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your pretty pictures seem to be incomplete.
So let`s see when You start violating it...shall we...:>>>if we consider an ideal situation in which an enclosure with perfectly reflecting walls contains radiation with a certain amount of energy, then at equilibrium, this "photon gas" will have a Planck distribution of energies.[1] This will be true even though the walls are perfectly reflecting due to the very small amount of interaction between the photons themselves. The equilibration process will take a considerable amount of time, but the distribution of energies and radiation density will ultimately approach a Planck distribution. The next step is to realize that, as a result of the second law of thermodynamics, any enclosure at thermal equilibrium must also have a Planck distribution of radiation.[1] If this were not true, then we could bring that system in contact with the above ideal system, both at the same temperature, and by connecting them through an optical filter, we can have a net amount of radiation pass from one body to the other. For example, suppose in the second system, the density of photons at narrow frequency band around wavelength λ were higher than that of a black body at that temperature. If a filter that passed only that frequency band was inserted in an opening that connected the two bodies, then there would be a net transfer of photons, and their energy, from the second system to the first. This is in violation of the second law of thermodynamics, which states that there can be no net transfer of energy between two bodies at the same temperature.<<<<
IanC you have just shown how immature and ignorant you are..
No you little lowlife you have gone back and forth insulting and the pretending we insult you over and again for weeks now. You were condescending again and the fact you try and deny this is all the more pathetic..
Your words your previous post to me:
"the electricity is decreased to keep the heated bar at 150F, by 75w the last time I explained it, and the container is cooled because it is only receiving 675w instead of 750w."
Your words above:
I am somewhat amused that you think Spencer would change the power input into the heated bar according to a thermostat set for 150F when the whole concept of the experiment is to find the change in temperature when the second bar is added.
Recognize them? You better dumbass they are your words...
So...whats your excuse this time? You just said one thing and then tried to make it seem ridiculous I would think that one thing..So, which is this time dancing bear does he change the input level or not? Or does he change it then not change to suit your silly posturing ass?
Got an explanation didpshit? NO?????? Didn't think you would.. Now dance boy!
Really? Hows that work? What planet does something get hotter with less energy input?
the container temp is cooled (at least less energy is needed to keep it at 0F than when the heater was receiving full power) when the power is reduced to keep the bar at 150F.
Yes yes and what exactly does this address regarding my post or contention, other than the fact your are once again talking out of your ass? nice way to dodge and divert dancing bear..
do you understand the concept of input=output? I think it is funny that you insulted me for bringing up this scenario previously (and explaining it), while you think you are brilliant and original for bringing it up yourself! I suppose you didnt understand it the first time and you probably dont understand it now. hahahahahaha
Yes I do it seems you don't though because you made a contention then denied making it or was it I made it then you denied I made it? Does it matter? After all you knew that too didnt ya phony.. Sure ya did because you know everything don't ya... You even talked about this before but didn't if its wrong... Sure pal sure..
IanC, you want to explain to me how it is a self proclaimed expert on this didn't recognize fermats theorem? How about the fact you didn't know the two-slit experiment? Remember what you told me about that experiment genius? Oh I sure do.. YOU told me you can't learn about physics watching waves in a liquid... LOL..
Now dance, and squirm and google up terms to try and save face phony, but from here on out I will make sure I remind you how, when and where you got outed as a fake..
Really? Hows that work? What planet does something get hotter with less energy input?
ho hum, more ad homs
your pretty pictures seem to be incomplete.
I couldnt find the same flourescent light as in polarbear's picture. perhaps the unlabelled y axis is part of the problem.
flourescent lights do not emit like a blackbody. the curve is a different shape because it uses specific atomic bands of emission.
sodium lights are even less like a blackbody
Let me get this straight,...and now You are saying that we have to distinguish between "light bulb photons" and "Spencer`s metal bar example photons"...?all photons are the same...
Nobody can tell them apart
photons coming from a 2000 K source are the same as photons coming from a 200 K source
Tell me now how did You "calculate" the power to increase the temperature of the second bar...if You don`t even know the mass or the distance of the second bar....and exactly what temperature did You get for either bar in this calculation You pretend to have done...I showed that the electrical power would have to be decreased by 75w to stay the same temperature, therefore if you decrease it by any value from 1-74w the bar will still increase in temperature. increased temp with decreased power when the second bar is placed next to the heated bar. on this planet or any other. __________________
"Nothing shows a lack of mathematical understanding more than an exaggeratedly accurate calculation.”
Carl Friedrich Gauss (1777-1855)
Okay, let`s use a mass of 1 kg, the unspecified metal of your "metal bar" shall be lead as in Pb with an oxidized surface...looks pretty black...increased temp with decreased power when the second bar is placed next to the heated bar. on this planet or any other.
Then You tried to make the same erroneous case again using the "extra time" a photon spends in Your phantasy climatology world,..and then I did the math for You and the "extra time" was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
try to use to get the "extra Energy"...and when You realized, that Your "extra Spencer`s example Power" evaporated by a factor of 10^(-15)
Just a minute here IanC...
Where exactly did I say that sodium lights are a black body...???
Just a minute here IanC...
Where exactly did I say that sodium lights are a black body...???
when you jumped into the thread I was excited that there was finally someone who was going to discuss ideas without all the distortion and unfounded denial of wirebender and gslack. unfortunately that is not the case. I dont know if it because you couldnt be bothered to read anything I write or if you just like being an ignorant asshole like those other two.
on any of the three options, nor did I bring up any of the numerous complexities of multiple absorption/emission, transforming to different wavelengths etc.I made a point of not putting values
I brought up real world mechanisms that warm the surface by reducing the radiation loss. I have no problem with you pointing out that the second option doesnt account for much warming, I do have a problem with you implying that it was the only option and that you have 'refuted' my case somehow.
and claim:I made a point of not putting values ..
Well haven`t You noticed, that in the real world this second "Spencer example balck body imaginary bar" is air containing CO2 and that they do make a point to put numerical values on this "extra heat"...and post these all over the internet and news papersI brought up real world mechanisms that warm the surface by reducing the radiation loss.
being an ignorant asshole like those other two.
Okay, let`s use a mass of 1 kg, the unspecified metal of your "metal bar" shall be lead as in Pb with an oxidized surface...looks pretty black...
the distance between the 2 is 10 cm,...and each cylindrical bar is say 5 cm thick...one bar = @ +10 C, the other @ -10 C
Allright IanC...do the math...because ""Nothing shows a lack of mathematical understanding more than faking it"
Oh and please do tell me also what happens when the bars mover ever closer...
and what happens when they make contact...does the -10 C bar still heat the +10 C bar..???
as "contact each other" lets use the distance at which they would bridge to make an electrical contact at 1 Volt...
According to You it still should...after all atoms can`t really touch each other...or do "climatology atoms" touch each other ...?
IanC you have just shown how immature and ignorant you are..
No you little lowlife you have gone back and forth insulting and the pretending we insult you over and again for weeks now. You were condescending again and the fact you try and deny this is all the more pathetic..
Your words your previous post to me:
"the electricity is decreased to keep the heated bar at 150F, by 75w the last time I explained it, and the container is cooled because it is only receiving 675w instead of 750w."
Your words above:
I am somewhat amused that you think Spencer would change the power input into the heated bar according to a thermostat set for 150F when the whole concept of the experiment is to find the change in temperature when the second bar is added.
Recognize them? You better dumbass they are your words...
So...whats your excuse this time? You just said one thing and then tried to make it seem ridiculous I would think that one thing..So, which is this time dancing bear does he change the input level or not? Or does he change it then not change to suit your silly posturing ass?
Got an explanation didpshit? NO?????? Didn't think you would.. Now dance boy!
Really? Hows that work? What planet does something get hotter with less energy input?
the container temp is cooled (at least less energy is needed to keep it at 0F than when the heater was receiving full power) when the power is reduced to keep the bar at 150F.
Yes yes and what exactly does this address regarding my post or contention, other than the fact your are once again talking out of your ass? nice way to dodge and divert dancing bear..
Yes I do it seems you don't though because you made a contention then denied making it or was it I made it then you denied I made it? Does it matter? After all you knew that too didnt ya phony.. Sure ya did because you know everything don't ya... You even talked about this before but didn't if its wrong... Sure pal sure..
IanC, you want to explain to me how it is a self proclaimed expert on this didn't recognize fermats theorem? How about the fact you didn't know the two-slit experiment? Remember what you told me about that experiment genius? Oh I sure do.. YOU told me you can't learn about physics watching waves in a liquid... LOL..
Now dance, and squirm and google up terms to try and save face phony, but from here on out I will make sure I remind you how, when and where you got outed as a fake..
you said-Really? Hows that work? What planet does something get hotter with less energy input?
you guys were the ones who insisted that the bar would not heat up. I simply told you what would happen if the bar was kept at the same temperature. in that case the power input would have to be lowered, which in turn would cool the container by the same amount that the input was lowered. to answer your question that was so badly phrased. in Spencer's experiment, I showed that the electrical power would have to be decreased by 75w to stay the same temperature, therefore if you decrease it by any value from 1-74w the bar will still increase in temperature. increased temp with decreased power when the second bar is placed next to the heated bar. on this planet or any other.
You are such a fake ..
You are such a fake ..
Expert on fake, eh? I guess you go with what you know!!!
You are such a fake ..
Expert on fake, eh? I guess you go with what you know!!!
Sorry junior this thread has an age limit.. Now run along and play with the other kids..
Expert on fake, eh? I guess you go with what you know!!!
Sorry junior this thread has an age limit.. Now run along and play with the other kids..
Fake, fake, fake, fake fake. Nothing you post makes much sense. Your only value is as an object of ridicule.
Then You tried to make the same erroneous case again using the "extra time" a photon spends in Your phantasy climatology world,..and then I did the math for You and the "extra time" was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
try to use to get the "extra Energy"...and when You realized, that Your "extra Spencer`s example Power" evaporated by a factor of 10^(-15)
when you jumped into the thread I was excited that there was finally someone who was going to discuss ideas without all the distortion and unfounded denial of wirebender and gslack. unfortunately that is not the case. I dont know if it because you couldnt be bothered to read anything I write or if you just like being an ignorant asshole like those other two.
with respect to the extra time for escape of radiation- I gave three basic options to consider. 1.the photon escapes at the speed of light 2.the photon is absorbed and reemitted by the CO2 molecule and escapes, at less than the speed of light 3. the photon is absorbed and reemitted towards the surface and does not escape.
the first option leads to the coldest surface because the shedding of radiation is the most efficient.
the second option slows the escape of radiation, making the surface warmer. you are implying that is the only mechanism for warming. I am stating that at the very least there is some warming. you agree that it takes time to absorb and reemit the radiation therefore you agree that CO2 warms the earth. the amount doesnt matter in this scenario, just the concept.
the third option stops the loss of radiation completely and returns the energy to the surface. obviously this is where the bulk of the inefficiency of radiation shedding due to CO2 occurs.
I made a point of not putting values on any of the three options, nor did I bring up any of the numerous complexities of multiple absorption/emission, transforming to different wavelengths etc. I brought up real world mechanisms that warm the surface by reducing the radiation loss. I have no problem with you pointing out that the second option doesnt account for much warming, I do have a problem with you implying that it was the only option and that you have 'refuted' my case somehow.
Expert on fake, eh? I guess you go with what you know!!!
Sorry junior this thread has an age limit.. Now run along and play with the other kids..
Fake, fake, fake, fake fake. Nothing you post makes much sense. Your only value is as an object of ridicule.
Then You tried to make the same erroneous case again using the "extra time" a photon spends in Your phantasy climatology world,..and then I did the math for You and the "extra time" was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
try to use to get the "extra Energy"...and when You realized, that Your "extra Spencer`s example Power" evaporated by a factor of 10^(-15)
when you jumped into the thread I was excited that there was finally someone who was going to discuss ideas without all the distortion and unfounded denial of wirebender and gslack. unfortunately that is not the case. I dont know if it because you couldnt be bothered to read anything I write or if you just like being an ignorant asshole like those other two.
with respect to the extra time for escape of radiation- I gave three basic options to consider. 1.the photon escapes at the speed of light 2.the photon is absorbed and reemitted by the CO2 molecule and escapes, at less than the speed of light 3. the photon is absorbed and reemitted towards the surface and does not escape.
the first option leads to the coldest surface because the shedding of radiation is the most efficient.
the second option slows the escape of radiation, making the surface warmer. you are implying that is the only mechanism for warming. I am stating that at the very least there is some warming. you agree that it takes time to absorb and reemit the radiation therefore you agree that CO2 warms the earth. the amount doesnt matter in this scenario, just the concept.
the third option stops the loss of radiation completely and returns the energy to the surface. obviously this is where the bulk of the inefficiency of radiation shedding due to CO2 occurs.
I made a point of not putting values on any of the three options, nor did I bring up any of the numerous complexities of multiple absorption/emission, transforming to different wavelengths etc. I brought up real world mechanisms that warm the surface by reducing the radiation loss. I have no problem with you pointing out that the second option doesnt account for much warming, I do have a problem with you implying that it was the only option and that you have 'refuted' my case somehow.
Ian, I have been reading this and the various other threads for awhile now and the one thing that has become very plain is you are terribly ignorant of real world physics and the mathematical skills one needs to express the world of physics. I suggest you get yourself a CRC and a good book on math and study, study, study.
You are far, far out of your depth here and need a good deal of study to catch up.
Well he did not direct that at me, but at wirebender and Gslack...konradv
post number 312
Quote: Originally Posted by gslack
You are such a fake ..
Expert on fake, eh? I guess you go with what you know!!!
post # 314
Fake, fake, fake, fake fake. Nothing you post makes much sense. Your only value is as an object of ridicule.
This guy can`t do the math any more than IanC...actually I think if IanC had really tried to do the math as I asked him now how many times...he would have done better than this fraud.The major reason that people give for thinking that DLR can’t affect the temperature is (a mistaken understanding of) the second law of thermodynamics, and they might say something like:A colder atmosphere can’t heat a warmer surfaceThere are semantics which can confuse those less familiar with thermal radiation.
If we consider the specific terminology of heat we can all agree and say that heat flows from the warmer to the colder. In the case of radiation, this means that more is emitted by the hotter surface (and absorbed by the colder surface) than the reverse.
However, what many people have come to believe is that the colder surface can have no effect at all on the hotter surface. This is clearly wrong. And just to try and avoid upsetting the purists but without making the terminology too obscure I will say that the radiation from the colder surface can have an effect on the warmer surface and can change the temperature of the warmer surface.
Therefore, if your current belief is that radiation from a colder atmosphere cannot “change the temperature” of the hotter surface then you have to believe that all of the radiation from the atmosphere is reflected.
They all look the same to me” – The Energy of a Photon
This part is very simple. The energy of a photon, E:
E = hν = hc/λ
where ν = frequency, λ = wavelength, c = speed of light, h = 6.6 ×10−34 J.s (Planck’s constant).
You can find this in any basic textbook and even in Wikipedia. So, for example, the energy of a 10μm photon = 2 x 10−20 J.
Notice that there is no dependence on the temperature of the source. Think of individual photons as anonymous – a 10μm photon from a 2,000K source has exactly the same energy as a 10μm photon from a 200K source.
And the guy that said :The joule-second is a unit equal to a joule multiplied by a second, used to measure action or angular momentum. The joule-second is the unit used for Planck's constant.
In SI base units, the joule-second is.
has been using photons as if they were little pebbles that You can add up into a bigger one...You can find this in any basic textbook and even in Wikipedia. So, for example, the energy of a 10μm photon = 2 x 10−20 J.
At least IanC never PRETENDED to have done his own math...he actually told me, that he did not want to assign any specific numbers when I pressed him for it...There are semantics which can confuse those less familiar with thermal radiation.
If we consider the specific terminology of heat we can all agree and say that heat flows from the warmer to the colder. In the case of radiation, this means that more is emitted by the hotter surface (and absorbed by the colder surface) than the reverse.
However, what many people have come to believe is that the colder surface can have no effect at all on the hotter surface. This is clearly wrong. And just to try and avoid upsetting the purists but without making the terminology too obscure I will say that the radiation from the colder surface can have an effect on the warmer surface and can change the temperature of the warmer surface.
Therefore, if your current belief is that radiation from a colder atmosphere cannot “change the temperature” of the hotter surface then you have to believe that all of the radiation from the atmosphere is reflected.
the Energy for a photon at a 10 μm wavelength =1.98 * 10^ (-21)a 10μm photon from a 2,000K source has exactly the same energy as a 10μm photon from a 200K source.
See it was not just a simple oversight like You made, when You confused POWER and ENERGY...So, for example, the energy of a 10μm photon = 2 x 10−20 J
One rod is at +10 C, the other at -10 C...then we move them together till the rods make contact and the optical cavity is 10 by 1 cm with the 2 channels milled 0.5 cm deep.Blackbody radiation becomes a visible glow of light if the temperature of the object is high enough. The Draper point is the temperature at which all solids glow a dim red, about 798 K.[4][5] At 1000 K, the opening in the oven looks red; at 6000 K, it looks white. No matter how the oven is constructed, or of what material,
A closed box of graphite walls at a constant temperature with a small hole on one side produces a good approximation to ideal blackbody radiation emanating from the opening
You come back here claiming that all photons are the same...we all know where and from who You got that one from...
So now You are claiming that these "all the same photons" each having been emitted by a radiation source say at 10 C....
that 10 of these photons stack up to a new and higher temperature of 10 times X ("climatology photon degrees")...
You got that nonsense from this idiot who told You that the photons coming from a 2000 degree black body are the same
as 10 photons coming from a 200 degree source and therefore that enough these are able to raise the temperature of a 2000 K black body
Incandescent bulbs emit light in a manner closely resembling
Plank’s law of blackbody radiation. The law describes how a
body capable of absorbing all radiation contacting it (a blackbody)
will emit at a given range of wavelengths dependant on its
temperature. The inefficiency inherent in an incandescent bulb
is due to the fact that it emits both infrared and visible light at
temperatures between 2000 and 3000K. Specifically, the infrared
portion of the radiation consumes about 88% of the input electric
energy and becomes wasted heat (see Fig.1). Hence, recycling infrared
light into useful visible light would improve incandescent
efficiency.
The filter acts as a perfect transmitter for the useful visible
light and a perfect reflector for the undesirable infrared light.
The reflected light is re-absorbed which, in turn, helps to heat
up the filament. This infrared recycling process has two major
energy consequences. First, it reduces the amount of electricity
required to maintain a hot filament and thus improves electricto-
optical conversion efficiency. Second, it reduces the thermal radiation of the bulb as infrared photons cannot escape. Withthis approach, the energy efficacy of an incandescent light bulb
can be improved by as much as eight times.