Resolving the Paradox of Entropy and Energy

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SUMMARY

The discussion centers on the relationship between entropy and energy, specifically addressing misconceptions about how energy loss affects entropy in gases. Participants clarify that while the entropy of a specific system can decrease when energy is removed, the total entropy of the universe, including the surroundings, must always increase, in accordance with the second law of thermodynamics. The conversation also touches on the concept of disgregation, which is deemed an outdated term, and emphasizes the importance of understanding entropy as a state function that depends on thermodynamic states rather than the path taken.

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  • Understanding of the second law of thermodynamics
  • Familiarity with the concept of entropy as a state function
  • Basic knowledge of thermodynamic systems and surroundings
  • Awareness of classical thermodynamics principles
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  • Study the second law of thermodynamics in detail
  • Explore the concept of entropy in various thermodynamic processes
  • Review classical thermodynamics textbooks for practical examples
  • Investigate the role of heat transfer in entropy changes in systems
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Students of physics, thermodynamics researchers, and professionals in engineering fields who seek a deeper understanding of entropy and its implications in energy systems.

  • #31
kntsy said:
If the air conditioner is reversible engine but not carnot, will the entropy(total) still increase? I think YES, but college physics text says entropy(total) remain constant for reversible?

Anyone can answer this brainteaser?
 
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  • #32
brainstorm said:
I thought entropy was relative to the ultimate state of disorder in a given system with isolated inputs. So, if a system reaches thermal equilibrium at on temperature, it has reached maximum entropy because no more disgregation of heat will take place. If heat is added unevenly, entropy can decrease insofar as the heat is concentrated/aggregated within some subset(s) of the system. In that case, the heat will dissipate and eventually cause the system to reach thermal equilibrium, but this time at a higher temperature.

Is this an incorrect description/example of thermal-equilibrium progress as increasing entropy?

Yes that looks correct.
 
  • #33
Cosmossos said:
we know that when the energy drops the entropy increases

but the entropy should DECREASE AS LONG AS THE ENERGY INCREASES.
where is my mistake?

How did you come to these conclusions? Where do we know this from?
 
  • #34
brainstorm said:
I thought entropy was relative to the ultimate state of disorder in a given system with isolated inputs. So, if a system reaches thermal equilibrium at on temperature, it has reached maximum entropy because no more disgregation of heat will take place. If heat is added unevenly, entropy can decrease insofar as the heat is concentrated/aggregated within some subset(s) of the system. In that case, the heat will dissipate and eventually cause the system to reach thermal equilibrium, but this time at a higher temperature.

Is this an incorrect description/example of thermal-equilibrium progress as increasing entropy?
Entropy has meaning only in terms of the difference in entropy between two states. As heat flows into the gas, there is a positive change in entropy of the gas (and a smaller negative change in the entropy of the surroundings).

"Disorder" is not really a very accurate explanation for entropy. First of all, it is not clear what "disorder" means. Consider 2 moles of gas in equilibrium at temperature T. Then consider one mole of the same gas at temperature T + \Delta T and the other mole at T - \Delta T. Which of these two states has the most disorder? Why?

Second, a concept of "disorder" is misleading. Consider a mole of He gas and a mole of Argon gas each at state (P,V,T) in its own compartment insulated from their surroundings separated by a common insulated wall. Then consider the situation where the wall is removed and the gases mix. Is there a change in entropy? Do both states represent the same amount of disorder?

AM
 

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