Entropy change in irreversible processes

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SUMMARY

The discussion centers on the validity of the entropy equation S=delta(Q)/T in irreversible processes, particularly in the context of cooling a body. Participants clarify that while cooling is typically considered irreversible due to the second law of thermodynamics, it can be modeled as a reversible process under specific conditions, such as using constant temperature baths. The key to defining reversible versus irreversible processes lies in maintaining thermodynamic equilibrium throughout the transition, which requires careful control of heat and work rates. The conversation emphasizes the importance of quasistatic conditions for achieving reversibility.

PREREQUISITES
  • Understanding of the second law of thermodynamics
  • Familiarity with the concept of thermodynamic equilibrium
  • Knowledge of quasistatic processes
  • Experience with heat transfer methods, particularly constant temperature baths
NEXT STEPS
  • Study the principles of quasistatic processes in thermodynamics
  • Explore the applications of constant temperature baths in reversible processes
  • Research the implications of the second law of thermodynamics on entropy
  • Learn about the mathematical derivation of entropy changes in reversible and irreversible processes
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Students and professionals in thermodynamics, physicists, engineers, and anyone interested in the principles of entropy and heat transfer in irreversible processes.

rajathjackson
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The equation for entropy S=delta(Q)/T is derived from reversible processes such as Carnot cycle. The delta(Q) in the equation is the reversible heat added or taken out from the system. So, why is this equation valid in the case of processes like cooling of a body which is irreversible?
 
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I think that the entropy is always measurable: it is zero if the system is at equilibrium, that is the transformation is reversible and if it is greater than zero the transformation is irreversible. I've realized it so, but I'm not sure.
 
rajathjackson said:
The equation for entropy S=delta(Q)/T is derived from reversible processes such as Carnot cycle. The delta(Q) in the equation is the reversible heat added or taken out from the system. So, why is this equation valid in the case of processes like cooling of a body which is irreversible?
To get the change in entropy of a system as a result of the transition from one thermodynamic equilibrium state to another, you need to dream up a reversible process path between the two states, and calculate the integral of dq/T for that path. This has nothing to do with the actual process path that caused the body to transition from the initial state to the final state (unless the actual process path just happened to be a reversible path).

What makes you think that the process of cooling a body is irreversible? Cooling of a body can be carried out reversibly. Can you dream up a reversible path that takes a body between a higher temperature and a lower temperature?

Chet
 
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Chestermiller said:
What makes you think that the process of cooling a body is irreversible? Cooling of a body can be carried out reversibly. Can you dream up a reversible path that takes a body between a higher temperature and a lower temperature?

Chet

But, according to 2nd law of thermodynamics heat doesn't flow from a cold to a hot body without any external work done on it. So, cooling of a body is an irreversible process because a body doesn't get hot on its own. We need to heat it up with an external source of energy.
 
rajathjackson said:
But, according to 2nd law of thermodynamics heat doesn't flow from a cold to a hot body without any external work done on it. So, cooling of a body is an irreversible process because a body doesn't get hot on its own. We need to heat it up with an external source of energy.
When you were talking about cooling a body, I naturally assumed you were talking about transferring heat to a colder body. But, if you want to transfer heat from a cold body to a hotter body, that can be done reversibly also. Just because you need an external source to do it (like, for example, a sequence of constant temperature baths) does not mean that it can't be done reversibly. And you can even accomplish this without doing any external work. But, of course, you will also be making a change in the constant temperature baths.

Chet
 
Chestermiller said:
When you were talking about cooling a body, I naturally assumed you were talking about transferring heat to a colder body. But, if you want to transfer heat from a cold body to a hotter body, that can be done reversibly also. Just because you need an external source to do it (like, for example, a sequence of constant temperature baths) does not mean that it can't be done reversibly. And you can even accomplish this without doing any external work. But, of course, you will also be making a change in the constant temperature baths.

Chet

So. if even processes in which we use an external source of energy is reversible which all processes are called irreversible and on the basis of what are they called so?
 
rajathjackson said:
So. if even processes in which we use an external source of energy is reversible which all processes are called irreversible and on the basis of what are they called so?
Are you asking for the definition of a reversible process?

Chet
 
Chestermiller said:
Are you asking for the definition of a reversible process?

Chet

My understanding was that processes which can't proceed in the reverse direction without any external intervention are called as irreversible processes. But, you mentioned that even if there is an external energy source that process can be reversible. So, what is it that actually makes a process irreversible or reversible?
 
rajathjackson said:
My understanding was that processes which can't proceed in the reverse direction without any external intervention are called as irreversible processes. But, you mentioned that even if there is an external energy source that process can be reversible. So, what is it that actually makes a process irreversible or reversible?
At each location along a reversible path between the initial and final thermodynamic equilibrium states of a system, the system is only slightly removed from thermodynamic equilibrium. So the path is essentially a continuous sequence of thermodynamic equilibrium states.

To achieve such a path, the rate at which heat is added to the system, dq/dt, must be controlled such that the temperature at the boundary of the system (with the surroundings) is changing very slowly, and the rate at which work is being done by the system on the surroundings, dw/dt, must be controlled such that the velocity at the boundary is very small. The latter constraint is often referred to as quasistatic. If these constraints are not met, either by allowing the temperature at the boundary to be changing very rapidly, or by allowing the velocity at the boundary to be substantial, the process will be irreversible.

A reversible process can be carried out in such a way that both the system and the surroundings can be returned to their original thermodynamic states without causing a perceptible change in either (or anything else). To accomplish this, it is also necessary to control what happens in the surroundings so that it too is close to thermodynamic equilibrium during the forward and reverse paths. This can be guaranteed most easily by exclusively employing within the surroundings only two types of devices: 1. an infinite array of constant temperature baths at a continuous selection of temperatures (to control the rate of transfer of the heat) and 2. an array of tiny weights that can be added or removed gradually from a piston on top of the cylinder (to control the rate of work being done). These tools are sufficient.

Chet
 
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