Clausius' Theorem: Solving the Mystery of T ∑(dQi/Ti) =< 0

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SUMMARY

The discussion centers on Clausius' Theorem and its application in thermodynamics, specifically within the context of a Carnot cycle. Participants analyze the equation T ∑(dQi/Ti) ≤ 0, which indicates that the total work produced per cycle is less than or equal to zero. The conversation highlights the relationship between heat transfer (dQi) and temperature (Ti), emphasizing that heat leaving a reservoir at a lower temperature contributes to the negative work output. The conclusion asserts that the change in entropy of the working fluid remains zero in a complete cycle, reinforcing the second law of thermodynamics.

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Homework Statement
Derive the Clausius' theorem
Relevant Equations
Qh/Ql= Th/Tl
ΔW= ΣdQi
I see this in my book but there is something I don't get!
If we consider a Carnot cycle where heat Qh enters and heat Ql leaves,
We know Qh/Ql=Th/Tl
And we define ΔQ_rev then :

∑(ΔQ_rev/T) = (Qh/Th) - (Ql/Tl) =0
I insert an image:
Clasius Theorem.jpg

Which shows the heat dQi entering the reservoir at Ti from a reservoir at temperature T via a Carnot (Ci).
We know:

heat to reservoir at Ti / Ti = heat from the reservoir at T/ T
So : dQi/Ti = (dQi+dWi)/T
and rearranging:
dWi=dQi((T/Ti) -1)

The system in the image seems to convert heat to work but it cannot convert 100% of heat to work according to Kelvin's statement of the second law and hence we must insist that this is not the case. Hence:

Total work produced per cycle = ΔW + ∑(dWi) =< 0

But there is something I don't understand!

ΔW = ∑(dQi)
and dWi=dQi((T/Ti) -1)

So the total work produced per cycle = T ∑(dQi/Ti)

But why we say that this is less or equal than zero ?
T ∑(dQi/Ti) =< 0

T > 0.
Ti must be more than zero of course (Ti > 0)
How can we say that dQi =< 0 ?
Is that because it leaves from the reservoir with lower temperature ?
 
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$$W=Q_H\left(1-\frac{T_i}{T_H}\right)$$
$$0<T_i<T_H$$
 
Chestermiller said:
$$W=Q_H\left(1-\frac{T_i}{T_H}\right)$$
$$0<T_i<T_H$$
Would you explain please? :smile:
 
Pouyan said:
Would you explain please? :smile:
This just follows from the equations you wrote. What part don’t you understand?
 
Chestermiller said:
This just follows from the equations you wrote. What part don’t you understand?
T ∑(dQi/Ti) =< 0

Is that correct to think dQi <=0 because it leaves the reservoir?
 
Pouyan said:
T ∑(dQi/Ti) =< 0

Is that correct to think dQi <=0 because it leaves the reservoir?
In a cycle, the change in entropy of the working fluid is zero, and the Q's represent heat flows to the working fluid. So, if one of the Q's is negative, it means that the heat is flowing from the working fluid to a reservoir.
 
Last edited:

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