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

• Pouyan
In summary: This means that the work produced by the system is less than or equal to zero, which is in line with the second law of thermodynamics. In summary, the equations and image show a Carnot cycle where heat is converted to work, but due to the second law of thermodynamics, the total work produced is always less than or equal to zero. This is because one of the heat flows, represented by dQi, is negative and the work produced, represented by ∑(dWi), is also negative. This is in line with the principle that the total change in entropy in a cycle must be zero.
Pouyan
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:

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 ?

$$W=Q_H\left(1-\frac{T_i}{T_H}\right)$$
$$0<T_i<T_H$$

Chestermiller said:
$$W=Q_H\left(1-\frac{T_i}{T_H}\right)$$
$$0<T_i<T_H$$

Pouyan said:
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:

1. What is Clausius' Theorem?

Clausius' Theorem is a fundamental concept in thermodynamics that states that the sum of the changes in heat (dQ) divided by the temperature (T) for all processes in a closed system must be less than or equal to zero. In other words, the total entropy of a closed system must either remain constant or increase.

2. Who discovered Clausius' Theorem?

Rudolf Clausius, a German physicist and mathematician, first formulated the theorem in the mid-19th century. He is also known for his work on the second law of thermodynamics and the concept of entropy.

3. How is Clausius' Theorem used in thermodynamics?

Clausius' Theorem is used to determine the direction of heat flow in a closed system. It helps to predict whether a process is spontaneous or not, and whether it will result in an increase or decrease in entropy.

4. What is the significance of Clausius' Theorem?

Clausius' Theorem is significant because it provides a fundamental understanding of the behavior of energy and heat in closed systems. It also helps to explain why certain processes occur spontaneously and others do not.

5. Can Clausius' Theorem be violated?

No, Clausius' Theorem is a fundamental law of thermodynamics and has been experimentally verified numerous times. Any violation of the theorem would contradict the second law of thermodynamics, which states that the total entropy of a closed system must always increase or remain constant.

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