Thermodynamic temperature scale and Carnot cycle

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SUMMARY

The forum discussion revolves around the thermodynamic temperature scale and the Carnot cycle, specifically addressing equations I.21 and I.22 from the MIT course "Statistical Mechanics I: Statistical Mechanics Of Particles" by Prof. Kardar. The user initially misinterprets the relationship between the efficiency of Carnot engines and temperature ratios, leading to a contradiction. The resolution clarifies that the function f(T) cannot be replaced with T but can be substituted with 1/T, allowing the equations to align correctly.

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chimay
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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.
 
Last edited:
Hi Hill,
thank you for your reply. It makes sense now.
 
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