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

  • Level: Graduate 
  • Thread starter Thread starter filip97
  • Start date Start date
  • Tags Tags
    Carnot Function
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
7 replies · 2K views
filip97
Messages
30
Reaction score
0
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.
 
Physics news on Phys.org
So you will have to try something else. Something thermodynamic involving ##Q##.
 
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)
 
  • Like
Likes   Reactions: JD_PM and BvU