Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
You never provided an article that said nor implied the CMB did not penetrate the air to a warm dish.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.
All models of EM radiation have been verified by experiments to parts per billion accuracy.models without physical verification mean exactly jack.
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.
You are certainly right. SSDD believes that the presence of a warmer body will keep the colder body from radiating anything toward it even thought the surface charges are vibrating like mad.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?
[
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.
You are certainly right. SSDD believes that the presence of a warmer body will keep the colder body from radiating anything toward it even thought the surface charges are vibrating like mad.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?
Where is the logical fallacy?Keep yanking your own chain if it gets you though life. Face it guy...you couldn't support your argument...you engaged in logical fallacies one after another and showed how easily you are fooled by instrumentation...nothing more. The Second law of thermodynamics still says what it says and the SB law still says what it says. Nothing has changed.
Really...Which expression within that equation allows you to calculate net? I see T-Tc...that expression describes a gross change, not a net change.[
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.
And I say the equation represents the NET change in energy between the object and environment. The object will still be radiating.
From my link:
If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form
But the formula states, e.g. 200F- 100F = 100F, it's all one way. There's no heat flowing upstream.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.
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.
Where is the logical fallacy?Keep yanking your own chain if it gets you though life. Face it guy...you couldn't support your argument...you engaged in logical fallacies one after another and showed how easily you are fooled by instrumentation...nothing more. The Second law of thermodynamics still says what it says and the SB law still says what it says. Nothing has changed.
This derivation shows what ChesBayJJ, Tod, etc, myself, and 372 thousand scientists agree with.
For a substance at temperature T₁
and the background at temperature T₂.
Emission: Rₑ = e σ T₁⁴
Absorption: Rₐ =e σ T₂⁴
The net rate of EM energy exchange:
Rnet = Rₑ - Rₐ = e σ T₁⁴ – e σ T₂⁴ = e σ (T₁⁴ – T₂⁴)
ehs
Really...Which expression within that equation allows you to calculate net? I see T-Tc...that expression describes a gross change, not a net change.[
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.
And I say the equation represents the NET change in energy between the object and environment. The object will still be radiating.
From my link:
If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form
ehs
Really...Which expression within that equation allows you to calculate net? I see T-Tc...that expression describes a gross change, not a net change.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.
And I say the equation represents the NET change in energy between the object and environment. The object will still be radiating.
From my link:
If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form
Wuwei just gave it to you but I see you are in denial
Where is the logical fallacy?Keep yanking your own chain if it gets you though life. Face it guy...you couldn't support your argument...you engaged in logical fallacies one after another and showed how easily you are fooled by instrumentation...nothing more. The Second law of thermodynamics still says what it says and the SB law still says what it says. Nothing has changed.
This derivation shows what ChesBayJJ, Tod, etc, myself, and 372 thousand scientists agree with.
For a substance at temperature T₁
and the background at temperature T₂.
Emission: Rₑ = e σ T₁⁴
Absorption: Rₐ =e σ T₂⁴
The net rate of EM energy exchange:
Rnet = Rₑ - Rₐ = e σ T₁⁴ – e σ T₂⁴ = e σ (T₁⁴ – T₂⁴)
You see the expression after the second equals sign...that expression speaks the SB law...the rest is just the math necessary to derive the SB law...you can write net all you like, but the expression that is the SB law does not calculate net. Once more, your math skills are laughable. You can't read an equation, and you don't seem to grasp that the final form of the equation is the working equation regardless of the calculations required to reach that final expression... Just more mental masturbation on your part...
Where is the logical fallacy?Keep yanking your own chain if it gets you though life. Face it guy...you couldn't support your argument...you engaged in logical fallacies one after another and showed how easily you are fooled by instrumentation...nothing more. The Second law of thermodynamics still says what it says and the SB law still says what it says. Nothing has changed.
This derivation shows what ChesBayJJ, Tod, etc, myself, and 372 thousand scientists agree with.
For a substance at temperature T₁
and the background at temperature T₂.
Emission: Rₑ = e σ T₁⁴
Absorption: Rₐ =e σ T₂⁴
The net rate of EM energy exchange:
Rnet = Rₑ - Rₐ = e σ T₁⁴ – e σ T₂⁴ = e σ (T₁⁴ – T₂⁴)
You see the expression after the second equals sign...that expression speaks the SB law...the rest is just the math necessary to derive the SB law...you can write net all you like, but the expression that is the SB law does not calculate net. Once more, your math skills are laughable. You can't read an equation, and you don't seem to grasp that the final form of the equation is the working equation regardless of the calculations required to reach that final expression... Just more mental masturbation on your part...
.the rest is just the math necessary to derive the SB law.
Yes. The first two expressions are the input and output EM energies.
.you can write net all you like, but the expression that is the SB law does not calculate net.
The third expression is a subtraction of output minus input (emission minus absorption). That is the definition of net. How can you say that the final expression is not net?
Sorry guy....when you use the SB law to calculate P, you use the equation after the last equals sign...nothing else. That is the mathematical expression of the SB law....there is no expression there to express energy the radiator is receiving from the back ground because it is receiving no energy from the background...refer to the second law...not possible for energy to flow spontaneously from a cool object to a warm object. There is no net energy flow.
The first two expressions of emission and absorption are subtracted. I told you many times. A vacuum has nothing to do with thermal emission or absorption. A non-GHG between object and background would give the same results.The expression after the second equals sign is the SB equation one uses for objects not in a vacuum and in the presence of other matter...it is the same as the equation I gave you...it is the mathematical expression of the SB law...so again, which expression within that formula are you using to calculate net?
Sorry guy....when you use the SB law to calculate P, you use the equation after the last equals sign...nothing else. That is the mathematical expression of the SB law....there is no expression there to express energy the radiator is receiving from the back ground because it is receiving no energy from the background...refer to the second law...not possible for energy to flow spontaneously from a cool object to a warm object. There is no net energy flow.
You point out the first two expressions being used to for the final form. The second expression is
Absorption: Rₐ =ε σ T₂⁴
Now you are saying absorption of energy from the colder body is zero. That doesn't make any sense.
Nope. Nobody believes that.Zero absorption of energy from the colder body is precisely what the second law predicts....it is not possible for energy to move spontaneously from a cooler body to a warmer body.
The expression after the second equals sign is the SB equation one uses for objects not in a vacuum and in the presence of other matter...it is the same as the equation I gave you...it is the mathematical expression of the SB law...so again, which expression within that formula are you using to calculate net?
Please note": P = NET RADIATED POWER
Sorry guy....when you use the SB law to calculate P, you use the equation after the last equals sign...nothing else. That is the mathematical expression of the SB law....there is no expression there to express energy the radiator is receiving from the back ground because it is receiving no energy from the background...refer to the second law...not possible for energy to flow spontaneously from a cool object to a warm object. There is no net energy flow.
You point out the first two expressions being used to for the final form. The second expression is
Absorption: Rₐ =ε σ T₂⁴
Now you are saying absorption of energy from the colder body is zero. That doesn't make any sense.
Zero absorption of energy from the colder body is precisely what the second law predicts....it is not possible for energy to move spontaneously from a cooler body to a warmer body.
