No Evidence

[Wuwei]

Yeah..I heard your model ad nauseam...the observable, measurable, repeatable evidence proves your understanding wrong

Proves which step is wrong?

 

[polarbear]

Plug in the data for the particular cube if you're interested in finding how much power the ice cube radiates.

This is a help for our simple friends here who are science challenged.
Stefan Boltzmann Law Calculator - Omni

He will see that ice emits about 300 W/m²

Okay, so it emits 300 W/m^2 but why is it not possible to let it do that for 3600 seconds and use the ~ 25 800 calories that came from the ice to boil 25 liters of water ?

 

[Toddsterpatriot]

They traveled in all directions, without regard to the temperature of the target.

and did what exactly? did it warm anything? And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

and did what exactly?

Was absorbed.

did it warm anything?

Depends.

And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

View attachment 232661

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates.

So no observed evidence!

he doesn't dare do the experiment himself...the results would challenge his beliefs and we can't have that.

What’s funny is my coffee never gets hotter in the cup. I’ve put ice near the cup all kinds of furniture and the coffee always cools to room temperature.

What’s funny is my coffee never gets hotter in the cup.

Stop putting your coffee in places where it radiates more power than it absorbs.

 

[Wuwei]

Okay, so it emits 300 W/m^2 but why is it not possible to let it do that for 3600 seconds and use the ~ 25 800 calories that came from the ice to boil 25 liters of water ?

If the nearby water is above 0 C it will be radiating more energy to the ice than it is absorbing from the ice.

Edit. Tod beat me to it.

 

[polarbear]

Okay, so it emits 300 W/m^2 but why is it not possible to let it do that for 3600 seconds and use the ~ 25 800 calories that came from the ice to boil 25 liters of water ?

If the nearby water is above 0 C it will be radiating more energy to the ice than it is absorbing from the ice.

Edit. Tod beat me to it.

Yeah I know that, but why don`t they have that problem in a computer model ?
They don`t even have 300 W/m^2. With less than 1/100 th of that they can warm up an entire ocean.
Explain that.
As far as the water not getting any warmer only because it is radiating more energy than it gets from the ice being the problem,... that should not matter if I let you use as many ice square meters as you want to make up for the number of watt seconds the warmer water can loose in 3600 seconds by radiative heat loss.
Use 4 times as many squaremeters of ice and you got 1200 watts. That more than makes up for the ~ 1100 W/m^2 the hotter water would loose in the same time.
Your problem is that you don`t know the difference between usable energy and just plain energy expressed in any of the SI units. And these idiotic computer models do the same thing pretending there is no difference..

 

[Wuwei]

Okay, so it emits 300 W/m^2 but why is it not possible to let it do that for 3600 seconds and use the ~ 25 800 calories that came from the ice to boil 25 liters of water ?

If the nearby water is above 0 C it will be radiating more energy to the ice than it is absorbing from the ice.

Edit. Tod beat me to it.

Yeah I know that, but why don`t they have that problem in a computer model ?
They don`t even have 300 W/m^2. With less than 1/100 th of that they can warm up an entire ocean.
Explain that.

I don't understand your first question. I presume you are referring to an atmospheric model.

It is sunshine that heats the ocean. There is no other incoming source of energy to do that other than some underwater vents or proximity to the hot magma.

 

[jc456]

Nope you said it should while it emits

In this wacky universe, many things happen simultaneously.
So, if the only thing in the system we're studying is ice, yes, ice emitting into the
cold vacuum of space would cool.

In the system you mentioned, the ice at 0C (32F) or below is emitting, but it's also absorbing faster from the 21C (70F) counter and surroundings.

what happens to photons it supposedly emits? does it heat the room?

They traveled in all directions, without regard to the temperature of the target.

and did what exactly? did it warm anything? And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

and did what exactly?

Was absorbed.

did it warm anything?

Depends.

And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

View attachment 232661

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates.

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates

one can't measure it? Observe it?

 

[polarbear]

Okay, so it emits 300 W/m^2 but why is it not possible to let it do that for 3600 seconds and use the ~ 25 800 calories that came from the ice to boil 25 liters of water ?

If the nearby water is above 0 C it will be radiating more energy to the ice than it is absorbing from the ice.

Edit. Tod beat me to it.

Yeah I know that, but why don`t they have that problem in a computer model ?
They don`t even have 300 W/m^2. With less than 1/100 th of that they can warm up an entire ocean.
Explain that.

I don't understand your first question. I presume you are referring to an atmospheric model.

It is sunshine that heats the ocean. There is no other incoming source of energy to do that other than some underwater vents or proximity to the hot magma.

Ooooh I see...the "extra heat" that has been hiding in the ocean came from the sun shine and its only the "extra heat" on dry land that came from the CO2....and why do you need the sun if you can steel heat with CO2 from something colder to warm something which is warmer (as the settled science says it can)
Aaah there is that pesky usable energy problem again, but no matter all it takes is the sun + maybe that 97% consensus of idiots and that problem is "solved"....even though it now took something way hotter to get the usable energy required instead of the 300 W/m^2 unusable energy.

 

[Toddsterpatriot]

In this wacky universe, many things happen simultaneously.
So, if the only thing in the system we're studying is ice, yes, ice emitting into the
cold vacuum of space would cool.

In the system you mentioned, the ice at 0C (32F) or below is emitting, but it's also absorbing faster from the 21C (70F) counter and surroundings.

what happens to photons it supposedly emits? does it heat the room?

They traveled in all directions, without regard to the temperature of the target.

and did what exactly? did it warm anything? And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

and did what exactly?

Was absorbed.

did it warm anything?

Depends.

And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

View attachment 232661

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates.

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates

one can't measure it? Observe it?

If you wanted to confirm the number you get from the formula.....feel free.

 

[Wuwei]

Yeah I know that, but why don`t they have that problem in a computer model ?
They don`t even have 300 W/m^2. With less than 1/100 th of that they can warm up an entire ocean.
Explain that.
As far as the water not getting any warmer only because it is radiating more energy than it gets from the ice being the problem,... that should not matter if I let you use as many ice square meters as you want to make up for the number of watt seconds the warmer water can loose in 3600 seconds by radiative heat loss.
Use 4 times as many squaremeters of ice and you got 1200 watts. That more than makes up for the ~ 1100 W/m^2 the hotter water would loose in the same time.
Your problem is that you don`t know the difference between usable energy and just plain energy expressed in any of the SI units. And these idiotic computer models do the same thing pretending there is no difference..

Suppose you completely surround a spherical coffee cup with ice in very close proximity, the coffee cup surface area is roughly the same size as the ice surface. The radiation densities are obvious from the BB law.

If you want to increase the surface area exposed by the ice, you have to make a much bigger ice sphere around the cup so the cup will be exposed to all that surface of ice. The ice surface area goes up by the radius squared, but the radiation energy from that ice drops by the inverse square law, so you are back to where you started from as far as total wattage of ice radiation hitting the cup.

In short, the more ice surface you try to use, the further away it has to be. There is a maximum to the flux density from radiation. This is one application of Liouville's theorem.

 

[Wuwei]

Ooooh I see...the "extra heat" that has been hiding in the ocean came from the sun shine and its only the "extra heat" on dry land that came from the CO2....and why do you need the sun if you can steel heat with CO2 from something colder to warm something which is warmer (as the settled science says it can)
Aaah there is that pesky usable energy problem again, but no matter all it takes is the sun + maybe that 97% consensus of idiots and that problem is "solved"....even though it now took something way hotter to get the usable energy required instead of the 300 W/m^2 unusable energy.

I know you are trying to be sarcastic, but even then your rambling doesn't make sense. All straw-men you are jeering about do not apply to anything I believe.

 

[polarbear]

Yeah I know that, but why don`t they have that problem in a computer model ?
They don`t even have 300 W/m^2. With less than 1/100 th of that they can warm up an entire ocean.
Explain that.
As far as the water not getting any warmer only because it is radiating more energy than it gets from the ice being the problem,... that should not matter if I let you use as many ice square meters as you want to make up for the number of watt seconds the warmer water can loose in 3600 seconds by radiative heat loss.
Use 4 times as many squaremeters of ice and you got 1200 watts. That more than makes up for the ~ 1100 W/m^2 the hotter water would loose in the same time.
Your problem is that you don`t know the difference between usable energy and just plain energy expressed in any of the SI units. And these idiotic computer models do the same thing pretending there is no difference..

Suppose you completely surround a spherical coffee cup with ice in very close proximity, the coffee cup surface area is roughly the same size as the ice surface. The radiation densities are obvious from the BB law.

If you want to increase the surface area exposed by the ice, you have to make a much bigger ice sphere around the cup so the cup will be exposed to all that surface of ice. The ice surface area goes up by the radius squared, but the radiation energy from that ice drops by the inverse square law, so you are back to where you started from as far as total wattage of ice radiation hitting the cup.

In short, the more ice surface you try to use, the further away it has to be. There is a maximum to the flux density from radiation. This is one application of Liouville's theorem.

I am not so stingy as to limit you to the confines of a "spherical coffee cup" ice igloo inner surface.
You can use a 1 by 1 meter ice cube which has 6 sides and all the gold plated mirrors and optics it takes to make sure none of your 1800 watts worth of ice cube photons go astray. I`ll even let you use charcoal water to absorb as much as possible. In addition to that you will only have to heat a little 2 cm diameter capsule of water with these photons. So don`t give me that crap "you have to make a much bigger ice sphere....but the radiation energy from that ice drops by the inverse square law so you are back to where you started from as far as total wattage of ice radiation hitting the cup." Next thing you tell me it`s not possible for any optics & reflectors to redirect the photons emitted by the 6 sides of that cube towards the location of a much smaller absorber.....or anything else you can cling to, other than admitting that the photons you need to heat warm water are not a usable energy if they come from a cooler object.
That whole CO2 "settled science" crumbles without the sun as an energy source. Of course CO2 absorbs IR but it also does that to the same extent to incoming IR as it can to outgoing.
Which is the point when you are the deniers and insist that a colder object can heat a warmer one without any additional IR from somewhere else and claim the "how" is all laid out in the StB equation. If that were so then why was there a need for you to try & swindle your way out of that problem with your limited ingenuity to overcome the m^2 obstacle you invented for me.

 

[jc456]

what happens to photons it supposedly emits? does it heat the room?

They traveled in all directions, without regard to the temperature of the target.

and did what exactly? did it warm anything? And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

and did what exactly?

Was absorbed.

did it warm anything?

Depends.

And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

View attachment 232661

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates.

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates

one can't measure it? Observe it?

If you wanted to confirm the number you get from the formula.....feel free.

I know I won't confirm the formula, because there won't be any reading on a test set. Cause no one has shown that one can measure the output of an ice cube.

 

[Toddsterpatriot]

They traveled in all directions, without regard to the temperature of the target.

and did what exactly? did it warm anything? And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

and did what exactly?

Was absorbed.

did it warm anything?

Depends.

And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

View attachment 232661

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates.

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates

one can't measure it? Observe it?

If you wanted to confirm the number you get from the formula.....feel free.

I know I won't confirm the formula, because there won't be any reading on a test set. Cause no one has shown that one can measure the output of an ice cube.

I know I won't confirm the formula, because there won't be any reading on a test set.

Radiation from an ice cube is somehow undetectable? Or is it non-existent?

Cause no one has shown that one can measure the output of an ice cube.

Link?

 

[Wuwei]

the photons you need to heat warm water are not a usable energy if they come from a cooler object.

Of course ice cube energy is not usable with only hot objects around. It violates the 2nd law. However it still radiates around 300 W/m². What is your point?

 

[polarbear]

the photons you need to heat warm water are not a usable energy if they come from a cooler object.

Of course ice cube energy is not usable with only hot objects around. It violates the 2nd law. However it still radiates around 300 W/m². What is your point?

Of course ice cube energy is not usable with only hot objects around.
My point is that it has been used by you and your buddies over and over again to do just that.
Even when somebody points out that observatories find it necessary to cool their heat sensors in order to detect low wattage IR sources that radiate less than the detector would if it were not cooled to the point where the incoming IR is a usable energy.
That`s when some other idiots who have no clue start yapping about IR cameras at room temperature which need the usable energy from an electrical source to sufficiently intensify the signal coming from lower temperature objects. Its the solid state version of a photo multiplier just not nearly as sensitive.
None of that technology would be necessary if the arguments made by the "science" that can conjure up usable energy from anything that is above zero Kelvin were true.

 

[sealybobo]

1. There is not a single piece of observed, measured evidence which supports the AGW hypothesis over natural variability.

2. There is not a single piece of observed, measured evidence which establishes a coherent link between the absorption of infrared radiation by a gas and warming in the atmosphere.

3. The hypothesized warming due to mankind's burins of hydrocarbon fuels, which is the foundation of the anthropogenic global warming hypothesis has never been empirically measured, quantified, and then attributed to so called green house gasses.


I have been asking for just a single piece of observed, measured evidence which supports the claims of climate alarmists for decades now and have never received the first piece.

I see alarmists claiming that such evidence exists all the time...sometimes they even post what passes for evidence in their minds like THIS. There is certainly observed, and measured data there, but none of it supports the AGW hypothesis over natural variability, and none of it even begins to establish a coherent relationship between the absorption of infrared radiation by a gas and warming in the atmosphere...those things are certainly assumed in the example linked to, but there certainly is no evidence to support the assumption. And there is no paper there in which the hypothesized warming due to our production of CO2 has been empirically measured, quantified, and then attributed to so called greenhouse gasses. Again, it is assumed, but assumptions based on lose correlation over a very short period of geological time are less than worthless in any scientific examination of an entity as large, variable, and chaotic as the global climate.

So there you go...I have stuck my chin out...I have made 3 very deliberate, and concise statements regarding the state of climate science and the evidence that mankind is having an effect on the global climate.

It is the complete absence of evidence challenging the 3 statements above that explain why I am a skeptic.

Prove me wrong. Don't tell me about the evidence that exists......don't tell me about the evidence you might believe you have produced...Step up to the plate and produce the evidence that I have quite clearly declared does not exist...

And when you can't, ask yourself why it is that you believe what you do regarding man made climate change.

Should we believe a Republican or 97% of the scientists out there?

The 97% consensus on global warming

 

[Wuwei]

the photons you need to heat warm water are not a usable energy if they come from a cooler object.

Of course ice cube energy is not usable with only hot objects around. It violates the 2nd law. However it still radiates around 300 W/m². What is your point?

Of course ice cube energy is not usable with only hot objects around.
My point is that it has been used by you and your buddies over and over again to do just that.

Absolutely never. You and your buddies misunderstood the physics.

Hot coffee in a room at room temperature will cool. Hot coffee in the same room but with ice cubes close by will cool slower. Hot coffee surrounded by even more ice cubes will cool even slower. That is the effect of radiating ice cubes.

 

[jc456]

and did what exactly? did it warm anything? And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

and did what exactly?

Was absorbed.

did it warm anything?

Depends.

And again, I asked a few posts back, have any observed data that shows the ice cube radiated?

View attachment 232661

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates.

Plug in the data for the particular cube if you're interested in finding how much power
the ice cube radiates

one can't measure it? Observe it?

If you wanted to confirm the number you get from the formula.....feel free.

I know I won't confirm the formula, because there won't be any reading on a test set. Cause no one has shown that one can measure the output of an ice cube.

I know I won't confirm the formula, because there won't be any reading on a test set.

Radiation from an ice cube is somehow undetectable? Or is it non-existent?

Cause no one has shown that one can measure the output of an ice cube.

Link?

well I have been in here for quite some time and no one has ever shown radiation off of an ice cube. You all tried with a Popsicle once, but that was blown up. So sure, post a link with someone measuring radiation off an ice cube.

Link?
you want me to post a link that doesn't exist? hmmmm you post the one that does exist.

 

[jc456]

the photons you need to heat warm water are not a usable energy if they come from a cooler object.

Of course ice cube energy is not usable with only hot objects around. It violates the 2nd law. However it still radiates around 300 W/m². What is your point?

Of course ice cube energy is not usable with only hot objects around.
My point is that it has been used by you and your buddies over and over again to do just that.

Absolutely never. You and your buddies misunderstood the physics.

Hot coffee in a room at room temperature will cool. Hot coffee in the same room but with ice cubes close by will cool slower. Hot coffee surrounded by even more ice cubes will cool even slower. That is the effect of radiating ice cubes.

no it won't, it will cool faster. the ice cube will cause the heat from the coffee to emit more heat flows to cold. the colder the quicker it cools. you should read up. Darwin's law or something.

I will do the experiment tonight. It is easy to confirm.