Trenberth's Energy Budget

[wirebender]

wirebender thinks computers are cooled by radiation rather than conduction and convection.

Ian, it doesn't matter whether the means of moving the heat is radiation, convection, or conduction, the laws of nature still apply. Whether it is in a vaccum or not doesn't allow it to break the 2nd law of thermodynamics. Because it is in the open atmosphere, the heat sink and the heat source will never reach equilibrium and become as a single radiating source as would happen in the vaccum of spencer's imagination.

Once more Ian, here is the math that proves you wrong. Do tell exactly what you believe is wrong with it. As you can see, the bottom line is both plates ending up at 53 degrees F; considerably cooler than the plates in spencer's imagination. Which physical law do you believe is being misapplied there, or what mathematical error do you believe has been made. The laws of nature are called the laws of nature for a reason Ian and not the laws of systems, or laws of vacuum vs laws of open space. You can not break the laws of nature and create energy simply because you suck all the air out of a chamber. If you could, our energy woes would be a thing of the past.

You have been willingly duped because your faith is stronger than your analytical mind. You want to believe that bar will get warmer because you are emotionally attached to the idea that CO2 can somehow cause the atmosphere to heat up. Give up the belief that the cooler bar will somehow make the heated bar warmer and you have to give up your fantasy about CO2.

Re: Vacuum Chamber with plates.

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan’s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts

(***That’s ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)

The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2
——–
If another identical “non-heated” and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:

The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.

The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.

The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

And the equilibrium temperature for both plates will be 284.69 K or 53 deg F.


The only way to make that heated bar warmer Ian, is to either pump in more electricity through the wire or to make it smaller so that its radiating surface is smaller. Placing a heat sink next to it serves to increase the radiating surface thus bleeding heat off faster and making it cooler. You can't increase the amount of energy in the system by increasing the radiating surface; that only serves to make the system cooler and it doesn't matter a whit whether it is happening in the open atmosphere or in a vaccum chamber.

 

[wirebender]

Now please debate HIS points as they are and not what you try and make them out to be..

I am afraid that is never going to happen gslack. Remember the stink he made over me not directing him to the math I did on the board earlier relating to the same argument except the material was CO2 and earth instead of a powered plate and a non powered plate as if he would destroy my argument if he could just get to that math. Well, there is some math right here, of a more simple nature as it doesn't involve time and distance and yet, I don't see him doing anything but dodging it. Later, after sufficient time has passed I will wager that if this thread comes up, he will ask for directions back here and pretend that if he could just see the math, he could destroy the argument.

It is clear by his avoidance of this relatively simple math that he is not the student of physics that he claims to be. The straight forward formulas above are used day in and day out by engineers of all stripes who must deal with heat and they accurately predict the temperature materials will reach given the amount of energy going in, the amount of radiating surface, and the nature of the materials. If they didn't work, every project would be guesswork and you can bet that there is no guesswork involved in designs intended to bleed off heat.

I doubt that you find many designs out in the real world that have to handle energy created by the heat sink. Can you imagine suddenly having to deal with energy that you weren't getting from the electric company or your own generators by simply adding to your radiating surface? Add some extra radiating surface, create some free energy. We could build radiators the size of sky scrapers, plug them into a couple of AAA's and power entire cities.

Perpetual motion incarnate; and Ian believes in it.

 

[gslack]

Now please debate HIS points as they are and not what you try and make them out to be..

I am afraid that is never going to happen gslack. Remember the stink he made over me not directing him to the math I did on the board earlier relating to the same argument except the material was CO2 and earth instead of a powered plate and a non powered plate as if he would destroy my argument if he could just get to that math. Well, there is some math right here, of a more simple nature as it doesn't involve time and distance and yet, I don't see him doing anything but dodging it. Later, after sufficient time has passed I will wager that if this thread comes up, he will ask for directions back here and pretend that if he could just see the math, he could destroy the argument.

It is clear by his avoidance of this relatively simple math that he is not the student of physics that he claims to be. The straight forward formulas above are used day in and day out by engineers of all stripes who must deal with heat and they accurately predict the temperature materials will reach given the amount of energy going in, the amount of radiating surface, and the nature of the materials. If they didn't work, every project would be guesswork and you can bet that there is no guesswork involved in designs intended to bleed off heat.

I doubt that you find many designs out in the real world that have to handle energy created by the heat sink. Can you imagine suddenly having to deal with energy that you weren't getting from the electric company or your own generators by simply adding to your radiating surface? Add some extra radiating surface, create some free energy. We could build radiators the size of sky scrapers, plug them into a couple of AAA's and power entire cities.

Perpetual motion incarnate; and Ian believes in it.

Going by their hypothesis, we could put a low pressure tank of CO2 as a addon to an electric generator and BAM! turn 10 generators into 1000...

 

[IanC]

Re: Vacuum Chamber with plates.

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan’s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts

(***That’s ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)

The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2
——–
If another identical “non-heated” and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:

The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.

The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.

The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

And the equilibrium temperature for both plates will be 284.69 K or 53 deg F.

this is the thought experiment that wirebender and gslack want me to respond to. so I will

Vacuum Chamber with plates.

I am assuming that the chamber is cooled to some specific constant temperature

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.
The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

the heater inside the plate is hotter than the outside of the plate because of the thermal impedance of the material of the plate. the outside plate temp is determined on how fast the internal energy can be shed to the vacuum of constant temperature. any change to the power source, thermal impedance of the plate material or temperature of the vacuum will change the equilibrium (changing the shape of the plate would also affect the temp if it changed the surface area).

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2
Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:
P = e*BC*A*T^4
Where P = net radiated power (Watts), e = emissivity, BC = Stefan’s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K
P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts
(***That’s ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)
The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2

your definition, live with it

If another identical “non-heated” and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:
The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

at the beginning the second plate (if at the same temp as the vacuum) will make little difference. but as the second plate heats up it will start to radiate in all directions, including back at the heated plate. this changes the ability of the heated plate to shed energy. the heated plate still wants to get rid of the same amount of heat therefore more heat will be dispersed through the area that can escape to the vacuum. increased heat means increased temperature.

We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.
The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.
The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

this is where your example really goes off the rails, hahahahaha. what sort of gullible idiot would think that both plates would be at the same temperature at equilibrium? hahahahaha. what kind of moron would believe that the areas exposed to the cold vacuum loses heat at the same rate as the areas next to warmer surfaces? could you imbeciles get more things wrong?


it is easy to see that the heated plate must warm up to compensate for the loss of exposure to the cooled vacuum due to the thermal impedance of the second plate, and that all four surfaces of the two plates (I am ignoring the edges) have different temperatures. the heated inside facing area is the warmest, the heated outside facing area is next, the inside facing unheated area is third and the unheated outside facing area is the coolest of them all.

 

[gslack]

HAAHAHAHAHAHAHAHAHAHAHAHAAHAHAHAAAAAAA!

Ian you really honestly, and truly believe that adding a heat sink alone (no other work added) to a hot object will make that object get hotter??? Seriously???

Ian I think you have some fundamental flaw in your logic... Follow me for a second here..

Remember these in your claimed studies Ian?

1. You cannot win.

2. You cannot break even.

3. You cannot get out of the game.

A british scientist coined the phrases above, can't remember his name off-hand.. Do you know what the mean? they are a very simple way to remember the first 3 laws of thermodynamics.. Notice I said the first 3 laws... Thats important cause there is another one that came along later..

1. means you can't get something for nothing. conservation of energy...

2. Means in this instance heat cannot flow from colder body to a warmer body on its own. Or energy exhibits entropy... meaning it flows away from its source..

3. means absolute zero (molecules cease to move) cannot be reached.

and the newest one..pay very close attention here..

4. The Zeroth law.. Its an observation where when two objects are separately in thermodynamic equilibrium with a third object, they are in equilibrium with each other. basically how thermometers work..

Now your contention breaks 1, 2 , and 4 ...Like it or not, believe it or not its how it is regardless.. So your argument would require first that energy out would have to be greater than energy in, second that heat would have to suddenly ignore its entropic nature, and third it would have to ignore equilibrium...

 

[IanC]

HAAHAHAHAHAHAHAHAHAHAHAHAAHAHAHAAAAAAA!

Ian you really honestly, and truly believe that adding a heat sink alone (no other work added) to a hot object will make that object get hotter??? Seriously???

Ian I think you have some fundamental flaw in your logic... Follow me for a second here..

Remember these in your claimed studies Ian?

1. You cannot win.

2. You cannot break even.

3. You cannot get out of the game.

A british scientist coined the phrases above, can't remember his name off- hand.. Do you know what the mean? they are a very simple way to remember the first 3 laws of thermodynamics.. Notice I said the first 3 laws... Thats important cause there is another one that came along later..

1. means you can't get something for nothing. conservation of energy...

2. Means in this instance heat cannot flow from colder body to a warmer body on its own. Or energy exhibits entropy... meaning it flows away from its source..

3. means absolute zero (molecules cease to move) cannot be reached.

and the newest one..pay very close attention here..

4. The Zeroth law.. Its an observation where when two objects are separately in thermodynamic equilibrium with a third object, they are in equilibrium with each other. basically how thermometers work..

Now your contention breaks 1, 2 , and 4 ...Like it or not, believe it or not its how it is regardless.. So your argument would require first that energy out would have to be greater than energy in, second that heat would have to suddenly ignore its entropic nature, and third it would have to ignore equilibrium...

wow! a coherent post with a minimum of ad hom. good job.

my explanation doesnt involve any extra heat, just a redistirbution. the same amount of power goes in and the same amount of radiation goes out. lets look at another scenario. a symetrical object has a heater embedded in it but instead of being in the centre it is offset half way to the surface on one of the planes. the escaping energy will be larger on the side that is closer to the surface and less on the other side. hopefully you can agree with that.

next, cut the object in half in the same plane that the the heater was offset but keep the two pieces close but not touching. is there now a difference in heat flow? yes the heat flow by conduction is stopped and the heated side will increase its heat flow by that much and the non-heated side will decrease its heat flow by that much. after equilibrium has been reached we are now at the two body state in Spencer's and wire bender's examples, although coming from a different direction.

what is the temperature of the exposed sides of the heated block (plate, hemishere, etc)? I dont know exactly but I do know that there is more heat radiating from the exposed sides than is flowing through the previously cut side. when we remove the unheated half what happens? all the sides have open access to the cold vacuum and radiate equally. obviously the power of the heater divided by six sides is less than (five sides and one side with less than unity heat flow). therefore the single heated piece is colder than when there were two pieces in the vacuum.

 

[wirebender]

this is the thought experiment that wirebender and gslack want me to respond to. so I will

Dishonesty right off the bat Ian. You are the one who brought spencer's failed thought experiment into the discussion

I am assuming that the chamber is cooled to some specific constant temperature

Assuming? You brought the thought experiment into the discussion, are you saying now that you don't even know the parameters spencer specified?

the heater inside the plate is hotter than the outside of the plate because of the thermal impedance of the material of the plate.

That makes absolutely no difference Ian. The temperature of the heated bar is determined by the electricity going into the heating element and the surface area of the bar. That's it. You might get a different temperature if you use different materials but that isn't part of the experiment. The bar reaches its maximum temperature based on the amount of power going in and the amount of surface area available to radiate off the heat.

the outside plate temp is determined on how fast the internal energy can be shed to the vacuum of constant temperature. any change to the power source, thermal impedance of the plate material or temperature of the vacuum will change the equilibrium (changing the shape of the plate would also affect the temp if it changed the surface area).

None of those things change during the course of the experiment.

your definition, live with it

Spoken like a man who doesn't have a clue as to what he is looking at. If there is a problem with the equation, then state it. If there is no problem with the equation, then you have lost the discussion. Simple as that. Clearly, you don't grasp the math and are arguing based on an article of faith.

at the beginning the second plate (if at the same temp as the vacuum) will make little difference. but as the second plate heats up it will start to radiate in all directions, including back at the heated plate.

No, Ian, it will not radiate back to the heated plate. See the second law of thermodynamics. It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.

I have asked before and will ask again, which part of NOT POSSIBLE and WILL NOT is it that you don't grasp. There is nothing in the second law about two way energy flow with a net result. The statement is pretty straight forward. It IS NOT POSSIBLE for heat to flow from a cooler body to a warmer body without work having been done to accomplish the flow. Sorry Ian, but that is a law of nature and it defeats your argument hands down.

this changes the ability of the heated plate to shed energy. the heated plate still wants to get rid of the same amount of heat therefore more heat will be dispersed through the area that can escape to the vacuum. increased heat means increased temperature.

It acts as a heat sink Ian. Do you know what the term means and implies? No more heat is going to go through the heated plate than at the onset of the experiment. The only thing that has happened is that by adding a second bar, you have inserted a heat sink into the equation which increases the radiating surface area which in turn will lower the temperature of the heated bar. Adding a heat sink does not increase the amount of energy being fed into the system. The amount of energy remains constant in the experiment.

this is where your example really goes off the rails, hahahahaha. what sort of gullible idiot would think that both plates would be at the same temperature at equilibrium? hahahahaha. what kind of moron would believe that the areas exposed to the cold vacuum loses heat at the same rate as the areas next to warmer surfaces? could you imbeciles get more things wrong?

Again, you fail to recognize the implications of the parameters that govern the experiment. It is in a vacuum Ian so of course they reach equilibrium, or within a fraction of a degree of equilibrium. Equilibrium would never be reached in the open atmosphere, but in a cooled vaccum, equilibrium would be readily reached. As to different surfaces being different temperatures; again, maybe a fraction of a degree but the bars are in a cooled vacuum.

Either point out a problem with the equations in question Ian, or forfiet the argument because if the equations are correct, then the results of the experiment are as the equations predict. The numbers don't lie Ian while faith will lead you off in all sorts of directions.

 

[wirebender]

HAAHAHAHAHAHAHAHAHAHAHAHAAHAHAHAAAAAAA!

Ian you really honestly, and truly believe that adding a heat sink alone (no other work added) to a hot object will make that object get hotter??? Seriously???

Ian I think you have some fundamental flaw in your logic... Follow me for a second here..

Remember these in your claimed studies Ian?

1. You cannot win.

2. You cannot break even.

3. You cannot get out of the game.

A british scientist coined the phrases above, can't remember his name off-hand.. Do you know what the mean? they are a very simple way to remember the first 3 laws of thermodynamics.. Notice I said the first 3 laws... Thats important cause there is another one that came along later..

1. means you can't get something for nothing. conservation of energy...

2. Means in this instance heat cannot flow from colder body to a warmer body on its own. Or energy exhibits entropy... meaning it flows away from its source..

3. means absolute zero (molecules cease to move) cannot be reached.

and the newest one..pay very close attention here..

4. The Zeroth law.. Its an observation where when two objects are separately in thermodynamic equilibrium with a third object, they are in equilibrium with each other. basically how thermometers work..

Now your contention breaks 1, 2 , and 4 ...Like it or not, believe it or not its how it is regardless.. So your argument would require first that energy out would have to be greater than energy in, second that heat would have to suddenly ignore its entropic nature, and third it would have to ignore equilibrium...

What amazes me is his complete disregard for the laws of physics. Faith is a hell of a thing, isn't it? I can easily imagine his eyes sparkling with a zealot's zeal as he thinks up even more convoluted thought experiments that also can not break the laws of physics.

 

[IanC]

this is the thought experiment that wirebender and gslack want me to respond to. so I will

Dishonesty right off the bat Ian. You are the one who brought spencer's failed thought experiment into the discussion

I brought up Spencer's, you brought up an example from [Gord] and then I was accused of ducking the issue of the plates

I am assuming that the chamber is cooled to some specific constant temperature

Assuming? You brought the thought experiment into the discussion, are you saying now that you don't even know the parameters spencer specified?

your experiment did not define the temperature size or shape of the containment vessel

That makes absolutely no difference Ian. The temperature of the heated bar is determined by the electricity going into the heating element and the surface area of the bar. That's it. You might get a different temperature if you use different materials but that isn't part of the experiment. The bar reaches its maximum temperature based on the amount of power going in and the amount of surface area available to radiate off the heat.

are you talking about Spencer's experiment because I was specifically asked by gslack to comment on yours

None of those things change during the course of the experiment.




Spoken like a man who doesn't have a clue as to what he is looking at. If there is a problem with the equation, then state it. If there is no problem with the equation, then you have lost the discussion. Simple as that. Clearly, you don't grasp the math and are arguing based on an article of faith.



No, Ian, it will not radiate back to the heated plate. See the second law of thermodynamics. It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.

every object warmer than 0K radiates. the direction of heat flow is determined by the difference of temperature between the two objects being considered. why do you deny these basic physics facts?

I have asked before and will ask again, which part of NOT POSSIBLE and WILL NOT is it that you don't grasp. There is nothing in the second law about two way energy flow with a net result. The statement is pretty straight forward. It IS NOT POSSIBLE for heat to flow from a cooler body to a warmer body without work having been done to accomplish the flow. Sorry Ian, but that is a law of nature and it defeats your argument hands down.

next you will be telling us that objects stop radiating if they are next to something the same temperature. you are a loon

this changes the ability of the heated plate to shed energy. the heated plate still wants to get rid of the same amount of heat therefore more heat will be dispersed through the area that can escape to the vacuum. increased heat means increased temperature.

It acts as a heat sink Ian. Do you know what the term means and implies? No more heat is going to go through the heated plate than at the onset of the experiment. The only thing that has happened is that by adding a second bar, you have inserted a heat sink into the equation which increases the radiating surface area which in turn will lower the temperature of the heated bar. Adding a heat sink does not increase the amount of energy being fed into the system. The amount of energy remains constant in the experiment.

the same amount of heat is being shed from the heater, just as I have been saying all along. what part of 'heat flow depends on the difference in temperature' dont you understand?

this is where your example really goes off the rails, hahahahaha. what sort of gullible idiot would think that both plates would be at the same temperature at equilibrium? hahahahaha. what kind of moron would believe that the areas exposed to the cold vacuum loses heat at the same rate as the areas next to warmer surfaces? could you imbeciles get more things wrong?

Again, you fail to recognize the implications of the parameters that govern the experiment. It is in a vacuum Ian so of course they reach equilibrium, or within a fraction of a degree of equilibrium. Equilibrium would never be reached in the open atmosphere, but in a cooled vaccum, equilibrium would be readily reached. As to different surfaces being different temperatures; again, maybe a fraction of a degree but the bars are in a cooled vacuum.

somehow you think a heated plate and an unheated plate in a cold vacuum are going to be within a fraction of a degree with each other? are you high or something??

Either point out a problem with the equations in question Ian, or forfiet the argument because if the equations are correct, then the results of the experiment are as the equations predict. The numbers don't lie Ian while faith will lead you off in all sorts of directions.

your experiment's equations are wrong in many areas. the most startling is that the author thought that the two plates would be the same temperature at equilibrium and that he thinks that heat loss to a cold vacuum is the same as to a less cold object between the heated plate and the vacuum. how could anyone not recognize that the areas facing each other between two close objects is not able to shed radiation as easily as free access to cold vacuum.

next time could you specifically state which experiment you are discussing

 

[gslack]

HAAHAHAHAHAHAHAHAHAHAHAHAAHAHAHAAAAAAA!

Ian you really honestly, and truly believe that adding a heat sink alone (no other work added) to a hot object will make that object get hotter??? Seriously???

Ian I think you have some fundamental flaw in your logic... Follow me for a second here..

Remember these in your claimed studies Ian?

1. You cannot win.

2. You cannot break even.

3. You cannot get out of the game.

A british scientist coined the phrases above, can't remember his name off- hand.. Do you know what the mean? they are a very simple way to remember the first 3 laws of thermodynamics.. Notice I said the first 3 laws... Thats important cause there is another one that came along later..

1. means you can't get something for nothing. conservation of energy...

2. Means in this instance heat cannot flow from colder body to a warmer body on its own. Or energy exhibits entropy... meaning it flows away from its source..

3. means absolute zero (molecules cease to move) cannot be reached.

and the newest one..pay very close attention here..

4. The Zeroth law.. Its an observation where when two objects are separately in thermodynamic equilibrium with a third object, they are in equilibrium with each other. basically how thermometers work..

Now your contention breaks 1, 2 , and 4 ...Like it or not, believe it or not its how it is regardless.. So your argument would require first that energy out would have to be greater than energy in, second that heat would have to suddenly ignore its entropic nature, and third it would have to ignore equilibrium...

wow! a coherent post with a minimum of ad hom. good job.

my explanation doesnt involve any extra heat, just a redistirbution. the same amount of power goes in and the same amount of radiation goes out. lets look at another scenario. a symetrical object has a heater embedded in it but instead of being in the centre it is offset half way to the surface on one of the planes. the escaping energy will be larger on the side that is closer to the surface and less on the other side. hopefully you can agree with that.

next, cut the object in half in the same plane that the the heater was offset but keep the two pieces close but not touching. is there now a difference in heat flow? yes the heat flow by conduction is stopped and the heated side will increase its heat flow by that much and the non-heated side will decrease its heat flow by that much. after equilibrium has been reached we are now at the two body state in Spencer's and wire bender's examples, although coming from a different direction.

what is the temperature of the exposed sides of the heated block (plate, hemishere, etc)? I dont know exactly but I do know that there is more heat radiating from the exposed sides than is flowing through the previously cut side. when we remove the unheated half what happens? all the sides have open access to the cold vacuum and radiate equally. obviously the power of the heater divided by six sides is less than (five sides and one side with less than unity heat flow). therefore the single heated piece is colder than when there were two pieces in the vacuum.

Ian, that is perhaps the single best example of Bullshit walking I have seen.. Seriously you couldn't defend the concepts and claims in the first experiment so you make up another one that makes absolutely no attempt to address anything in the first experiment....WTH?

What kind of posturing, preening wannabe peacock have you turned into? What in the hell does that experiment show other than heat transfer is stronger the closer an object is to its heat source... Seriously you just took a shit load of time to come up with a few paragrapsh that not only does not address the problems in the first experiment, but do nothing but affirm that which we already know about heat transfer relative to the distance from its heat source..

WOW! You really are pulling shit out of your ass and shouting Eureka!

 

[gslack]

this is the thought experiment that wirebender and gslack want me to respond to. so I will

Dishonesty right off the bat Ian. You are the one who brought spencer's failed thought experiment into the discussion



Assuming? You brought the thought experiment into the discussion, are you saying now that you don't even know the parameters spencer specified?



That makes absolutely no difference Ian. The temperature of the heated bar is determined by the electricity going into the heating element and the surface area of the bar. That's it. You might get a different temperature if you use different materials but that isn't part of the experiment. The bar reaches its maximum temperature based on the amount of power going in and the amount of surface area available to radiate off the heat.



None of those things change during the course of the experiment.




Spoken like a man who doesn't have a clue as to what he is looking at. If there is a problem with the equation, then state it. If there is no problem with the equation, then you have lost the discussion. Simple as that. Clearly, you don't grasp the math and are arguing based on an article of faith.



No, Ian, it will not radiate back to the heated plate. See the second law of thermodynamics. It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.


I have asked before and will ask again, which part of NOT POSSIBLE and WILL NOT is it that you don't grasp. There is nothing in the second law about two way energy flow with a net result. The statement is pretty straight forward. It IS NOT POSSIBLE for heat to flow from a cooler body to a warmer body without work having been done to accomplish the flow. Sorry Ian, but that is a law of nature and it defeats your argument hands down.



It acts as a heat sink Ian. Do you know what the term means and implies? No more heat is going to go through the heated plate than at the onset of the experiment. The only thing that has happened is that by adding a second bar, you have inserted a heat sink into the equation which increases the radiating surface area which in turn will lower the temperature of the heated bar. Adding a heat sink does not increase the amount of energy being fed into the system. The amount of energy remains constant in the experiment.



Again, you fail to recognize the implications of the parameters that govern the experiment. It is in a vacuum Ian so of course they reach equilibrium, or within a fraction of a degree of equilibrium. Equilibrium would never be reached in the open atmosphere, but in a cooled vaccum, equilibrium would be readily reached. As to different surfaces being different temperatures; again, maybe a fraction of a degree but the bars are in a cooled vacuum.

Either point out a problem with the equations in question Ian, or forfiet the argument because if the equations are correct, then the results of the experiment are as the equations predict. The numbers don't lie Ian while faith will lead you off in all sorts of directions.

your experiment's equations are wrong in many areas. the most startling is that the author thought that the two plates would be the same temperature at equilibrium and that he thinks that heat loss to a cold vacuum is the same as to a less cold object between the heated plate and the vacuum. how could anyone not recognize that the areas facing each other between two close objects is not able to shed radiation as easily as free access to cold vacuum.

next time could you specifically state which experiment you are discussing

Playing obtuse again Ian? Shameless... The experiment has been the one you brought into the discussion from the start, you brought in another experiment to try and bolster the previous one which did nothing but state the obvious and uncontested in this discussion... in BOTH cases the experiments where brought in by you..

Deliberate obfuscation makes you look like a desperate fake trying to save face...

 

[IanC]

gslack- why did you edit out almost all of my answers to wirebender in that last quote? are you taking dishonesty tips from the SkepticalScience blog?

 

[wirebender]

I brought up Spencer's, you brought up an example from [Gord] and then I was accused of ducking the issue of the plates

It is clear by now that you don't grasp the math even though it is little more than algebra but DAMN MAN, get with the program. The math, courtesy of Gord, is addressing spencer's experiment, and yes, you ducked the issue of the plates. It is abundantly clear now why you ducked.

your experiment did not define the temperature size or shape of the containment vessel

Once more Ian, get with the program. I have no experiment. The math you were presented with was addressing spencer's thought experiment as he defined the parameters. Geez guy, is it all that far over your head? I am embarassed for you again over that stink you made regarding the math that I did here that you couldn't find. You acted as if you could destroy my argument if you could just see the math. Now here you are looking at an experiment and don't even recognize the same experiment when it is stated in mathematical terms. The math regarding spencers experiment is simpler than what I did as it doesn't involve distance or time. Come on guy.

every object warmer than 0K radiates. the direction of heat flow is determined by the difference of temperature between the two objects being considered. why do you deny these basic physics facts?

I understand it perfectly Ian and have been stating exactly that for lo these hundreds of posts. You have just agreed with my statements and apparently fail to comprehend the implications of what you just said.

Every object in spencer's experiment is radiating, even the chamber because he set the parameters at 0 degrees F. The direction of heat flow is away from the heated plate. The heated plate is radiating an EM field and because it is warmer than every other object in the experiment, it determines the direction of propagation and that direction is away from itself. No other object there can radiate energy towards the warmed plate.

Now apply what you just said to the earth and atmosphere and see how wrong you have been all this time. You stated correctly that the direction of heat flow is determined by the difference of temperature between the two objects being considered. The earth is warmer than the atmosphere and therefore the direction of heat flow is away from the earth. The atmosphere can not radiate a fraction of a watt of energy towards the earth because as the 2nd law of thermodynamics states, it IS NOT POSSIBLE for heat to flow from a cool object to a warmer object without some work having been performed to make it happen.

The cool bar can not radiate energy to the warmer bar and the cool atmosphere can not radiate energy to the warmer earth.

Why is it that you can state the obvious but can not make the connection to the real world?

next you will be telling us that objects stop radiating if they are next to something the same temperature. you are a loon

It is called equilibrium Ian and is a well known and long understood phenomenon. Two objects at the same temperature radiate as a single object. The do not warm each other.

the same amount of heat is being shed from the heater, just as I have been saying all along. what part of 'heat flow depends on the difference in temperature' dont you understand?

Of course the same amount of heat is being shed Ian but it is being shed faster because of the heat sink and because of that, its temperature is going to drop. Are you really this far behind the curve?

somehow you think a heated plate and an unheated plate in a cold vacuum are going to be within a fraction of a degree with each other? are you high or something??

Not high Ian, just doing the math. The math was done for you but clearly it is over your head so you don't grasp it. Hell, you didn't even realize that the math you were given was a mathematical statement about your damned experiment.

The bars are in a vacuum Ian, of course they will reach equilibrium or very close to it. As I so patiently explained to you before, there is no conduction or convection allowed in the experiment because the bars are in a vacuum. Conduction and convection are far and away more efficient means of transporting heat but neither are available for the bars. They must radiate into the vacuum and as a result, will reach equilibrium quite quickly.

Of course they could never achieve equilibrium in the open atmosphere because of convection and conduction. Do you believe that the fins on your heat sink ever even approaches the temperature of the processor in your computer? Do you think that the radiator in your car ever approaches the temperature of your engine block? It is because of convection and conduction. If you put your processor and its heat sink into a vacuum, they will achieve equilibrium or very close to it.

I don't know if you could find the info today, but there used to be volumes on the topic of vacuum tubes for electronics and one of the major problems with them was heat and the fact that everying inside the tube was operating at, or very near the same temperature.

your experiment's equations are wrong in many areas. the most startling is that the author thought that the two plates would be the same temperature at equilibrium and that he thinks that heat loss to a cold vacuum is the same as to a less cold object between the heated plate and the vacuum. how could anyone not recognize that the areas facing each other between two close objects is not able to shed radiation as easily as free access to cold vacuum.

So here is where the rubber meets the road. State specifically what you believe is wrong with the equations. They are standard derivations of the Stefan-Boltzman Law for blackbody radiation. They are used hundreds of thousands of times per day by engineers of all sorts who have to deal with transporting heat away from whatever it is that they are working on. Those equations are used for everyting from determining what sort of heat sink is necessary to keep your computer processor cool to the size and shape of the radiator in your car.

If you can't state what is wrong with the equations (here is a hint, nothing is wrong with the equations) you have effectively lost the argument. I couldn't help but note that even dr spencer didn't try to argue with those equations. The fact that dr spencer didn't catch on to the fact that he was merely adding a heat sink to his heated bar and understand the consequences of that says a great deal about the way an academic's mind works vs that of an engineer who actually has to use the math in the real world and be able to predict whtih high accuracy what is going to happen, temperature wise, when a million dollars is spent to produce his design. If those formulas didn't work, nothing would ever be built.

So tell me Ian, what is wrong with the math?

next time could you specifically state which experiment you are discussing

And one final time Ian, get with the program. This whole thing has been only about spencer's failed thought experiment. The fact that you didn't recognize the math as a statement about the experiment that spencer proposed exposes you as a fraud Ian. Perhaps it would be best to simply bow out now and avoid further public humiliation. The fact that it went that far over your head says all that need be said about your scientific "skills".

 

[wirebender]

gslack- why did you edit out almost all of my answers to wirebender in that last quote? are you taking dishonesty tips from the SkepticalScience blog?

For the simple reason that you didn't say anything Ian. Flailing about and grasping for straws and generally demonstrating that the topic is over your head doesn't constitute any sort of an argument at all.

 

[IanC]

wirebender- this is what you posted

Re: Vacuum Chamber with plates.

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan’s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts

(***That’s ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)

The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2
——–
If another identical “non-heated” and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:

The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.

The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.

The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

And the equilibrium temperature for both plates will be 284.69 K or 53 deg F.

is it a different experiment by [Gord] or just his description of Spencer's example? you didnt specify.

it doesnt really matter because Spencer's description is correct and your copied description is incorrect.

you still havent described the mechanism by which you think both plates (or bars) will come to equilibrium at the same temperature even though one is heated and the other is not.

you still havent explained why you think the heat loss in the direction of the second bar is the same as the heat loss in the direction of the cooled vacuum, even though there is an obvious difference in temperatures.

the math in your example is not only wrong but farfetched. no wonder you refuse to bump your post where you 'proved that photons disappear in space, and did the math'. obviously it was just as infintile and incorrect as your 'proof' that Spencer was wrong.

 

[wirebender]

is it a different experiment by [Gord] or just his description of Spencer's example? you didnt specify.

Sorry Ian, this is just to far over your head. If you can't look at the math and determine which damned experiment it relates to, you really don't have anything to add to the discussion. You have become rocks. Congratulations.

you still havent described the mechanism by which you think both plates (or bars) will come to equilibrium at the same temperature even though one is heated and the other is not.

I have described it in detail Ian. Sorry it is so far over your head. I can't help you any further.

you still havent explained why you think the heat loss in the direction of the second bar is the same as the heat loss in the direction of the cooled vacuum, even though there is an obvious difference in temperatures.

I have described that in detail as well Ian. Once again, sorry this is so far over your head. You had me fooled at one time into thinking that you had a clue. It is obvious now that you don't.

the math in your example is not only wrong but farfetched.

The Stefan-Boltzman law dealing with black body radiation is far fetched? Interesting Ian. You should run along now. At this point all you can do is further embarass yourself. Maybe rocks will come along and give you some sugar though, or wave his pom poms.

no wonder you refuse to bump your post where you 'proved that photons disappear in space, and did the math'. obviously it was just as infintile and incorrect as your 'proof' that Spencer was wrong.

I am laughing at you Ian.

By the way, it has also become abundantly and undeniably clear that you don't have the slightest notion of what a photon is and don't have even an inkling of how particle wave duality might relate to the topic at hand.

I am sorry for you Ian. At one time, you were viewed as a respected, reasonably intelligent member of this board. You have been amazingly successful at completely tearing that reputation to ribbons.

 

[gslack]

gslack- why did you edit out almost all of my answers to wirebender in that last quote? are you taking dishonesty tips from the SkepticalScience blog?

Ian I did not edit out anything in your post.. The board software does that to reserve space in quoted posts. It always does this when you quote from a post with more and a few quoted sections in it..

For example your post I responded to, had multiple quoted lines in it. I am not sure the cut off number in the software as I am not a site admin, but in cases where the post has multiple quotes in it, even if they are just a few lines each, it will automatically cut off some in any reply with quotes to that post.

If you don't like it, join the club I have complained about it before, but seeing as i don't run the site or pay for the bandwidth I really have no grounds to dictate policy. Its happened to me numerous times for even fewer quoted lines...

if it bothers you can copy the entire posts text and paste into the quick reply box at the bottom of the page. The board will assume it is a new post and not edit it. Well as long as the admins don't see it as a repeated post..

Again, editing posts is something I do not do.. Matter of fact if you look you can see I have take offense when it is done to me by many posters here in the past.

 

[IanC]

wirebender said-

Now apply what you just said to the earth and atmosphere and see how wrong you have been all this time. You stated correctly that the direction of heat flow is determined by the difference of temperature between the two objects being considered. The earth is warmer than the atmosphere and therefore the direction of heat flow is away from the earth. The atmosphere can not radiate a fraction of a watt of energy towards the earth because as the 2nd law of thermodynamics states, it IS NOT POSSIBLE for heat to flow from a cool object to a warmer object without some work having been performed to make it happen.

The cool bar can not radiate energy to the warmer bar and the cool atmosphere can not radiate energy to the warmer earth.

Why is it that you can state the obvious but can not make the connection to the real world?


Quote: Originally Posted by IanC
next you will be telling us that objects stop radiating if they are next to something the same temperature. you are a loon
It is called equilibrium Ian and is a well known and long understood phenomenon. Two objects at the same temperature radiate as a single object. The do not warm each other.

this is the crux of the problem. wirebender does not understand the difference between radiation and heat flow. radiation is a function of the temperature of the radiating object (in very simple terms) and heat flow is the measurement of the size and direction of net radiation.

two objects of the same temperature both radiate at the other but the radiation of one replaces the radiation of the other and therefore there is no net heat transfer. if there is a difference in temperature between the two objects the warmer one radiates more and is slightly cooled while the cooler object still radiates towards the warmer one but not enough to fully replace the warmer object's radiation.

an object releases the same amount of radiation whether it is in cold outer space or next to a blast furnace. the heat flow is simply a matter of whether the incoming radiation is lesser or greater than the outgoing radiation. there is no magical disappearance of photons, they are always there and it is only the net amount of photons in a direction that counts.

 

[gslack]

wirebender- this is what you posted

Re: Vacuum Chamber with plates.

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan’s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts

(***That’s ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)

The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2
——–
If another identical “non-heated” and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:

The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.

The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.

The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

And the equilibrium temperature for both plates will be 284.69 K or 53 deg F.

is it a different experiment by [Gord] or just his description of Spencer's example? you didnt specify.

it doesnt really matter because Spencer's description is correct and your copied description is incorrect.

you still havent described the mechanism by which you think both plates (or bars) will come to equilibrium at the same temperature even though one is heated and the other is not.

you still havent explained why you think the heat loss in the direction of the second bar is the same as the heat loss in the direction of the cooled vacuum, even though there is an obvious difference in temperatures.

the math in your example is not only wrong but farfetched. no wonder you refuse to bump your post where you 'proved that photons disappear in space, and did the math'. obviously it was just as infintile and incorrect as your 'proof' that Spencer was wrong.

I just want to address a few things directly...

"you still havent described the mechanism by which you think both plates (or bars) will come to equilibrium at the same temperature even though one is heated and the other is not."

Its due to the laws of thermodynamics Ian... Particularly the last one I mentioned in a previous post.. The zeroth law... I suggest you read it again..

here is a better explanation than I gave I believe...

Zeroth Law, Thermal Equilibrium and Temperature

Zeroth Law

The zeroth law is a consequence of thermal equilibrium and allows us to conclude that temperature is a well-defined physical quantity. The zeroth law of thermodynamics states:

If a body A and a body B are both in equilibrium with each other; then a body C which is in thermal equilibrium with body B will also be in equilibrium with body Aand the temperature of body C is equal to the temperature of body A.
It is the zeroth law, because it preceeds the first and second laws of thermodynamics and is also a tacit assumption in both laws.

We use the zeroth law when we wish to compare the temperatures of two objects, A and B. We can do this by using a thermometer, C and placing it again object A it reaches thermal equilibrium with object A and measure the temperature of A. Placing the thermometer against object B until thermal equilibrium is reached we measure the temperature of object B. If they are the same temperature then they will be in thermal equilibrium with each other.

Now if you are somehow changing the distance of the plates from one another that would be effected by that new distance. I was under the assumption the plates were placed on top of one another. This was implied throughout the experiment I believe. if they are indeed in contact with one another the outside temps only effect the outside surface areas of the plates. Of course the closer to the heat source the warmer it will be no one made any claim to the contrary. however, the assumed term "all things being equal" will apply here.

The entire experiment is a thought experiment. The physical reality is its not even a good example of how a real physical experiment would be. First there is always a difference in the temperature of any heated surface the farther it gets from its point of contact with the heat source. Second standing plates are not a moving atmosphere with all the variables it entails. Third, the vacuum is only applicable to the relation of the Sun to the atmospheric surface. Once it hits that atmosphere the vacuum is no longer a factor. The earth sits in a vacuum, but the atmosphere is not a vacuum, and that is the part that is in contention in greenhouse effect theory. That would make the entire premise of the experiment pointless.

However you brought this experiment into the discussion and professed its accuracy. So we tried to show how it was not an accurate example of how the "greenhouse effect" works. You vehemently defended it and still do so now. This experiment forces many of the natural finer points and details to be ignored.

Now if you want us to ignore the finer points and details and allow this experiment to go through to its claimed result, you cannot very well fall back on those finer points to defend it when it falls flat.. The experiment assumes "all things being equal" from the start, that is the only way it can work even in a thought experiment. So either all things are indeed equal in it or all things are as they would be in the real world, either way it falls flat. But you must stick to one or the other, switching between them only confounds the thing and makes it impossible..

 

[IanC]

this is the thought experiment that wirebender and gslack want me to respond to. so I will

Dishonesty right off the bat Ian. You are the one who brought spencer's failed thought experiment into the discussion



Assuming? You brought the thought experiment into the discussion, are you saying now that you don't even know the parameters spencer specified?



That makes absolutely no difference Ian. The temperature of the heated bar is determined by the electricity going into the heating element and the surface area of the bar. That's it. You might get a different temperature if you use different materials but that isn't part of the experiment. The bar reaches its maximum temperature based on the amount of power going in and the amount of surface area available to radiate off the heat.



None of those things change during the course of the experiment.




Spoken like a man who doesn't have a clue as to what he is looking at. If there is a problem with the equation, then state it. If there is no problem with the equation, then you have lost the discussion. Simple as that. Clearly, you don't grasp the math and are arguing based on an article of faith.



No, Ian, it will not radiate back to the heated plate. See the second law of thermodynamics. It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.


I have asked before and will ask again, which part of NOT POSSIBLE and WILL NOT is it that you don't grasp. There is nothing in the second law about two way energy flow with a net result. The statement is pretty straight forward. It IS NOT POSSIBLE for heat to flow from a cooler body to a warmer body without work having been done to accomplish the flow. Sorry Ian, but that is a law of nature and it defeats your argument hands down.



It acts as a heat sink Ian. Do you know what the term means and implies? No more heat is going to go through the heated plate than at the onset of the experiment. The only thing that has happened is that by adding a second bar, you have inserted a heat sink into the equation which increases the radiating surface area which in turn will lower the temperature of the heated bar. Adding a heat sink does not increase the amount of energy being fed into the system. The amount of energy remains constant in the experiment.



Again, you fail to recognize the implications of the parameters that govern the experiment. It is in a vacuum Ian so of course they reach equilibrium, or within a fraction of a degree of equilibrium. Equilibrium would never be reached in the open atmosphere, but in a cooled vaccum, equilibrium would be readily reached. As to different surfaces being different temperatures; again, maybe a fraction of a degree but the bars are in a cooled vacuum.

Either point out a problem with the equations in question Ian, or forfiet the argument because if the equations are correct, then the results of the experiment are as the equations predict. The numbers don't lie Ian while faith will lead you off in all sorts of directions.

your experiment's equations are wrong in many areas. the most startling is that the author thought that the two plates would be the same temperature at equilibrium and that he thinks that heat loss to a cold vacuum is the same as to a less cold object between the heated plate and the vacuum. how could anyone not recognize that the areas facing each other between two close objects is not able to shed radiation as easily as free access to cold vacuum.

next time could you specifically state which experiment you are discussing

test

my apologies. I did not realize the message board did that.