2nd law

tmiddles wrote:

IBdaMann wrote:
Scenario: An 8x8x3 meter room ....5cm diameter steel ball .

Not factoring in air in the room and conduction convection, both the ball and the walls will, as they have thermal energy and a temperature, radiate out. The radiance leaving the walls and ball results in a loss of thermal energy.

Both the walls and the ball are also hit by radiance.

For the ball it is hit by some of the radiance from the walls and this is absorbed and converted to thermal energy.

The walls will be hit by radiance from the walls and the ball and absorb that radiance.

The walls will see a net increase in thermal energy as the ball gives more radiant energy than it absorbs. The ball will see a net decrease as it gives more than it gets.

IBdaMann wrote:

tmiddles wrote:...
The walls will see a net increase in thermal energy ... The ball will see a net decrease...

I wanted...
Could you elaborate on the above, focusing on the just the thermal energy? ...

tmiddles wrote:
So IBD are we once again playing the game where you ignore my questions and only ask your own?

Oh and you still haven't responded here: https://www.climate-debate.com/forum/do-i-have-the-co2-calamity-math-right-help-from-an-expert-please-d10-e2720-s160.php#post_52389

tmiddles wrote:Oh and you still haven't responded here:

tmiddles wrote: So the difference in this net radiative transfer

tmiddles wrote:

IBdaMann wrote:
...., water entering the cup has to enter through the hole..

ok, is that important?

IBdaMann wrote:Just imagine the ball, the walls and the thermal energy.

tmiddles wrote:

IBdaMann wrote:Just imagine the ball, the walls and the thermal energy.

So do you also get 0.1206 Watts ?

IBdaMann wrote:

tmiddles wrote:

IBdaMann wrote:Just imagine the ball, the walls and the thermal energy.

So do you also get 0.1206 Watts ?

... of what? ...photons?

tmiddles wrote:That's the rate at which the ball is dropping in thermal energy content and the walls are increasing in thermal energy content.

IBdaMann wrote:
No, I didn't do the calculation but that's really immaterial.

IBdaMann wrote:
You are not discussing...

tmiddles wrote:

A 5cm diameter ball has a surface area of 78.54cm. With a temperature of 42C the ball will lose thermal energy, through radiance, as per Stefan-Boltzmann:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

As the room basically provides an ambient radiance at 18C the absorption by the ball, and conversion to thermal energy, will be using the same surface area.
__________P(in)=σeA(T2^4), 291.15K
0.0000000567 * 1.00 * .007854 * 715668499.46 = +0.3187W

So the difference in this net radiative transfer is a loss of 0.1206 Watts by the ball and a gain of that same amount by the room. That's what happens with the thermal energy.

tmiddles wrote: It is "what happens", it is the resulting change in thermal energy.

tmiddles wrote: Do you find that I did it correctly?

IBdaMann wrote:

tmiddles wrote: Do you find that I did it correctly?

...Supposing I turn it over to you and go with what you computed. Does it somehow change what you describe as having happened?

tmiddles wrote:I show two things are happening:
1- The ball loses thermal energy that is gained by the walls -0.4393W
2- The walls lose thermal energy that is gained by the ball +0.3187W

tmiddles wrote:
A 5cm diameter ball has a surface area of 78.54cm. With a temperature of 42C the ball will lose thermal energy, through radiance, as per Stefan-Boltzmann:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

As the room basically provides an ambient radiance at 18C the absorption by the ball, and conversion to thermal energy, will be using the same surface area.
__________P(in)=σeA(T2^4), 291.15K
0.0000000567 * 1.00 * .007854 * 715668499.46 = +0.3187W

So the difference in this net radiative transfer is a loss of 0.1206 Watts by the ball and a gain of that same amount by the room.

tmiddles wrote:
A 5cm diameter ball has a surface area of 78.54cm. With a temperature of 42C the ball will lose thermal energy, through radiance, as per Stefan-Boltzmann:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

As the room basically provides an ambient radiance at 30C the absorption by the ball, and conversion to thermal energy, will be using the same surface area.
__________P(in)=σeA(T2^4), 303.15K
0.0000000567 * 1.00 * .007854 * 8445595755.5 = +0.3761W

So the difference in this net radiative transfer is a loss of 0.0632 Watts by the ball and a gain of that same amount by the room.

tmiddles wrote:
A 5cm diameter ball has a surface area of 78.54cm. With a temperature of 42C the ball will lose thermal energy, through radiance, as per Stefan-Boltzmann:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

As the room basically provides an ambient radiance at 42C the absorption by the ball, and conversion to thermal energy, will be using the same surface area.
__________P(in)=σeA(T2^4), 315.15K
0.0000000567 * 1.00 * .007854 * 986436754.6 = +0.4393W

So the difference in this net radiative transfer is 0.00 Watts by the ball and a gain of that same amount by the room. There is an equilibrium

tmiddles wrote:
Semantics aside let's look at three scenarios.
A: Room is 18C the ball loses thermal energy to the room at 0.1206 Watts
B: Room is 30C the ball loses thermal energy to the room at 0.0632 Watts
C: Room is 42C there is not change in the thermal energy of the ball or the room

tmiddles wrote:

tmiddles wrote:
Semantics aside let's look at three scenarios.
A: Room is 18C the ball loses thermal energy to the room at 0.1206 Watts
B: Room is 30C the ball loses thermal energy to the room at 0.0632 Watts
C: Room is 42C there is not change in the thermal energy of the ball or the room

Still waiting....

IBdaMann wrote:
I don't understand your scenarios. .

IBdaMann wrote:
Scenario: An 8x8x3 meter room with steel panel walls, roof and floor, all with uniform 18 degrees Celsius initial temperature, inside and out. Suspended in the exact middle/center of the room by a powerful repelling magnet in the floor is a 5cm diameter steel ball (the "body") with initial uniform temperature of 42 degrees Celsius.
Discussion: What happens exactly with thermal energy of the body (steel ball) and that of the room (i.e. the air+walls+roof+floor) as photons fly about every which way.
[note: because you are apt to misread, I am asking you to describe the thermal energy, not the photons]

A: Room is 18C the ball loses thermal energy to the room at 0.1206 Watts
B: Room is 30C the ball loses thermal energy to the room at 0.0632 Watts
C: Room is 42C there is not change in the thermal energy of the ball or the room

A
A 5cm diameter ball has a surface area of 78.54cm. With a temperature of 42C the ball will lose thermal energy:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

The room provides an ambient radiance at 18C the absorption by the ball, and conversion to thermal energy, will be using the same surface area.
__________P(in)=σeA(T2^4), 291.15K
0.0000000567 * 1.00 * .007854 * 715668499.46 = +0.3187W

There is a loss of 0.1206 Watts by the ball and a gain of that same amount by the room.

B
A 5cm diameter ball at 42C, the ball will lose thermal energy:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

The room at 30C:
__________P(in)=σeA(T2^4), 303.15K
0.0000000567 * 1.00 * .007854 * 8445595755.5 = +0.3761W

There is a loss of 0.0632 Watts by the ball and a gain of that same amount by the room.

C
A 5cm diameter ball at 42C, the ball will lose thermal energy:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

The room at 42C:
__________P(in)=σeA(T2^4), 315.15K
0.0000000567 * 1.00 * .007854 * 986436754.6 = +0.4393W

A loss of 0.00 Watts by the ball and a gain of that same amount by the room.

tmiddles wrote:

IBdaMann wrote:
I don't understand your scenarios. .

Please let me know if I've departed in any way from your scenario.

IBdaMann wrote:
Scenario: An 8x8x3 meter room with steel panel walls, roof and floor, all with uniform 18 degrees Celsius initial temperature, inside and out. Suspended in the exact middle/center of the room by a powerful repelling magnet in the floor is a 5cm diameter steel ball (the "body") with initial uniform temperature of 42 degrees Celsius.
Discussion: What happens exactly with thermal energy of the body (steel ball) and that of the room (i.e. the air+walls+roof+floor) as photons fly about every which way.
[note: because you are apt to misread, I am asking you to describe the thermal energy, not the photons]

Let's stay focused on that as it's an excellent scenario to work with.

I've presented a more concise and focused solution to your senario below. Have I found the correct solutions?
Do you have a problem with my adding two variations in wall temperature?

A: Room is 18C the ball loses thermal energy to the room at 0.1206 Watts
B: Room is 30C the ball loses thermal energy to the room at 0.0632 Watts
C: Room is 42C there is not change in the thermal energy of the ball or the room

A
A 5cm diameter ball has a surface area of 78.54cm. With a temperature of 42C the ball will lose thermal energy:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

The room provides an ambient radiance at 18C the absorption by the ball, and conversion to thermal energy, will be using the same surface area.
__________P(in)=σeA(T2^4), 291.15K
0.0000000567 * 1.00 * .007854 * 715668499.46 = +0.3187W

There is a loss of 0.1206 Watts by the ball and a gain of that same amount by the room.

B
A 5cm diameter ball at 42C, the ball will lose thermal energy:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

The room at 30C:
__________P(in)=σeA(T2^4), 303.15K
0.0000000567 * 1.00 * .007854 * 8445595755.5 = +0.3761W

There is a loss of 0.0632 Watts by the ball and a gain of that same amount by the room.

C
A 5cm diameter ball at 42C, the ball will lose thermal energy:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

The room at 42C:
__________P(in)=σeA(T2^4), 315.15K
0.0000000567 * 1.00 * .007854 * 986436754.6 = +0.4393W

A loss of 0.00 Watts by the ball and a gain of that same amount by the room.

Into the Night wrote:Photons are not thermal energy.

tmiddles wrote:
A: Room is 18C the ball loses thermal energy to the room at 0.1206 Watts
B: Room is 30C the ball loses thermal energy to the room at 0.0632 Watts
C: Room is 42C there is not change in the thermal energy of the ball or the room

IBdaMann wrote:
I don't understand your scenarios. I get as far as the word "net" and I get confused.

tmiddles wrote:

Into the Night wrote:Photons are not thermal energy.

Where do you see me use the word "photons" in my solutions? My answers are in watts.

tmiddles wrote:
Do you have a solution to these problems ITN?

tmiddles wrote:
Are you saying these are the wrong answers?

tmiddles wrote:

tmiddles wrote:
A: Room is 18C the ball loses thermal energy to the room at 0.1206 Watts
B: Room is 30C the ball loses thermal energy to the room at 0.0632 Watts
C: Room is 42C there is not change in the thermal energy of the ball or the room

And no IBD did not comment on if I'd gotten the correct answers at all. He said simply:

IBdaMann wrote:
I don't understand your scenarios. I get as far as the word "net" and I get confused.

So I represented the problems with a focus on the equations used and answers without any troubling vocabulary.

tmiddles wrote:
Keep in mind this is IBD's scenario I've solved for. No one else, including you, has presented a solution.

Into the Night wrote:
You are trying to describe thermal energy in terms of radiance...

tmiddles wrote:
A: Room is 18C the ball loses thermal energy to the room at 0.1206 Watts
B: Room is 30C the ball loses thermal energy to the room at 0.0632 Watts
C: Room is 42C there is not change in the thermal energy of the ball or the room

Into the Night wrote:Making up random numbers as data is a fallacy.

tmiddles wrote: Well this was IBD's scenario so take it up with him.

IBdaMann wrote:
step 6) a detailed scenario is proposed as an illustration.
step 7) tmiddles performs some irrelevant arithmetic and demands confirmation that his number crunching is correct.

tmiddles wrote: IBD you asked what happens with the thermal energy. I have answered you, in watts. What were you asking for if not that?

IBdaMann wrote:
....
Not any units of measure. None are needed.
....
In the scenario I detailed, you show thermal energy flowing only in one direction (warmer to cooler). .

tmiddles wrote:

IBdaMann wrote:
....
Not any units of measure. None are needed.
....
In the scenario I detailed, you show thermal energy flowing only in one direction (warmer to cooler). .

So did I get the correct result for all three scenarios? Why is that so hard for you to answer?

IBdaMann wrote:
You did not answer the problem. I needed to see the correct final answer which is your observation ...

tmiddles wrote: ... so lets get some numbers on the table.

Do you not know how to solve for your own scenario?

IBdaMann wrote:
. Didn't I just give you the correct answer? (hint: yes).

tmiddles wrote:

IBdaMann wrote:
. Didn't I just give you the correct answer? (hint: yes).

Answer to what?

tmiddles wrote:
....deleted duplication of previous message...Mantra 29

IBdaMann wrote:

tmiddles wrote: Well this was IBD's scenario so take it up with him.

@ Into the Night, we really need to capture this tactic and get it into the list. This is such a beautiful example that we can't let this go. I'd like to get your thoughts on the optimal wording.

Step 1) tmiddles argues a violation of physics as an angle to show Global Warming, in this case, that SOME thermal energy flows from cooler to warmer.

Step 2) somebody politely points out that he is violating physics, in this case the 2nd LoT.

Step 3) tmiddles protests with a vague scenario that he claims demonstrates that he is not violating physics.

step 4) someone politely points out that he is still violating physics.

step 5) tmiddles asks a flurry of specific and irrelevant questions demanding that they all be explained.

step 6) a detailed scenario is proposed as an illustration.

step 7) tmiddles performs some irrelevant arithmetic and demands confirmation that his number crunching is correct.

step 8) when asked to refocus on the physics violation, he bows out, in this case he tells you to talk to me because *I* was the one who tried to address his requests for specificity.


This was a big waste of time.


.

tmiddles wrote:
...deleted TMSa7...

tmiddles wrote:
...deleted TMSa7...

tmiddles wrote:
...deleted TMSa7...Mantra 22...

IBdaMann wrote:

tmiddles wrote:

IBdaMann wrote:
. Didn't I just give you the correct answer? (hint: yes).

Answer to what?

Refresh my memory. What were we talking about?.

tmiddles wrote:

IBdaMann wrote:

tmiddles wrote:

IBdaMann wrote:
. Didn't I just give you the correct answer? (hint: yes).

Answer to what?

Refresh my memory. What were we talking about?.

I don't know what you were talking about.

I'll say again thermodynamics is an applied science.

tmiddles wrote: The reason is in part incompitence but mainly because you can't get the right answer without using net radiance.

IBdaMann wrote:
Yes, I gave you the correct answer to that.
.

tmiddles wrote:

IBdaMann wrote:

tmiddles wrote:

IBdaMann wrote:

tmiddles wrote:

IBdaMann wrote:
Didn't I just give you the correct answer? (hint: yes).

Answer to what?

Refresh my memory. What were we talking about?

I don't know what you were talking about. I'll say again thermodynamics is an applied science.

Yes, I gave you the correct answer to that.

no clue what you are talking about.

tmiddles wrote: IBD are you able to solve a thermodynamic problem? Do you know how? Is the solution not in watts?

IBdaMann wrote:
Refresh my memory then. What were we talking about?

IBdaMann wrote:

tmiddles wrote: ...a thermodynamic problem...Is the solution not in watts?

... No.

tmiddles wrote:

IBdaMann wrote:

IBdaMann wrote:

tmiddles wrote:

IBdaMann wrote:

tmiddles wrote:

IBdaMann wrote:

tmiddles wrote:
[quote]IBdaMann wrote:
Didn't I just give you the correct answer? (hint: yes).

Answer to what?

Refresh my memory. What were we talking about?

I don't know what you were talking about. I'll say again thermodynamics is an applied science.

Yes, I gave you the correct answer to that.

no clue what you are talking about.

Refresh my memory then. What were we talking about?

I don't know what you were talking about. I'll say again thermodynamics is an applied science. Real answers, in watts, can and should be calculated.

tmiddles wrote:

IBdaMann wrote:

tmiddles wrote: ...a thermodynamic problem...Is the solution not in watts?

... No.

What is it in then?

IBdaMann wrote:I gave you the correct answer.

IBdaMann wrote:

tmiddles wrote:

IBdaMann wrote:

tmiddles wrote: ...a thermodynamic problem...Is the solution not in watts?

... No.

What is it in then?

It depends on the problem.

tmiddles wrote:
A 5cm diameter ball has a surface area of 78.54cm. With a temperature of 42C the ball will lose thermal energy, through radiance, as per Stefan-Boltzmann:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

As the room basically provides an ambient radiance at 18C the absorption by the ball, and conversion to thermal energy, will be using the same surface area.
__________P(in)=σeA(T2^4), 291.15K
0.0000000567 * 1.00 * .007854 * 715668499.46 = +0.3187W

So the difference in this net radiative transfer is a loss of 0.1206 Watts by the ball and a gain of that same amount by the room.

tmiddles wrote:
A 5cm diameter ball has a surface area of 78.54cm. With a temperature of 42C the ball will lose thermal energy, through radiance, as per Stefan-Boltzmann:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

As the room basically provides an ambient radiance at 30C the absorption by the ball, and conversion to thermal energy, will be using the same surface area.
__________P(in)=σeA(T2^4), 303.15K
0.0000000567 * 1.00 * .007854 * 8445595755.5 = +0.3761W

So the difference in this net radiative transfer is a loss of 0.0632 Watts by the ball and a gain of that same amount by the room.

tmiddles wrote:
A 5cm diameter ball has a surface area of 78.54cm. With a temperature of 42C the ball will lose thermal energy, through radiance, as per Stefan-Boltzmann:
Stefan-Boltzmann equation:___P(out)=σeA*(T1^4)
(5.67×10−8J/s⋅m2⋅K4)(1.00)(.007854m2)315.15K^4=-0.4393W
0.0000000567 * 1.00 * .007854 * 986436754.6 = -0.4393W

As the room basically provides an ambient radiance at 42C the absorption by the ball, and conversion to thermal energy, will be using the same surface area.
__________P(in)=σeA(T2^4), 315.15K
0.0000000567 * 1.00 * .007854 * 986436754.6 = +0.4393W

So the difference in this net radiative transfer is 0.00 Watts by the ball and a gain of that same amount by the room. There is an equilibrium