Carnot function -- How can I prove f(t2,t1)=f(t2−t1,0)

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The discussion revolves around proving the relationship f(t2,t1) = f(t2−t1,0) within the context of empirical temperature scales and thermodynamics. The quotient of heats Q2/Q1 is expressed as f(t2,t1), and the function is argued to be independent of the specific temperature scales used. The conversation references the monotonic nature of temperature functions and the linearity required for homomorphism between different temperature scales. The Cauchy equation is introduced to support the argument that the function must be linear, leading to the conclusion that the function behaves consistently across different scales. The discussion emphasizes the foundational principles of thermodynamics, particularly the second law, in deriving these relationships.
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If we have that quotient of heats ##Q_2/Q_1=f(t_2,t_1)##, where ##t_1,t_2## are emirical temperatures. Is this function satisfies :

##f(t_2,t_1)=f(t_2-t_1,0)##

I try prove it with Taylor series of two variables, but i can't prove anything.
 
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So you will have to try something else. Something thermodynamic involving ##Q##.
 
How ?
 
Check your notes and/or textbook. It is your exercise.
 
No, this is my intuitive thinking about empirical temperature scales because temperature is variable of interval type(levels of measurement).
 
filip97 said:
intuitive thinking
What is the context of your exercise? A curriculum on intuition, or something in termodynamics ?
 
I was reading Thermodynamics by Enrico Fermi first they give this function ##\frac{Q_2}{Q_1}=f(t_2,t_1)=\frac{g(t_2)}{g(t_1)}##, where ##g## is monotone increasing function. Idea is if we have scales of temperature ##r,s## (monotone increasing), but and their inverses is also scales of empirical temperatures ##r^{-1},s^{-1}##. ##f(t_2,t_1)=f(s(t_2),s(t_1))##, because function ##f##, doesn't depend of scales, and depend only of temperatures (Carnot's theorem). If we has two objective temperature scales they must has homomorphism between scales, and they must be linear, because temperature is interval data e.g.

##s(t)=\alpha x+\beta, \alpha>0, \beta## is some parameter. I must pack properties that homomorphism is linear function between ##s## and ##r##. Let it be ##s(t_2-t_1)=s(t_2)-s(t_1)##

We have ##f(t_2,t_1)=f(t_2-t_1,0)=f(s(t_2-t_1),s(0))=f(s(t_2-t_1)-s(0),0)=f(s(t_2),s(t_1))=f(s(t_2)-s(t_1),0)\implies
s(t_2-t_1)-s(0)=s(t_2)-s(t_1)##, set ##t_2=t_1##, this is Caushy equation and this solution is ##s(t)=ct##. As I says
scale is and function ##F(t)=r(s^{-1}(t)),r(t_2-t_1)=r(t_2)-r(t_1), s(t_1)=x, s(t_2)=y\implies F(x-y)=F(x)-F(y) \implies F(x)=\alpha x+\beta##(Pfanzagl, Theory of Measurement, page 98)
 
Usually this calculation occurs in textbooks, where thermodynamics is inductively derived by analyzing thermodynamic cycles. Here one uses the 2nd Law of thermodynamics in the formulation that there cannot be a perpetuum mobile of the 2nd kind (Clausius, Planck).

A masterpiece is the textbook

R. Becker, Theory of Heat, Springer Verlag (1967)
 
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