the material we're interested in. So the rate of heat transfer to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

Calculation Example
Okay, let's go through an example. Let's say you're going to a water park, and you're going to take a Styrofoam cooler with you. The cooler has a total surface area of 1.2 meters squared, and the walls are 0.03 meters thick. The temperature inside the cooler is 0 Celsius, and at the hottest part of the day it's 38 degrees Celsius. During this time of day, how much heat energy per second is lost by the cooler? And how much heat energy is lost in three hours at the water park assuming the temperature stays at 38 degrees? (Note: The thermal conductivity of Styrofoam is 0.01.)"

So you're saying their equation is incorrect, that the computation is not T2-T1, but T2-T1 + imaginary heat radiated from the cooler to the warmer

Heat Transfer Through Conduction:

Thanks for the link.
We're discussing radiation at the moment.
Not conduction.

 

[SSDD]

[

Going to my example, an IR sensor would detect energy flowing in all directions from both steel balls. That would be direct, measured evidence of two way energy flow. And what do you mean there is no net energy flow? Are you saying the cold steel ball wouldn't get warmer and the warm steel ball wouldn't get colder?

No, an IR sensor at ambient temperature would not detect energy flowing in all directions.

What do I mean? It's easy.

that is the SB equation for radiating bodies, not in a vacuum, and in the presence of other matter. Set T and Tc to the same number....what does P then equal?

 

[SSDD]

So does heat radiates from cooler to warm?

Plug in the temperature of the cool object.
That tells you how much it radiates.

Now plug in the temperature of the warm object.
That tells you how much it radiates.

You see, they both radiate at the same time.

But the formula states, e.g. 200F- 100F = 100F, it's all one way. There's no heat flowing upstream.

But the formula states, e.g. 200F- 100F = 100F, it's all one way.

The formula does not state one way.

There's no heat flowing upstream.

Who said anything about heat? We're discussing radiation. Power emitted.

Heat Transfer Through Conduction: Equation & Examples - Video & Lesson Transcript | Study.com

Q over t is the rate of heat transfer - the amount of heat transferred per second, measured in Joules per second, or Watts. kis the thermal conductivity of the material - for example, copper has a thermal conductivity of 390, but wool has a thermal conductivity of just 0.04. T1 is the temperature of one object, and T2 is the temperature of the other. Since it's a temperature difference, you can actually use Celsius or Kelvin, whichever is most convenient. And d is the thickness of the material we're interested in. So the rate of heat transfer to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

Calculation Example
Okay, let's go through an example. Let's say you're going to a water park, and you're going to take a Styrofoam cooler with you. The cooler has a total surface area of 1.2 meters squared, and the walls are 0.03 meters thick. The temperature inside the cooler is 0 Celsius, and at the hottest part of the day it's 38 degrees Celsius. During this time of day, how much heat energy per second is lost by the cooler? And how much heat energy is lost in three hours at the water park assuming the temperature stays at 38 degrees? (Note: The thermal conductivity of Styrofoam is 0.01.)"

So you're saying their equation is incorrect, that the computation is not T2-T1, but T2-T1 + imaginary heat radiated from the cooler to the warmer

Heat Transfer Through Conduction:

Thanks for the link.
We're discussing radiation at the moment.
Not conduction.

Doesn't matter whether you are talking about conduction, radiation, free electrons flowing down a wire, or a bowling ball rolling down a hill...it is all energy exchange and all obeys the second law of thermodynamics. You miss the boat with your terribly flawed belief that radiation has some special exemption from obeying the second law of thermodynamics.

 

[ChesBayJJ]

Going to my example, an IR sensor would detect energy flowing in all directions from both steel balls. That would be direct, measured evidence of two way energy flow. And what do you mean there is no net energy flow? Are you saying the cold steel ball wouldn't get warmer and the warm steel ball wouldn't get colder?

No, an IR sensor at ambient temperature would not detect energy flowing in all directions.

What do I mean? It's easy.

that is the SB equation for radiating bodies, not in a vacuum, and in the presence of other matter. Set T and Tc to the same number....what does P then equal?

Both steel balls will be radiating in all directions. Do you dispute this?

And this

Stefan-Boltzmann Law

States your equation describes a hot body, T, radiating into cooler surroundings Tc. But back to your comment. I don't disagree

Set T and Tc to the same number....what does P then equal?​

The answer must be zero. But both steel balls will still radiate, in my example. Going to my link, it implies the emitter is the same temperature as its surroundings. But the emitter will still emit if it is above absolute zero.

 

[SSDD]

[

Going to my example, an IR sensor would detect energy flowing in all directions from both steel balls. That would be direct, measured evidence of two way energy flow. And what do you mean there is no net energy flow? Are you saying the cold steel ball wouldn't get warmer and the warm steel ball wouldn't get colder?

No, an IR sensor at ambient temperature would not detect energy flowing in all directions.

What do I mean? It's easy.

that is the SB equation for radiating bodies, not in a vacuum, and in the presence of other matter. Set T and Tc to the same number....what does P then equal?

Both steel balls will be radiating in all directions. Do you dispute this?

And this

Stefan-Boltzmann Law

States your equation describes a hot body, T, radiating into cooler surroundings Tc. But back to your comment. I don't disagree

Set T and Tc to the same number....what does P then equal?​

The answer must be zero. But both steel balls will still radiate, in my example. Going to my link, it implies the emitter is the same temperature as its surroundings. But the emitter will still emit if it is above absolute zero.

P= the amount of radiation emitting from the radiator which is dependent upon the temperature difference between T and Tc. When T and Tc are the same and P=zero...that means something. The equation is making a statement about reality. Alter the statement and you must alter the equation...alter the equation and it no longer represents the physical law.

 

[Wuwei]

I have provided you with the article that stated explicitly that they received a resonant radio frequency that equated to CMB...it isn't my fault that you think that means that the resonant radio frequency was actually cmb.

You never provided an article that said nor implied the CMB did not penetrate the air to a warm dish.

models without physical verification mean exactly jack.

All models of EM radiation have been verified by experiments to parts per billion accuracy.

models and models and models and not the first piece of physical evidence. Is EM radiation an observable, measurable, testable physical entity? Wouldn't you think that you might have physical evidence supporting claims regarding a physical entity? You are laughable.

Yes EM radiation is an observable, measurable, testable physical entity? You are observing it right now as you read this. Scientists have shown it agrees with the mathematical models to parts per billion accuracy. Scientists observed the CMB on the earth surface too.