Understanding the Criteria for Reversibility in Thermodynamics

In summary, the two criteria for reversibility in classical thermodynamics are that there should be no dissipative forces (i.e. no friction) and there is no heat flow across a definite temperature gap, only across an infinitesimal temperature difference. This is because all processes must be reversible in a reversible machine, and it is not possible to reverse heat exchange if it occurs across a definite temperature gap without causing other changes in the surroundings. Additionally, in the definition of entropy, we talk about reversible heat, but when work is done against friction, the entropy change of the system increases by the heat supplied divided by the temperature of the body. The heat supplied in this case is not reversible. However, in the case of a hot stone being put in
  • #1
aim1732
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I need a clarification about the reversibility criteria in classical thermodynamics.There are two criteria in it to my understanding.one is that there should be no dissipative forces i.e no friction.The second is that there is no heat flow across a definite temperature gap--only across an infinitesimal temperature difference.The apparent reason given for this is that all processes need to be reversible in a reversible machine and you can not possibly reverse the heat exchange if it occurs a definite temperature gap without causing any other change in the surroundings.I need a little more explanation about this...as in what kind of changes?

Also in the definition of entropy,we talk about reversible heat.Yet when we do work against friction we say that the entropy of the thing we do work on(that gets heated up) rises by the heat supplied divided by the temperature of the body(provided its temperature does not rise much).Is the heat supplied here reversible?Another case is a hot stone being put in a cold reservoir.The heat exchanged here is not reversible;yet we use that heat to calculate the entropy change.

Any help is appreciated.
 
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  • #2
Sorry no one else has answerded you so far. I'll give it what I can.

A definite temperature change can occur reversibly if the change is envisioned to be performed by an infinite series of heat reservoirs exchanging heat with the system. For anyone reservoir, T is the same for the system as the reservoir, so dQ/T is + for one and - for the other, the two add to zero, and no change in entropy of the universe takes place. You don't need to add work to this process.

The Carnot cycle is another, simpler example of moving a system from one finite temperature to another within the cycle, but this requires work. Since all sections of the cycle are performed reversibly, at all times the entropy change of the hot & cold reservoirs plus the system is zero.

The question of your hot rock into a cooler liquid is a good one. As I see it, a reversible path can be found. What you do is take the rock and immerse it in a reservoir at T infinitesmally below the temperature of the rock. This is a reversible change since again T is "the same" for the rock and the reservoir, so again ΔS = 0 for both. You repeat this process on the rock an infinite number of times until its final temperature equals the final temperature of the rock + water, which you can easily compute ahead of time from heat lost = heat gained and the heat capacities of the rock & water.

Then you perform the same stunt with the water, contacting it with an infinite series of reservoirs, each differentially higher in temperature than the previous, until the water too is brought up to the final temperature.

For both processes, entropy of each system (rock and water separately) can be computed, so the total change in entropy of your universe (the rock and the water together) can be determined. It would be

ΔS = Cr∫dT/T from Tir to Tf + Cw∫dT/T for Tiw to Tf
where r stands for rock, w for water, i for initial and f for final.

I'd appreciatye someone else competent looking at this!
!
 

FAQ: Understanding the Criteria for Reversibility in Thermodynamics

1. What is the Second Law of Thermodynamics?

The Second Law of Thermodynamics states that in any isolated system, the total entropy (or disorder) will always increase over time. This means that energy tends to disperse, and systems tend to move towards a state of equilibrium.

2. How does the Second Law relate to reversibility?

The Second Law of Thermodynamics also states that all natural processes are irreversible, meaning they cannot be undone or run backwards. This is because in a closed system, energy will always disperse and become less concentrated, making it impossible to return to its original state.

3. Can the Second Law of Thermodynamics be violated?

No, the Second Law of Thermodynamics is a fundamental law of physics and cannot be violated. However, it is possible to temporarily decrease entropy in a localized system by inputting energy or work, but this ultimately increases the total entropy of the entire system.

4. What is an example of the Second Law in action?

An example of the Second Law of Thermodynamics is the cooling of a hot cup of coffee. The coffee will naturally cool down over time, as heat is dispersed to its surroundings. This is because heat naturally moves from areas of higher concentration (the hot coffee) to areas of lower concentration (the cooler room).

5. How does the Second Law of Thermodynamics relate to the concept of energy efficiency?

The Second Law of Thermodynamics implies that it is impossible to achieve 100% energy efficiency in any system. This is because some energy will always be lost in the form of heat due to the natural tendency for energy to disperse. However, by designing systems and processes with this law in mind, we can strive to increase energy efficiency and minimize energy loss.

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