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

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    Clausius Theorem
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Homework Help Overview

The discussion revolves around Clausius' theorem and its implications in the context of a Carnot cycle, specifically focusing on the relationship between heat transfer and work output. Participants are examining the equation T ∑(dQi/Ti) ≤ 0 and its interpretation within thermodynamic principles.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are exploring the implications of the equation T ∑(dQi/Ti) ≤ 0, questioning the conditions under which dQi may be considered less than or equal to zero, particularly in relation to heat leaving a reservoir. They are also discussing the relationship between heat transfer and work output in the context of the Carnot cycle.

Discussion Status

The discussion is active, with participants seeking clarification on the reasoning behind the inequality T ∑(dQi/Ti) ≤ 0. Some have reiterated the importance of understanding the signs of dQi in relation to heat flow, while others are attempting to connect the mathematical expressions to the physical concepts involved.

Contextual Notes

Participants are working under the assumption that the temperatures involved are positive and are referencing the second law of thermodynamics. There is an ongoing exploration of the definitions and implications of heat transfer in the context of the Carnot cycle.

Pouyan
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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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