# Is the ratio of heat transfers always given by the same function?

In deriving the thermodynamic temperature scale, it uses the concept that heat transfer between two reservoirs is the function of the reservoirs' temperatures. i.e. $\frac{Q_1}{Q_2}=Φ(T_1,T_2)$. And then further express that $Φ(T_1,T_2)=\frac{ψ(T_1)}{ψ(T_2)}$.

Is it a hidden assumption that the function $Φ$ doesn't change its form for different temperatures? i.e. for different reservoirs we just plug in different temperatures $Φ(T_1,T_2)$, $Φ(T_3,T_4)$ but it can never be some other functions (say $f(T_3,T_4)$, $g(T_5,T_6)$, etc.) for different temperatures. Why?

• Hi Kelvin; I removed your second question because each post should only consist of one question. I'd encourage you to post that second question separately. Dec 31, 2014 at 4:23

As you have written it for a given and apparently fixed by your assumption $T_1$ and $T_2$ there is a $Φ(T_1,T_2)=ψ(T_1)/ψ(T_2)$. Of course for any given ratio of $Q_1$ and $Q_2$ one can say so as that is just a number but what Kelvin proved was that this ratio changes in a particular manner under the underlying assumption that it may depend only on the two temperatures. This is the crucial assumption. As one varies the temperatures the ratio of the absorbed and rejected heat must satisfy a compatibility criterion when one uses a 3rd intermediary heat sink and source. Now then it follows from simple continuity argument that the two-variable function $\Phi$ must be decomposable as a ratio of single variable functions. Fermi describes this very clearly, see page 40 in http://gutenberg.net.au/ebooks13/1305021p.pdf