I had suggested surrounding a internally heated body with a froven - a fridge/oven giving active heating and cooling to a set temperature.
Also body and inner surface of froven are black bodies with same albedo
The post with clarification added!:
thefordprefect says: 2013/04/04 at 11:52 AM
[JP:... If the body has a heat source then it will stay at the temperature it was at without the oven heating it from a higher temperature. If the oven is cooler than the body than it can not heat the body. Photon quanta from a cooler source do not warm up a warmer source, even if they might exist. It is not a "sudden" cessation of effect when the oven becomes cooler than the body - it is a smooth transition in the direction of q, of heating.]
This cannot be correct.
If the temperature of the froven is warmer than the body you suggest it heats the body.
If the froven is cooler than the body you suggest it has no effect.
If the body is radiating quanta from a 100°C source then the hotter froven will be radiating to the body quanta from its 100C+ walls.
If the body is radiating quanta from a 100°C source then the cooler froven will be radiating nothing from it 100C- as if it were at absolute zero thats one heck of a sudden step.
Do I understand correctly?
[JP: Not quite yet. If the body is warmer than the oven, then the body heats the oven. If the oven is warmer than the body, then the oven heats the body. This is a smooth transition in the direction of heating as a function of the temperature differential: -2 -1 0 1 2 etc. A smooth transition, not a sudden stop.]
thefordprefect says: Your comment is awaiting moderation. 2013/04/05 at 11:52 AM
Seem to have problems posting so I’ll try again:
[JP: Not quite yet. If the body is warmer than the oven, then the body heats the oven. If the oven is warmer than the body, then the oven heats the body. This is a smooth transition in the direction of heating as a function of the temperature differential: -2 -1 0 1 2 etc. A smooth transition, not a sudden stop.]
===========
you have stated definitely that there is no transfer of energy from cold to hot:
“but what I do know is that they do NOT work by cold heating hot – hahaha what a stupid idea.”
“[JP Reply: Trashed because we've already answered you. q from the shell to the planet is 0. ZERO. There is no heat loss from the shell to the planet. Even if the shell is emitting on the inside, there is no heat loss to the planet. The only direction the shell can lose heat is outwards, and hence it loses the equivalent of 800 W/m2 outwards.]”
“Radiated energy does not equate to net heat transfer or even net energy transfer. The equation of heat flow for radiation, from physics, from actual physics textbooks and from actual universities and actual physics degrees, is q ~ (T2^4 – T1^4). If T2 = T1, then q = 0, and nothing heats up, even though there’s all that radiation. ”
So firstly I hope you would agree that the quanta of energy leaving a surface cannot depend on the final destination of the quanta i.e. its temperature, material and surface – it only depends on the source material and temperature.
I also believe this describes your point of view:
The final destination of the radiation determines what happens to the quanta (rejected or absorbed)
where 100C- a very very very! small bit less than 100C 100C+ a very very very! small bit more than 100C w greater than y
y greater than x
and x greater than z
oven at 101C transfers zero quanta to body at 10000C (equivalent to back radiation)
body at 10000C transfers w quanta to oven at 101C
body at 100C transfers zero quanta to oven at 101C (equivalent to back radiation)
oven at 101C transfers x quanta to body at 100C
oven at 100C- transfers zero quanta to body at 100C (equivalent to back radiation)
body at 100C transfers x quanta to oven at 101C-
oven at 100C+ transfers x+1 quanta to body at 100C
body at 100C transfers zero quanta to oven at 100C+ (equivalent to back radiation)
body at 100C transfers y quanta to oven at 99C
oven at 99C transfers zero quanta to body at 100C (equivalent to back radiation)
oven at 10000C transfers w quanta to body at 100C body at 100C transfers zero quanta to oven at 10000C (equivalent to back radiation)
at 100C- to 100C+ oven temperature the body quanta changes from outputting x to receiving x+1 quanta
.
Somehow this does not seem to be a smooth or logical transition
Warmists would say quanta emitted from an object depends only on the object and its temperature. the final destination of the radiation is immaterial (well actually the quanta knows nothing until it hits the surface)
The sum of all quanta determines the rate of loss/gain of heat
oven at 101C transfers y quanta to body at 10000C (equivalent to back radiation)
body at 10000C transfers w quanta to oven at 101C
oven at 101C transfers y quanta to body at 100C
body at 100C transfers x quanta to oven at 101C (equivalent to back radiation)
oven at 100C- transfers x-1 quanta to body at 100C (equivalent to back radiation)
body at 100C transfers x quanta to oven at 101C-
oven at 100C+ transfers x+1 quanta to body at 100C
body at 100C transfers x quanta to oven at 100C+ (equivalent to back radiation)
oven at 99C transfers z quanta to body at 100C (equivalent to back radiation)
body at 100C transfers x quanta to oven at 99C
oven at 10000C transfers w quanta to body at 100C
body at 100C transfers x quanta to oven at 10000C (equivalent to back radiation)
Consider 100C- to 100C+ oven temperature - the100C body quanta output is x and at 100C- it receives x-1 quanta and at 100C+ it receives x+1 quanta
A smooth and logical transition.
I assume that I have this wrong somehow so perhaps using x,y,z you could explain your position
[This last post did not get past moderation!]
===========================
In pictures:
 |
| Assumed output from body and shell at 0K this is zero at 10K this is 100 quanta |
.
As the energy quanta increases from zero to 100 from body B with the temperature of B increasing from 0 to 10K
the energy quanta from shell A goes from 100 to zero as the temperature decreases from 10 to 0K
The temperature determines the quanta of energy released from the bodies
The warmist view would be that all energy from A gets absorbed by B and all energy from B gets absorbed by A irrespective of the temperature of each body
The Slayer version suggests that if the temperature of A is less than B then the transfer to B becomes zero/is reflected/cancels /nulled
and if the temperature of B is less than A then the transfer to A becomes zero/is reflected/cancels/nulled
This is shown in this diagram.
If one then looks at the net flow of quanta from A to B then the slayer version has a discontinuity where the temperatures are the same. The warmist version is a simple straight line which at B=A temperature the net transfer is zero.
===========================================
=========================================================================The whole thread
(on line of course at http://climateofsophistry.com/2013/03/08/the-fraud-of-the-aghe-part-11-quantum-mechanics-the-sheer-stupidity-of-ghe-science-on-wuwt/#comment-1559)
Joseph E Postma says: 2013/03/25 at 7:21 PM
When these devices detect “cool”, it is a LACK of signal that they detect, a “negative voltage”.
[referring to this post http://climateandstuff.blogspot.co.uk/2013/03/a-cool-object-reduces-energy-loss-from.html ]
Look at the drawing for a bolometer. There is an IR absober. It has IR only focussed on it (the camera lens is made from germanium – which is opaque to visible light – I have added the transmission property to the post indexed above). The IR absober temperature is measured and the this value is output to the video processor.
Can you please explain what you mean by “detect cool” I keep stressing there are no cool rays there is only Thermal radiation (between 2u and 15um) that can get through the lens. This thermal energy adds to the energy from the ambient conditions and changes the temperature of the IR absorber – a -273C temperature adds no energy, a -20C adds more and the absorber warms above the temperature that would have occurred at -273C. a source of 1500C (within the range of the cameras calibrated sensitivity) adds even more energy so the absorber is warmer still.
Whatever the temperature of the object that is focussed on the absorber (above -273C) energy is added to it and its temperature rises (it is not that -273C cools the object because that would require cooling rays to be focussed)
thefordprefect says: 2013/03/26 at 12:43 PM
…
The microbolometer is extremely simple in operation. It is warmed by ambient or a temperature controlled heater. It loses heat to ambient. It gains heat from whatever is focussed onto the IR receiver plate via the germanium lens+ power loss in the readout circuit+power from any heater+ heat from other sources in the camera etc.
It periodically calibrates itself against a standard plate at fixed temperature. Thus the internal sources of heat are known. The only variable is the IR (2um to 13um) coming through the lens
All this IR can do is to add to the energy hitting each cell, it cannot subtract. It may be coming from a 1500degC source or from a -200degC
Perhams section 2.3 of this doc will help?
Joseph E Postma says: 2013/03/26 at 8:16 PM
…
An IR receiver does not function by cold heating hot. Interpreting it that way is pure sophist BS.
thefordprefect says: 2013/03/27 at 7:20 AM
Mr Joseph E Postma
If you would care to explain how a uncooled thermal imaging camera works when photographing cold objects. This would add to the understanding about how cold objects add/do not add to the energy of hot objects.
Can you please provide an explanation.
Here’s another bolometer explanation (note that this one is cooled to minimise NOISE)
search: Detection of Light Lecture 10 bolometers
or search this one:
R. Westervelt Imaging Infrared Detectors II
Nowhere in the documentation I have read does it mention that you subtract heat from the bolometer when the source is below the temperature of the bolometer – energy is always added to the bolometer.
from the second doc.
“Bolometers operate by sensing the temperature rise associated with the
absorption of radiation. Bolometers are one of the oldest types of radiation
detector, and they have many advantages. Bolometers respond to absorbed
energy, and can be made sensitive to a very wide range of wavelengths. Because
they sense heat rather than photocarriers, bolometers are insensitive
to the photo carrier population and dynamics.”
Joseph E Postma says: 2013/03/27 at 8:02 AM
Well the first thing for you to learn is that cold things don’t heat up hot things. Try it yourself. Go around your house, the town, the country side, other countries, etc., and try to find something cold that heats up a hotter thing. Please report back on your findings. Actual findings – not just interpretations or assurances from other sources.
Whatever the process is, you will always have the limitations of the 1st Law of Thermodynamics, and then the 2nd and 3rd etc. Cold will never heat up hot. The reading suggestions easily fall under the physically-real framework provided earlier: what is the direction of ‘q’? What you need to do is develop an explanation of the bolometer which does not make the statements that cold heats up hot or that it is a demonstration of the GHE or that it violates the laws of thermodynamics – meaning precisely that cold does not heat up hot. One thing will be clear – a bolometer is not an example of cold heating up hot, at best, it will be an example of measuring the directionality of ‘q’. Does a bolometer violate the laws of thermodynamics? No of course not, nothing does. Therefore, precisely, they are not an example of cold heating up hot. And again, what the heck does a boloemter have to do with something HEATING ITSELF UP WITH IT OWN ENERGY!? Again, this is just another retarded analogy to something which doesn’t even do what the GHE is said to do in the first place(!) – a bolometer doesn’t heat itself up with its own reflected energy, so therefore this has nothing to do with the GHE and is in fact another disproof of the GHE…
Joseph E Postma says: 2013/03/27 at 9:46 AM
Well I wouldn’t encourage the creation of more sophistry myself Greg, but I’ll let you deal with it ;) I’ll quote Bryan’s response again because it has all the info necessary, and it basically means the thing measures the direction of ‘q’, as I said…
A bolometer senses the change in resistance of the sensor due to temperature change.
The sensor forms one part of an initially balanced Wheatstone bridge
No current will flow if the external object is at the same temperature as the sensor – the null point.
If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the Wheatstone bridge is out of balance and a current flows through the sensor.
If the object is at a higher temperature; the resistance increases as the temperature rises the Wheatstone bridge is again out of balance and current flows but in the opposite direction.
If calibrated against a known temperature the current can then represent temperature.
This change in resistance is linear near the null point but if moved too far from the null point the change in resistance is non linear and will give rise to errors.
thefordprefect says: 2013/03/27 at 1:00 PM
A bolometer is a devive for converting radiation into heat and then measuring the effect of the heat for example by a resistance change. The bolometer does not care what wavelength it is (providing it is within its 2 to 13um range) – it can only convert the radiation to heat.
We have a warm bolometer array (warmed from its use of power to digitise the resistance of each pixel) This will obviously be emitting as black body radiation from each pixel (it is designed that way) in all directions. Some of the pixels radiation will pass through the lens and be focussed on an object
I think most will agree that black body objects above absolute zero will radiate energy. So whatever the temperature of the object that the radiation from the pixel hits will radiate back to that pixel. No radiation of this type can cause cooling. There is zero radiation at -273°C there is some radiation at -272°C. There is no such animal as -274°C – there is no negative radiation!
Mr Postma the brings into play a numerical value q which depends on the temperature between the two objects. The question is where does this differencing occur.
The pixel will warm the remote object but by a negligible amount (compare the pixel size to the object). There canl be no differencing in the space between the pixel and object. It can only occur at the bolometer.
Assume the object is 0.3 metres away from the camera then energy leaving the bolometer pixel will not know what temperature the object is until 2ns later (2ns=2*10^-9s the time it takes to travel 2×0.3m) and similarly for the radiation leaving the object. Obviously each source of radiation does not know where it will land so the energy in transit can only depend only on the temperature of the pixel or the object.
The bolometer is at a stable temperature so:
energy lost [conduction, convection and radiation]=energy gained from the electronics and ambient)
BUT this is not true! The effect of the stuff in front of lens has not been considered.
Put a lens cap on – there is thermal radiation from the cap equivalent to its temperature.
Point it at ice at 0°C – there is radiation being emitted at +273Kelvin.
Point the lens at absolute zero – that would work, no radiation now!
Absolute zero is not an option but the camera does self calibrate by inserting a warmed plate in front of the bolometer array but again this is radiating this time at a known temperature
Consider the case when the bolometer is colder than the object then obviously the energy from the object gets focussed on the pixel and it warms – q is obviously positive
Consider the case where the pixel is hotter than the object BUT the object is above -273°C and therefore emitting radiation. The radiation still gets focussed on the pixel. This is of course positive radiation there is no negative radiation. The bolometer does not care what wavelength of radiation is hitting it.
Lets assume that because the pixel is hotter than the objet the object’s radiation gets rejected. But then all objects below the temperature of the pixel will get rejected and so there would be no image. This is not what happens.
Lets assume that the cooler radiation cancels an equivalent amount of energy in the pixel (q is negative). But how does this happen there is no difference in the effect of long or short wave radiation to a bolometer it is all just energy. There is no physical mechanism for cancellation. This is not what happens
Greg House says: 2013/03/27 at 7:16 PM
@thefordprefect
You have forgotten an important scientific issue, let me repeat it. What I would like to know, is this: how much warmer gets the sensor when the IR-Thermometer is pointed to the sky, let’s say, at night? Because, you know, I would disregard the whole thermodynamics immediately, if the GHE were alleged to be like 0.000001C. Let us just clarify that before we get deeper into the issue about how it is possible that colder bodies do not warm warmer bodies but IR-sensor still detects IR from colder bodies. Let us address the core issue: how much. I am looking forward to scientifically proven numbers.
thefordprefect says: 2013/03/27 at 7:31 PM
Joseph E Postma says: 2013/03/27 at 6:53 PM
thefordprefect – read Bryan’s answer.
——————
I assume you mean the one where he describes a whetstone bridge for measuring the resistance change caused by temperature difference?
This is jus a means of measuring the resistance change caused by thermalisation of IR. (This is certainly not the only means and definitely not the method used in an imaging array)
What I need explained to me is what happens if the pixel is hotter than the object focussed on it. Basically what causes its TEMPERATURE to change. not how that change is measured.
Cheers
Joseph E Postma says: 2013/03/27 at 7:39 PM
q ~ (T2^4 – T1^4)
The direction and magnitude of q is what changes the resistance and the current. “Thermalisation” does not mean that cold is heating up hot. That’s not what thermalisation means. The sensor is only heated proper when the balance of q is positive coming into it, else q is negative and that is what happens when the pixel is hotter. Alan and Bryan probably know more about this than me, so hopefully they can comment. But what is plainly true is that nothing violates the laws of thermodynamics, and hence, precisely, a bolometer can not be an example of something cold heating up hot. Nor do they heat themselves up with their own radiation so their operation isn’t even a relevant concern in regards to the GHE in any case.
thefordprefect says: 2013/03/27 at 7:59 PM
Joseph E Postma says: 2013/03/27 at 7:39 PM
q ~ (T2^4 – T1^4)
…
sensor is only heated proper when the balance of q is positive coming into it, else q is negative and that is what happens when the pixel is hotter.
———————————–
Apologies I still do not understand!
are you suggesting that if the object is cooler then the pixel is cooled? q is negative?
If so please tell me how a pixel is cooled below its ambient without assuming negative radiation
Joseph E Postma says: 2013/03/27 at 8:05 PM
q can be negative but the sensor, I believe, is attached to a large heat sink, so, it doesn’t cool even though q is negative. As more and more radiation comes in, from warmer and warmer sources, q becomes less and less negative until it is positive (when the source is hotter than the sensor). The heat sink, being very large, still would’t change. Hoping for Alan or Bryan to comment in more detail.
thefordprefect says: 2013/03/27 at 8:40 PM
Joseph E Postma says: 2013/03/27 at 8:05 PM
q can be negative but the sensor, I believe, is attached to a large heat sink, so, it doesn’t cool even though q is negative.
———————
tfp.
The resistive sensor is conductively isolated as from the heat sink. For the bolometer resistance to change the sense element must change in temperature. In some bolometers there is even a mirror under the sense element to reflect IR back to the sensor.
————————
jp. As more and more radiation comes in, from warmer and warmer sources, q becomes less and less negative until it is positive (when the source is hotter than the sensor). The heat sink, being very large, still would’t change.
——————.
tfp.
you use the term “radiation comes in” this as I stated in my post 2013/03/27 at 1:00 PM can only have a positive thermal effect. If you open a fridge door it feels cold because of a LACK of heat radiation (o.k. and because of cold air escaping) not because of cold radiation!
At 0K there is no radiation and no heating. at 1K there is radiation and therefore heating.
The bolometer measures radiation at all wavelengths (filtered usually into 2um to 14um bandwidth).
It converts this radiation to heat The wavelength is not material to this type of bolometer.
All changes in temperatures that the bolometer reaches are effectively referenced to zero incoming radiation i.e. the source is at 0K.
All other temperatures of source add proportionally more radiation to the pixel.
The pixel temperature will therefore always be above that with no incoming radiation (from 0K source)
The temperature of the pixel will always depend on balance of heat loss and heat input.
Heat input from the source beyond the lens is the only variable (internal heating is calibrated out).
The focused source radiation will control the pixel temperature – higher temperature of source = higher temperature of pixel.
Joseph E Postma says: 2013/03/27 at 8:52 PM
prefect – all you have to do is use the laws of thermodynamics. Cold doesn’t heat up hot. The type of “pixels” I use in my field do not detect radiation by increasing in temperature – they convert photons to electrons. Bolometers might do something different than that and I’m not an expert on them. But the only thing you do need to know, is that cold doesn’t heat up hot. Not only that, a bolometer does not heat up via its own radiation, and so however they function has nothing to do with the GHE in any case. Therefore this is a useless discussion. We’re discussing how many teeth are in a horses mouth when the claim is that tigers can eat airplanes. Bolometers don’t heat themselves up with their own radiation, therefore they offer no support for the GHE. If a bolometer is pointed at a mirror, it does not spontaneously begin reporting higher and higher temperature.
A bolometer senses the change in resistance of the sensor due to temperature change.
The sensor forms one part of an initially balanced Wheatstone bridge
No current will flow if the external object is at the same temperature as the sensor – the null point.
If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the Wheatstone bridge is out of balance and a current flows through the sensor.
If the object is at a higher temperature; the resistance increases as the temperature rises the Wheatstone bridge is again out of balance and current flows but in the opposite direction.
If calibrated against a known temperature the current can then represent temperature.
This change in resistance is linear near the null point but if moved too far from the null point the change in resistance is non linear and will give rise to errors.
Joseph E Postma quotes: 2013/03/27 at 8:52 PM
If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the …
————————————
But the question is – Why does the temperature go down?
Bolometers are in thermal balance with all energy inputs so power into the array is part of this balance as is the focused radiation from the source
I think any scientist would agree that cold cannot heat up hot, However in the world of radiation balance the starting temperature should be the ambient temperature with added radiation at zero (-273°C) – “zero temp” – any change from this source temperature adds more energy and the temperature rises from its “zero temp”.
So if there is no ghg then the earth sees TSI plus near zero radiation from space this will give be its “zero temp”.
Add ghg and the radiation from above is perhaps from a temperature of -50°C (223K) This adds to the energy heating the earth (at perhaps -18C). The temperature will now be heated above its “zero temp”.
This may seem like cold heating hot but really its cold heating “zero temp”.
But what is important is that the new radiative balance allows the earth to warm to a new higher temperature above “zero temp”
There is no breaking of any laws.
Joseph E Postma says:
2013/03/27 at 9:46 PM
“Add ghg and the radiation from above is perhaps from a temperature of -50°C (223K) This adds to the energy heating the earth (at perhaps -18C). The temperature will now be heated above its “zero temp”.”
No, that is not correct at all. And it has nothing to do with bolometers either because bolometers do not heat themselves up or report higher temperatures from their own radiation coming back to them. Being surrounded by ice or mirrors with your own radiation coming back does not heat you up or add to your heating. This is of course what bolometers prove as well.
thefordprefect says: 2013/03/27 at 10:50 PM
[trashed]
tfp, bolometers don’t violate the laws of thermo. Cold doesn’t heat hot. CCD’s absorb photons and create current just fine without being caused to heat up etc.
thefordprefect says: 2013/03/27 at 11:03 PM
Max, although this is for astro purposes this is helpful on bolometer detectors and why they are sometimes cooled (noise reduction)
Max™ says:
2013/03/28 at 12:58 AM
Joe also went over the actual mechanism: there is a sink tied to the device to stabilize it, the detector pixels are isolated, and the default with no radiation incoming is treated as 0 K, any level of radiation coming in will reduce that emitted by the device, this can be used to work out the temperature of a distant source
—————
What is the mechanism for reducing the radiation emitted by the device. There is no cancellation the bolometer does not respond to phase of radiation.
Please explain to me how an object above absolute zero but below bolometer temperature reduces the outgoing radiation.
But please keep it simple.
Joseph E Postma says: 2013/03/28 at 7:14 AM
tfp said: “Please explain to me how an object above absolute zero but below bolometer temperature reduces the outgoing radiation.”
This is why I’m trashing your comments. This has already been answered repeatedly. q ~ (T2^4 – T1^4); q ~ (T2 – T1), etc etc. Outgoing q is reduced as T1 increases; as long as q is net outgoing, no actual heating (i.e. temperature increase) of the warmer source occurs from the colder source. Reducing cooling does not mean increasing temperature.
This is a direct physical analogy:
a = 1/m*(F2-F1). a = acceleration, m = mass, F2 = force 2, F1 = force 1. If F2 is some value, and F1 starts at zero, as F1 increases towards F2, the acceleration decreases, but, the direction of acceleration doesn’t change . The direction of the acceleration only changes when F1 becomes greater than F2.
thefordprefect says: 2013/03/28 at 3:35 PM
Joseph E Postma says: 2013/03/28 at 7:14 AM
This is why I’m trashing your comments. This has already been answered repeatedly. q ~ (T2^4 – T1^4); q ~ (T2 – T1), etc etc. Outgoing q is reduced as T1 increases; as long as q is net outgoing, no actual heating (i.e. temperature increase) of the warmer source occurs from the colder source. Reducing cooling does not mean increasing temperature.
———————-
Right so you are talking temperatures. But radiation is not temperature until thermalized.
The real situation is:
radiation+radiation hits pixel. Pixel converts sum of radiation to a single temperature. temperature causes resistive change. resistive change is read out by electronics.
The pixel is not sensitive to IR wavelength. It warms by thermalisation of the radiation (frequency and phase of waves is irrelevant. The amount of radiation falling on the pixel per unit time causes a temperature rise
So how does radiation cause a cooling?
This paper posted by Max gives the answer at the end
Thanks Max
Joseph E Postma says: 2013/03/29 at 7:16 PM
@tfp @2013/03/28 at 3:35 PM
Yes I read about Picktet many years ago.The results are wonderful confirmation of the laws of thermodynamics applying to radiation, and the falsity of the claims of the GHE. The results can also be understood with the idea that “radiation is conduction at a distance” as said elsewhere. I mean how amazing is the result: radiation from a cold source focused on a warmer target does not add to the temperature of the warmer target, but causes it to cool! Cold never warms up hot. Only when the thermal load is warmer than the target does it warm it up.
thefordprefect says: 2013/03/29 at 10:41 PM
Joseph E Postma says: 2013/03/29 at 7:16 PM
From the document that you say “The results are wonderful confirmation of the laws of thermodynamics applying to radiation”
IV. QUALITATIVE EXPLANATION OF PICTET’S EXPERIMENT
As the radiation and reflection of cold may appear paradoxical, it may not be amiss to offer a qualitative explanation of Pictet’s experiment in modern terms.
Let us note that every object—even a cold object—continually emits radiation. Each object also continually receives radiation that has been emitted by the objects surrounding it. The energy emitted per unit time by a given object depends both upon the object’s temperature and the properties of its surface. A highly polished metal mirror is a very poor absorber of infrared radiation; it is also, consequently, a poor radiator. Thus we may safely ignore any emission or absorption of radiant heat by the mirrors themselves. The sole function of the mirrors then consists in the reflection of the radiation that is incident upon them.
Consider first the version of the experiment involving a flask of boiling water. This flask emits radiation in all directions, as shown schematically in Fig. 5(a). A negligible part of this radiation is received directly by the air thermometer, which has but a small cross-sectional area and which is placed at a considerable distance from the flask. In Fig. 5(b), we imagine the mirrors to be placed so that the focus of mirror A lies in the flask, and the focus of B in the bulb of the thermometer. Now we direct our attention toward that part of the radiation from the flask which happens to strike mirror A. This radiation will, after two reflections, impinge on the bulb of the thermometer. This is radiation that the thermometer did not receive prior to the introduction of the mirrors. Of course, the thermometer also receives radiation from other objects in the room. The introduction of mirror B effectively eliminates the right third or half of the room as a source of radiation for the thermometer. The effect of the mirrors is therefore to replace a part of the ambient radiation that formerly impinged on the thermometer by the more intense radiation from the flask. The thermometer therefore grows a little warmer.
The version of the experiment involving a flask of snow may be explained in a similar way. Before the mirrors are introduced, the air thermometer receives radiation from all the objects in the room surrounding it. After the introduction of the mirrors, the radiation from the right third or half of the room is cut off from the thermometer by mirror B. This relatively high-temperature radiation is replaced by the radiation from the flask of snow. The thermometer now receives less energy per unit time than previously. The thermometer, initially at room temperature, now radiates away more energy than it absorbs, and so suffers a decrease in temperature.
——————-
It is all in this last paragraph – “This relatively high-temperature radiation is replaced by the radiation from the flask of snow”
i.e. the warm object still receives additional radiation from the snow But this is less than it would have received from the warm room. so it cools.
The radiation from the snow is not being rejected – it is simply less than from the room.
If you agree with this then we both agree with the operation of a bolometer!!!!
[JP Reply: So yes then...the directionality of q, and cold radiation doesn't heat up a warmer target.]
thefordprefect says: 2013/03/30 at 8:21 AM
The smallest amount of energy (i.e. one quantum) that an object can absorb from IR light with a wavelength of 8Āµm
Energy quantum = hĪ½
h is known as “Planck’s constant”, and has a value of 6.63 x 10^-34 Joule seconds (Js)
so we need to know the frequency Ī½
Ī½Ī»= c
Ī½ = c/Ī»
Ī½ = (3.00 x 108 m/s)/(8*10-6 m) s-1
plugging into Planck’s equation:
E = (6.63 x 10^-34 Js)*( (3.00 x 10^8 m/s)/(8*10^-6 m) s-1)
E (1 quanta) =2.49×10^-20 J at 8Ī¼m IR
I small calorie raises the temp of water at 19.5C by 1°C
4.182 joules=1cal
I quanta of IR at 8Ī¼m will RAISE 1 gm of water by (1 / 4.18)x2.49×10^-20 °C
=5.95×10^-21 °C!!
NB. The absolute temperature of the water does not come into this analysis. The energy is added to whatever other energy is present and the temperature INCREASES,
If the IR disappears the temperature falls
[JP Reply: Well that's a really simple-minded analysis using highschool physics. Of course, the radiation from an ice-cube doesn't cause water to warm up, so, the source of the radiation is indeed a factor. The Laws of Thermo aren't really learned at a mathematical level until about 3rd-year undergrad in a major in physics.]
thefordprefect says: 2013/03/31 at 6:03 AM
It is all in this last paragraph – “This relatively high-temperature radiation is replaced by the radiation from the flask of snow”
i.e. the warm object still receives additional radiation from the snow But this is less than it would have received from the warm room. so it cools.
The radiation from the snow is not being rejected – it is simply less than from the room.
If you agree with this then we both agree with the operation of a bolometer!!!!
————————————
But q you have stated is the difference in temperatures. Here they are talking about radiation. Radiation does not have a plus and minus. So there is NO directionality of radiation. zero radiation does nothing. Added radiation gets thermalized whatever its value or source.
A hot object in a -273°C universe will be colder than the same object after receiving on quanta of energy from a hot object or one at -272°C
A hot object at 100°C in a 100°C universe will not cool add an object at 99°C that blocks 10% of the hot objects view of the 100°C universe will cause the hot object to cool. I would postulate that a cold -273°C point source blocking 0% of the 100°C universe would NOT cause the hot object to cool.
[JP Reply: q ~ (T2^4 - T1^4), so yes, q does have a direction even with radiation.]
Joseph E Postma says: 2013/04/01 at 9:01 AM
@tfp 2013/04/01 at 8:38 AM
Then, on problem 1026, just below the one you refer to, it describes the Willis’ shells game idea, and again, the source doesn’t become heated, but this time the shell just radiates the equal amount of energy that the source provides.
Both the problem 1023 and 1026 are consistent in that the presence of backradiation does not increase the source temperature – that is very explicitly stated in the results. Back radiation can exist, but q can be equal to zero. q being equal to zero does not mean that backradiation doesn’t exist. Backradiation can exist and not cause any heating.
an unscientific piece of crap.
Joseph E Postma says: 2013/04/01 at 9:01 AM
It shows backradiation which does not cause any increase in temperature. Please look at what it says and sketches carefully. The backradiation does not increase the temperature of the source. The only thing the shield does is decrease the rate of emission – and this does not increase the temperature of the source. The result couldn’t be any more clear: T1 = T/4root(2), and so there was no change to T, the source temperature. Answers a) and b) say nothing about the source becoming warmer, with the presence of backradiation, as would be expected from thermodynamics.
————–
So if what you say is correct then all your diagrams showing 235watts passing unchanged from the core to the shell must be incorrect.
A core generating 400 watts receives 200 watts back radiation from the shell the shell transmits 200 watts radiation to space is the what you are claiming the Berkley question shows, The only way this could be true is if the generator and back radiation cancel and we are then back to positive and negative radiation.
This still has the problem that you do not admit to the back radiation?
I think you will find the Berkley model is showing a non generating core and the effect of a single “multi” layer insulator
If the source were still generating 400 watts the 400 watts must be emitted to space I think all would agree.
If you substitute that in the equations then J1=400 and J-J1=400 and therefore J=800
[JP Reply: In problem 1026 the same amount of energy is leaving the shell to outerspace that it is receiving from below from the sphere. You do see that, correct? You do see that problem 1026 exists and you've read it? The result is that the same amount of energy is emitted from the shell as it receives from below. Also, backradiation doesn't cause heating of the sphere in either problem. You see that too? There does seem to be a contradiction between problems 1023 and 1026 in terms of the total energy output - there is no contradiction in that backradiation or radiant trapping does not cause the source to become warmer. The correct answer for thermal equilibrium is when the outer shell/wall/etc reaches the same temperature as the source, with it emitting outwardly all the radiation it receives from below, as we've discussed here. "Positive and negative" radiation would just be the usual 'q' term we've discussed at length. Admitting to backradiation? q ~ T2^4 - T1^4.
thefordprefect says: 2013/04/01 at 9:55 PM
M Postma
Question 1023 refers to a fixed temperature body.
It is not a power source.
So your response to my question is not valid:
A core generating 400 watts receives 200 watts back radiation from the shell the shell transmits 200 watts radiation to space is the what you are claiming the Berkley question shows, The only way this could be true is if the generator and back radiation cancel and we are then back to positive and negative radiation.
This still has the problem that you do not admit to the back radiation?
…
If the source were still generating 400 watts the 400 watts must be emitted to space I think all would agree.
If you substitute that in the equations then J1=400 and J-J1=400 and therefore J=800
[JP Reply: In problem 1026 the same amount of energy is leaving the shell to outerspace that it is receiving from below from the sphere. You do see that, correct? You do see that problem 1026 exists and you've read it? The result is that the same amount of energy is emitted from the shell as it receives from below.
=======
tfp: so this agrees with my 1023 comment j1=400 J=800
=======
Also, backradiation doesn't cause heating of the sphere in either problem.
=======
tfp the core is constant temp according to 1023.
=======
...there is no contradiction in that backradiation or radiant trapping does not cause the source to become warmer.
===========
If the core is generating 400 watts and 200 watts (according to 1023) is coming back into it It must still be generating 400 watts so the shell has to emit 400 watts to prevent thermal runaway.
Remember the core is a generator not a fixed temperature
I am not sure why problems 1023 and 1026 seem to describe the same idea but give different results - in terms of halving the energy. They do give the same result in that backradiation or radiant trapping does not cause the source to warm up. Let me solve that problem for you, because it certainly can and is causing confusion.]
thefordprefect says: 2013/04/02 at 6:20 AM
1}} Joseph E Postma says: 2013/04/01 at 9:01 AM @tfp 2013/04/01 at 8:38 AM
It shows backradiation which does not cause any increase in temperature. Please look at what it says and sketches carefully. The backradiation does not increase the temperature of the source. The only thing the shield does is decrease the rate of emission – and this does not increase the temperature of the source. The result couldn’t be any more clear: T1 = T/4root(2), and so there was no change to T, the source temperature. Answers a) and b) say nothing about the source becoming warmer, with the presence of backradiation, as would be expected from thermodynamics.
…Both the problem 1023 and 1026 are consistent in that the presence of backradiation does not increase the source temperature – that is very explicitly stated in the results. Back radiation can exist, but q can be equal to zero. q being equal to zero does not mean that backradiation doesn’t exist. Backradiation can exist and not cause any heating.
—————-
2}} thefordprefect says: 2013/04/01 at 9:55 PM
If the core is generating 400 watts and 200 watts (according to 1023) is coming back into it It must still be generating 400 watts so the shell has to emit 400 watts to prevent thermal runaway.
Remember the core is a generator not a fixed temperature
[JP Reply: Right, the shell emits the equivalent of 400 Watts outwards (equivalent depending on its radius/surface area), and that's what the core generates, so, no backradiation heating and all energy balances out.]
=========
You cannot have it both ways
in 1 you say back radiation exists from the inside of the shell and halves the energy loss from the core (400w to 200w) and the remaining 200 w goes to space from the outside of the shell
in 2 you say there is no backradiation to the core reducing the energy loss so 400 watts from the core hits the shell and the shell then radiates 400w to space. If there is back radiation then the 400watts from core to shell gets split between back and fore radiation
I assume therefore that you are saying that this is wrong?
If so can you please explain what the difference is between the inside of the hot shell and the outside of the hot shell that prevents the inside radiating.
does multilayer insulation work
Does MLI rely on backradiation. If not then how does it work radiatively.
If back radiation does not exist then presumably the questions in the document are wrong.
If back radiation does exist then what happens to it when it approaches an object hotter than the one it left.
I am very confused and would really appreciate a summary of the physics you believe is happening.
cheers.
thefordprefect says: 2013/04/02 at 6:25 AM
also why does the inside of a shell not radiate like the outside of a shell. The heat does not know which side it is leaving?
If it does leave the inside where does it go.
The steel green house is based on earth dimensions so the shell to a reasonable accuracy is the same size as the core.
Joseph E Postma says: 2013/04/02 at 6:42 AM
I’ll jut answer this one. tfp: “Try not to use T^4 equations, because we are talking radiation, which is not the same!”
Radiated energy does not equate to net heat transfer or even net energy transfer. The equation of heat flow for radiation, from physics, from actual physics textbooks and from actual universities and actual physics degrees, is q ~ (T2^4 – T1^4). If T2 = T1, then q = 0, and nothing heats up, even though there’s all that radiation. It works the same way in conduction.
A temperature does not spontaneously heat itself up. You have the UCB links Max provided which precisely show that the sphere does not heat itself up. Please work those questions out for yourself on paper as well. If you can’t do that, then you will never be able to understand thermodynamics. You could understand thermodynamics if you simply adopted the knowledge and intuition that a temperature does not heat itself up.
thefordprefect says: 2013/04/02 at 8:15 AM
Temperatures are not allwiki:
Thermospheric temperatures increase with altitude due to absorption of highly energetic solar radiation. Temperatures are highly dependent on solar activity, and can rise to 2,000 °C (3,630 °F).
How does this gell with your T^4 equations. q to earth must be staggeringly big.!!!
wiki:Even though the temperature is so high, one would not feel warm in the thermosphere, because it is so near vacuum that there is not enough contact with the few atoms of gas to transfer much heat. A normal thermometer would read significantly below 0 °C (32 °F), because the energy lost by thermal radiation would exceed the energy acquired from the atmospheric gas by direct contact. In the anacoustic zone above 160 kilometres (99 mi), the density is so low that molecular interactions are too infrequent to permit the transmission of sound.
[JP Reply: The thermosphere doesn't violate the Stefan-Boltzmann Law, and nothing about this violates T^4. It fits exactly and precisely into it. If you don't like T^4, you must really hate the Stefan-Boltzmann Law.]
thefordprefect says: 2013/04/02 at 10:13 AM
Greg House says: 2013/04/02 at 9:05 AM
If you mean that back radiation affects the temperature of the source (the IPCC position and the one of some so called “skeptics”) I suggest you present a real experimental proof for that.
==========
I linked the experiment I did – the post got trashed
But just in case here is the link again. I would suggest posting there as I think my replies would get trashed here!
http://climateandstuff.blogspot.co.uk/2013/04/the-copper-greenhouse-new-test.html
[JP Reply: You get trashed comments for asking the same question which has been answered dozens of times.]
Greg House says: 2013/04/02 at 11:11 AM
thefordprefect says (2013/04/02 at 10:13 AM) “…”
[JP Reply: You get trashed comments for asking the same question which has been answered dozens of times.]
=======================================================
By the way, since he refereed to IR thermometers as a proof for back radiation warming, I asked him 3 times on this thread the same question about how much exactly IR from a colder target warms the sensor. Still no answer. I wonder why.
thefordprefect says: 2013/04/03 at 5:44 AM
Greg House says: 2013/04/02 at 11:11 AM
By the way, since he refereed to IR thermometers as a proof for back radiation warming, I asked him 3 times on this thread the same question about how much exactly IR from a colder target warms the sensor. Still no answer. I wonder why.
———-
Consider an ir thermometer at an ambient of 22°C
Point the IR thermometer at -273°C it reads THE ZERO temperature. no incoming radiation sensor temperature =x
Point the IR thermometer at -272°C it reads a higher temperature – its is now receiving radiation sensor temperature =x+a little bit
Point the IR thermometer at +272°C it reads a higher temperature – its is now receiving radiation sensor temperature =x+a lot
There is a continuous temperature increase with source temperature increase
the energy of the radiation adds.
If you have 2 resistors exactly the same size 1 dissipating 1 watt at 60°C above ambient and another dissipating 2 watts at 120°C above ambient then if you join the 2 together I would suggest that the overall temperature increases to above 120°C (in fact to around 180°C)
What would you suggest the outcome was?
…
thefordprefect says: 2013/04/04 at 8:33 AM
Repeat of lost post
thefordprefect says: 2013/04/03 at 10:47 AM
Note 8um radiation gives 2.49e^-20 Joules (watts*seconds) of energy for 1 quantum
Your rather good example of water needs looking at.
Firstly I only talk energy as this is what 1 quatum is.
….
so adding together two pots at different temperatures gives you more available work than any one pot in fact the work is simply the sum of the energy available from both pots
There still seems to be some confusion about cold radiation heating hot bodies.
You seem to agree that you can add quanta of energy from hot and cold sources (the hot and cold pots of water).
[JP: Yes, you can add one pot of water to another pot of water. This shouldn't be a questionable idea. The result of this addition is not a warmer temperature than the original hot pot.]
I’m confused if you agree the shell emits radiation outwards and inwards (even if the inwards back radiation has no effect).
[JP: It may emit inwards, and it has no effect, no net 'q' and no net transfer of energy.]
a body at 0K emits no radiation (no contentious issues there!)
a body at 1K emits some radiation (no contentious issues there I hope!)
A body internally heated with FIXED power to a stable temperature of T0 in a infinite box at 0K and a complete vacuum receives no external radiation and will loose heat as quanta (no conduction, no convection) at a rate determined by its temperature and surface area. This rate of loss of heat will be exactly the same as the energy supplied per unit time (otherwise the temperature will be changing) lets call this x watts.
so x watts produces a temperature of T0 on the body
add a warm (T0+Ts) shell, lets call it a spherical oven, which bathes the body with a total of y watts (evenly spread around the body). Its temperature is above the body at T0 so the energy loss of the body is now x-y watts and it will warm above T0 lets call it Tt (this shouldn’t be controvertial since the oven is above the body temperature).
Now we turn off the oven heater and let it cool. (we’ll ignore the radiation hitting the oven from the body!)
[JP: Why ignore the radiation from the body which you put inside the oven? This is a meaningless and arbitrary change to the system for no reason, and no valid result can be determined from this change relative to the conditions which were previously established.]
I believe what you are saying is that as the oven cools to below the body temperature the radiation from the shell stops adding to the internal heating of the body and there is a sudden change in cooling slope. Is this correct?.
This seems very odd to me unless you can help my understanding?
thefordprefect says: 2013/04/04 at 10:08 AM
There still seems to be some confusion about cold radiation heating hot bodies.
You seem to agree that you can add quanta of energy from hot and cold sources (the hot and cold pots of water).
[JP: Yes, you can add one pot of water to another pot of water. This shouldn't be a questionable idea. The result of this addition is not a warmer temperature than the original hot pot.]
===But the total “quanta” available has been added together it has not been lost.
But you suggest that radiation from the shell to the core just gets lost. from the above shouldn’t it get added to the energy feeding the core?
[JP: The total energy added together from a cold and hot pot do not result in a higher temperature than the hot pot. Total energy is conserved and a temperature warmer than the hot pot is not created. It is not generally good to analogize fermionic material behavior with bosonic photon behavior. The energy emitted inside the shell to the core combines in superposition with the existing radiation field - this produces a blackbody spectrum inside the cavity and this is a quantum effect that is totally unlike any other material behaviour of atoms etc. If radiation didn't combine in this quantum fashion then you would have the classical "ultra violet catastrophe". With the two pots of water, the thermal energy adds together into a new thermal ensemble which will not be warmer than the original hot pot. You have all the total energy together but it did not produce a warmer temperature. Radiative quanta behave similarly to that in that adding quanta from a cold source plus a hot source does not make an even hotter radiation field. The only energy which gets lost is that on the outside of the shell.]
thefordprefect says: 2013/04/04 at 10:13 AM
The oven bit – I shouold really have said froven actively cooled and heated (fridge-oven).
As you turn the froven temperature below the body temperature I suggest that there will be no discontinuity but you suggest that the the body will suddenly loose all radiation from the froven. Is this correct?
[JP: No not correct. If the oven is cooled to below the body temperature, then the body will warm the oven. The coolant engine of the froven will need to work to remove the heat supplied by the body. When the froven cools to below the body temperature, then the directionality of 'q' changes direction, but this is not a discontinuity - it is a smooth transition from one direction to the other in heat energy.]
thefordprefect says: 2013/04/04 at 10:53 AM
thefordprefect says: 2013/04/04 at 10:13 AM
As you turn the froven temperature below the body temperature I suggest that there will be no discontinuity but you suggest that the the body will suddenly loose all radiation from the froven. Is this correct?
Good so far. I agree the froven will need to absorb the radiation from the body. The body of course does not know what surrounds it – it is in a vacuum, it can only radiate quanta according to the body temperature. so as the froven goes from warmer to colder than the body i suggest it stil radiates quanta to the body (fewer and fewer as the frovenn temperature decreases) and the body cools since it is receiving less radiation – this continues until the froven is at absolute zero and the body will then cool to T0 (as it was when the froven was not present). There will be no discontinuities in this cooling curve vs froven temperature..
Am I right in suggesting that you would believe that as the froven goes from warmer to colder than the body the quanta from the froven wall will suddenly cease to have an effect on the body which can only expel quanta according to its temperature.?
If so then as the froven goes from warmer to cooler than the body the rate of cooling of the body will increase to the same rate as it would be without the froven i.e. without incoming quata from the froven shell. there will be a dicontinuity in the cooling curve vs froven temperature.
thefordprefect says: 2013/04/04 at 11:52 AM
[JP:... If the body has a heat source then it will stay at the temperature it was at without the oven heating it from a higher temperature. If the oven is cooler than the body than it can not heat the body. Photon quanta from a cooler source do not warm up a warmer source, even if they might exist. It is not a "sudden" cessation of effect when the oven becomes cooler than the body - it is a smooth transition in the direction of q, of heating.]
This cannot be correct.
If the temperature of the froven is warmer than the body you suggest it heats the body. If the froven is cooler than the body you suggest it has no effect.
If the body is radiating quanta from a 100°C source then the hotter froven will be radiating to the body quanta from its 100C+ walls.
If the body is radiating quanta from a 100°C source then the cooler froven will be radiating nothing from it 100C- as if it were at absolute zero thats one heck of a sudden step. Do I understand correctly?
[JP: Not quite yet. If the body is warmer than the oven, then the body heats the oven. If the oven is warmer than the body, then the oven heats the body. This is a smooth transition in the direction of heating as a function of the temperature differential: -2 -1 0 1 2 etc. A smooth transition, not a sudden stop.]
thefordprefect says: Your comment is awaiting moderation. 2013/04/05 at 11:52 AM
Seem to have problems posting so I’ll try again:
[JP: Not quite yet. If the body is warmer than the oven, then the body heats the oven. If the oven is warmer than the body, then the oven heats the body. This is a smooth transition in the direction of heating as a function of the temperature differential: -2 -1 0 1 2 etc. A smooth transition, not a sudden stop.]
===========
you have stated definitely that there is no transfer of energy from cold to hot
“but what I do know is that they do NOT work by cold heating hot – hahaha what a stupid idea.”
“[JP Reply: Trashed because we've already answered you. q from the shell to the planet is 0. ZERO. There is no heat loss from the shell to the planet. Even if the shell is emitting on the inside, there is no heat loss to the planet. The only direction the shell can lose heat is outwards, and hence it loses the equivalent of 800 W/m2 outwards.]”
“Radiated energy does not equate to net heat transfer or even net energy transfer. The equation of heat flow for radiation, from physics, from actual physics textbooks and from actual universities and actual physics degrees, is q ~ (T2^4 – T1^4). If T2 = T1, then q = 0, and nothing heats up, even though there’s all that radiation. ”
so
Firstly I hope you would agree that the quanta of energy leaving a surface cannot depend on the final destination of the quanta i.e. its temperature, material and surface – it only depends on the source material and temperature.
I also believe this describes your point of view
The final destination of the radiation determines what happens to the quanta (rejected or absorbed)
oven at 101C transfers zero quanta to body at 10000C (equivalent to back radiation)
body at 10000C transfers w quanta to oven at 101C
oven at 100C- transfers zero quanta to body at 100C (equivalent to back radiation)
body at 100C transfers x quanta to oven at 101C-
oven at 100C+ transfers x quanta to body at 100C
body at 100C transfers zero quanta to oven at 100C+ (equivalent to back radiation)
body at 100C transfers y quanta to oven at 99C
oven at 99C transfers zero quanta to body at 100C (equivalent to back radiation)
oven at 10000C transfers w quanta to body at 100C
body at 100C transfers zero quanta to oven at 10000C (equivalent to back radiation)
at 100C- to 100C+ oven temperature the body quanta changes from outputting x to receiving x quanta.
Some how this does not seem to be a smooth or logical transition
Warmists would say
quanta emitted from an object depends only on the object and its temperature. the final destination of the radiation is immaterial (well actually the quanta knows nothing until it hits the surface)
The sum of all quanta determines the rate of loss/gain of heat
oven at 101C transfers y quanta to body at 10000C (equivalent to back radiation)
body at 10000C transfers w quanta to oven at 101C
oven at 101C transfers y quanta to body at 100C
body at 100C transfers x quanta to oven at 101C (equivalent to back radiation)
oven at 100C- transfers x-1 quanta to body at 100C (equivalent to back radiation)
body at 100C transfers x quanta to oven at 101C-
oven at 99C transfers z quanta to body at 100C (equivalent to back radiation)
body at 100C transfers x quanta to oven at 99C
oven at 10000C transfers w quanta to body at 100C
body at 100C transfers x quanta to oven at 10000C (equivalent to back radiation)
at 100C- to 100C+ oven temperature the body quanta does not change from outputting x and receiving x quanta.
where
100C- a very very very! small bit less than 100C
100C+ a very very very! small bit more than 100Cw greater than y
y greater than x
and x greater than z
A smooth and logical transition.
I assume that I have this wrong somehow so perhaps using x,y,z you could explain your position
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