I Thermodynamic temperature scale and Carnot cycle

AI Thread Summary
The discussion centers on the thermodynamic temperature scale and the Carnot cycle, specifically addressing equations I.21 and I.22 from a lecture on statistical mechanics. The user expresses confusion over deriving the relationship between the efficiencies of two Carnot engines and the corresponding temperatures. A key point made is that while one might be tempted to equate efficiency with temperature ratios directly, this leads to an incorrect conclusion. The clarification provided indicates that the function f(T) cannot simply be replaced with T, but rather with 1/T, resolving the initial misunderstanding. This exchange highlights the intricacies of thermodynamic principles in relation to Carnot engines.
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This thread is about the concept of thermodynamic temperature scale and its derivation through thought experiments involving Carnot engines.
Hi all,
recently I started following the MIT course "Statistical Mechanics I: Statistical Mechanics Of Particles" by MIT (here).
In the second lesson Prof. Kardar introduces the concept of thermodynamic temperature analyzing the behavior of two Carnot engines that share a thermal reservour at temperatre T_2. The lecture notes can be found here.

My doubt is about Eq. I.21 and I.22 at pag. 10. It seems to me that from
1-\eta(T1,T2) = \frac{1-\eta(T1,T3)}{1-\eta(T2,T3)}
I can conclude that
1-\eta(T1,T2) = \frac{f(T_1)}{f(T_2)}
by taking T_3 as a reference temperature (note that T_1>T_2>T_3). In the previous equation f is a generic function but since the definition of T is arbitrary we can say
1-\eta(T1,T2) = \frac{T_1}{T_2}.

Now the problem is that from the definition of efficiency of a Carnot engine
1-\eta(T1,T2) = \frac{Q_2}{Q_1}

and equating the last two equations it results
\frac{Q_2}{Q_1}= \frac{T_1}{T_2}
that is clearly wrong (see Eq. I.22).

Where is my mistake here?

Thank you in advance for your reply.
 
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##\frac {f(T_1)} {f(T_2)} < 1, T_1>T_2##. Thus, you cannot replace ##f(T)## with ##T##, but you can replace it with ##\frac 1 T##. Then, everything works just fine.
 
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Hi Hill,
thank you for your reply. It makes sense now.
 
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