Violating Second Law of thermodynamics

  • Thread starter bhave
  • Start date
  • #1
5
0
Background:

suppose that we have single body say sphere and nothing else in the universe. Now it will radiate heat depending on its temperature.since its energy is going down its temperature will decrease.Eventually wel have energy in the form of photons and sphere at zero kelvin. thus In practical case, if we reduce the incoming radiation and increase the out going radiations by sutaibly choosing absorptivity and emissivity we will achieve heat transfer from body at lower temperature to body at higher temperature.



let us consider two infinite parallel plates P1 and P2 with vaccum in between them so that only radiation takes place between them
plate 2 is held at temperature t2.

Let us assume the following


Plate 1
Absorptivity a1 = 0.1
Emissivity e1 = 0.9
Temperature t1


Plate 2
abosrptivity a2 = 0.9
emissivity e2 = 0.1
Temperature t2


Then Consider a steady state

I claim that the steady state temperature of Plate 1 is not equal to t2

for steady state the Incoming heat on Plate 1 must be equal to outgoing heat.

Incoming heat = abosrptivity * radiation from plate 2
= 0.1 * 0.1* sigma* t2*t2*t2*t2 ---------equ1

where sigma is stefan Boltzmann constant = 5.67 exp(-8)


Outgoing heat = emissivity * BlackBody radiation at that temperature
= 0.9 * sigma * t1*t1*t1*t1 ------------ equ 2


Thus equating above 2 equations

0.1 * 0.1 * sigma t2*t2*t2*t2 = 0.9 *sigma *t1*t1*t1*t1

Thus we get
0.325 * t2 = t1


Thus if initially both bodies were at same temperature then more heat will leave plate 1 and thus cause cooling without any input of work.

this way we would be violating second law of Thermodynamics.

Note

It is observed that reflection effects cancel each other and are energy associated with reflection is quite low.

The materials with above mensioned properties are available.

we even considered spectral distribution of Emissivity and Absorptivity and found that overall emissivity or overall absorbtivity values of 0.1 or 0.9 can be obtained.


Final word

Can radiation effects violate second law of thermodynamics.
 

Answers and Replies

  • #2
Bystander
Science Advisor
Homework Helper
Gold Member
5,203
1,224
bhave said:
(snip)Can radiation effects violate second law of thermodynamics.
Nope. Check the definitions for "absorbance" and for "emissivity."
 
  • #3
5
0
Bystander said:
Nope. Check the definitions for "absorbance" and for "emissivity."

Kindly help me coz i dont get the point ur trying to highlit
 
  • #4
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
Thus if initially both bodies were at same temperature then more heat will leave plate 1 and thus cause cooling without any input of work.
Obviously your analysis is flawed. The transfer of energy is proportional to the DIFFERENCE in temperature. If the plates are at the same temperature there is NO energy transfer.

It is observed that reflection effects cancel each other and are energy associated with reflection is quite low.
How did you observe this? You may want to look at it again.
 
  • #5
5
0
Integral said:
The transfer of energy is proportional to the DIFFERENCE in temperature.


transfer of energy is proportional to difference in temperature only in conduction or convection. But in radiation The amount of heat emitted is independant of other bodies sorrounding it.
 
  • #6
5
0
Integral said:
It is observed that reflection effects cancel each other and are energy associated with reflection is quite low.



How did you observe this? You may want to look at it again.
I worked it out and found that the terms that add heat to after reflection are GP of common ratio (1-a1)(1-a2) which is less than 1. So i got the expression of the form

a2 e1 T1*T1*T1*T1 = a1 e2 T2*T2*T2*T2

Thus giving slight change with final result equal to

T1 = 0.333 T2
 
  • #7
russ_watters
Mentor
20,159
6,682
bhave said:
transfer of energy is proportional to difference in temperature only in conduction or convection. But in radiation The amount of heat emitted is independant of other bodies sorrounding it.
You are incorrect. You may, if you wish, calculate the radiation and absorption separately and subtract, but it works out the same: radiation depends on temperature difference.
 
  • #8
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
bhave said:
transfer of energy is proportional to difference in temperature only in conduction or convection. But in radiation The amount of heat emitted is independant of other bodies sorrounding it.
This is incorrect.

the total energy transfered to one of your plates is the energy recieved - the energy emitted.

[tex] E_t - E_r \sim {T_1}^4 - {T_2}^4 [/tex]
 
Last edited:
  • #9
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
bhave said:
I worked it out and found that the terms that add heat to after reflection are GP of common ratio (1-a1)(1-a2) which is less than 1. So i got the expression of the form

a2 e1 T1*T1*T1*T1 = a1 e2 T2*T2*T2*T2

Thus giving slight change with final result equal to

T1 = 0.333 T2
That is one reflection how about the next? and the Next? ... the Next?

This must be resolved via an infinite series.
 
  • #10
Bystander
Science Advisor
Homework Helper
Gold Member
5,203
1,224
bhave said:
Kindly help me coz i dont get the point ur trying to highlit
The "point" is that YOU need to understand the definitions of "absorbance" and "emissivity." Since you haven't taken the time to follow my advice that you check them, I'll spell it out for you this time --- "absorbance" and "emissivity" are two different names for the same property, the coefficients CANNOT be different as you have postulated, your argument left the rails at that point.

Physics is a science in which the definitions have been cut, dried, stuffed, mounted, pickled, and preserved in formaldehyde "forever." It is NOT a science in which intuitive interpretation of definitions, and casual guesses at meanings of terms lead to meaningful trains of thought.

Suggestions that you "check" some aspect of your work are commonly offered in lieu of more severe criticisms of the care with which you have done the background work necessary to prepare your arguments.

You want radiation and the second law? Work your way through Prigogine. If that recommendation is totally meaningless to you, you are in waayyy over your head --- take the intro physics courses, then give it another try.
 
  • #11
5
0
Integral said:
This is incorrect.

the total energy transfered to one of your plates is the energy recieved - the energy emitted.

[tex] E_t - E_r \sim {T_1}^4 - {T_2}^4 [/tex]

what about emissivity and absorbtivity of two substances. More over the driving force in conduction and convection depends on temperature difference.

See if you dont have any other body in universe no conduction or convection take place from the sole body in universe. Since the energy is transferred from one body to another directly, you need another body for conduction or convection to take place.

but in case of radiation even if you have only one body in universe it will radiate heat.The emitted radiation depends on temperature.Note that the energy is not lost. the enegy still exists in the universe in the form of photons.


Thus we can calculate the emitted radiation independant of other body but this can not be done in case of conduction or convection.
 
  • #12
russ_watters
Mentor
20,159
6,682
bhave said:
but in case of radiation even if you have only one body in universe it will radiate heat.
In your thought experiment, there is more than one body.

And as Bystander noted, you are also misusing absorptivity and emissivity. Similar, but separate problem.
 
Last edited:

Related Threads on Violating Second Law of thermodynamics

Replies
17
Views
4K
  • Last Post
Replies
6
Views
3K
Replies
38
Views
5K
Replies
5
Views
2K
Replies
4
Views
2K
Replies
7
Views
1K
Replies
8
Views
1K
  • Last Post
Replies
2
Views
4K
Top