Total Work out of Three Bodies (Thermodynamics)

[Mathoholic!]

The problem is:

Having three bodies with thermal capacities (C) as sources of heat to a heat machine, what is the maximum work I can extract from this system, given that the bodies are at temperatures T₃, T₂ and T₁ (T₃>T₂>T₁), leaving them at an equal final temperature?

I tackled this problem by assuming that the amount of heat I can extract from each body is:

Q_i=C(T_i-T_f) i=1,2,3

And so:

W=ƩQ_i

Now, I'm not sure if I'm breaking the second law of thermodynamics by totally converting heat into work. This is where I'm kind of stuck.

So, if anyone can help me figure out how to solve this and other type-like problems, it'd be great.

Thank you :)

 

[Studiot]

Now, I'm not sure if I'm breaking the second law of thermodynamics by totally converting heat into work. This is where I'm kind of stuck.

The second law forbids the total conversion of heat into work in a cyclic process.

It does not forbid it for some part of a process that is not cyclic.

So post details of your proposed system and process for further comment.

 

[Mathoholic!]

¹Are you saying that I can only fully convert heat into work in a irreversible process?

I'm not sure I can give much more detail about the system. I've already given all there's to know about the problem.

I'm having trouble getting the definitions of reversible process and irreversible process clear in my head. It's all a bit fuzzy and I think that's what is keeping me from solving this problem.

 

[Studiot]

Cyclic is not the same as reversible.

You get back to the same conditions (set of thermodynamic values) in each case, but the paths are different.

There are no real reversible processs which must be under equilibrium at all stages.

Cyclic processes may include non equilibrium processes. Irreversible processes are non equilibrium.

 

[sardar]

I can provide some guidance on how to approach this problem. First, it is important to understand the second law of thermodynamics, which states that in any thermodynamic process, the total entropy of a closed system will either remain constant or increase. This means that it is impossible to convert all heat into work without any loss of energy.

In this case, the maximum work that can be extracted from the system is limited by the Carnot efficiency, which is given by:

η = 1 - T1/T3

where T1 and T3 are the lowest and highest temperatures of the three bodies, respectively. This efficiency represents the theoretical maximum amount of work that can be obtained from the system.

To find the maximum work that can be extracted, you can use the following equation:

Wmax = Q3 - Q1 = C(T3-Tf) - C(T1-Tf)

where Tf is the final temperature of the bodies after the heat extraction process. This equation takes into account the fact that heat is also being transferred from the highest temperature body (T3) to the lowest temperature body (T1).

To find the final temperature (Tf), we can use the fact that the total entropy of the system must remain constant or increase. This means that the total heat transfer (Q3 + Q2 + Q1) must equal the total change in entropy (ΔS) of the system. This can be expressed as:

Q3 + Q2 + Q1 = ΔS = Cln(Tf/T3) + Cln(Tf/T2) + Cln(Tf/T1)

Solving for Tf in this equation will give you the final temperature of the system.

In summary, the maximum work that can be extracted from the system can be found by using the Carnot efficiency and considering the total change in entropy of the system. It is important to remember that this represents the theoretical maximum and in reality, there will be some loss of energy due to the second law of thermodynamics. I hope this helps in solving similar problems in the future.