Interaction between blackbodies in relation to the second law

Click For Summary
SUMMARY

The discussion centers on the interaction between two blackbodies, B_1 and B_2, with temperatures T_1 and T_2, respectively, where T_1 is greater than T_2. It is established that B_2 can indeed receive energy from B_1 through radiation exchange, despite the second law of thermodynamics. However, the net energy transfer will always favor B_1, as it will emit more energy to B_2 than it receives. The mathematical representation of this energy exchange is given by the equation dQ(1/T_2 - 1/T_1) = dQ((T_1 - T_2)/(T_1T_2)) > 0, confirming that the net heat flow is from B_1 to B_2.

PREREQUISITES
  • Understanding of blackbody radiation principles
  • Familiarity with the second law of thermodynamics
  • Knowledge of thermal contact and energy exchange mechanisms
  • Basic proficiency in thermodynamic equations and concepts
NEXT STEPS
  • Study the Stefan-Boltzmann law for blackbody radiation
  • Explore the implications of the second law of thermodynamics in thermal systems
  • Investigate the mathematical modeling of heat transfer between bodies
  • Learn about the concept of thermal equilibrium and its applications
USEFUL FOR

Physicists, thermodynamics students, and engineers interested in heat transfer and energy exchange principles in thermal systems.

Shunyata
Messages
3
Reaction score
0
I am trying to understand the behavior of blackbodies in interaction with each other.

Conditions as follows.

[itex]1)[/itex] There are two blackbodies, say [itex]B_1[/itex] and [itex]B_2[/itex], with corresponding temperature [itex]T_1[/itex] and [itex]T_2[/itex].

[itex]2)[/itex] Initially [itex]T_1>T_2[/itex].

[itex]3)[/itex] [itex]B_1[/itex] and [itex]B_2[/itex] have thermal contact only through radiation exchange.

And finally my question,

Can [itex]B_2[/itex] give energy to [itex]B_1[/itex] or is this prevented by the second law?
 
Science news on Phys.org
Sure it can. In fact, it always will be. However, B1 will always be giving more energy to B2 than the reciprocal effect, so the net energy exchange will always involve B1 giving energy to B2.
 
Is it as simple as

[itex]dQ\Big( \frac{1}{T_2} - \frac{1}{T_1}\Big)=dQ\Big( \frac{T_1-T_2}{T_1T_2}\Big)>0,[/itex]

if [itex]dQ=dQ_1 - dQ_2[/itex] is the net-heat where positive direction is from [itex]B_1[/itex] to [itex]B_2[/itex] and [itex]dQ_1>dQ_2[/itex]?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 37 ·
2
Replies
37
Views
5K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K