# Proof that if we give heat to a pure substance which exists only in one phase, its temperature will necessarily increase

I was challenged to answer this question. I could solve it indirectly, by proving that $C_v$ is always positive (and since $C_p > C_v$, it holds for all cases).

Using the 2nd and 3rd postulates of thermodynamics (from "Thermodynamics and an Introduction to Thermostatistics", by Herbert Callen), we know that entropy is a crescent function of the energy and it always reach a maximum, then we can write:

$\left(\frac{\partial^2S}{\partial U^2}\right)_{V,N} < 0$

$\frac{\partial}{\partial U} \left(\frac{\partial S}{\partial U}\right)_{V,N} = \frac{\partial}{\partial U} \left(\frac{1}{T}\right)_{V,N} = \frac{-1}{T^2} \left(\frac{\partial T}{\partial U}\right)_{V,N} = \frac{-1}{T^2 C_V} < 0 \therefore C_V > 0$

I would like to know whether there is another simple way to show the same result, not using Callen's postulate approach.

• When adding energy to spin-systems, this will drive them to negative temperatures (ok, pathological case). – Pieter Dec 6 '17 at 18:13
• @Pieter Wouldn't this transition to a negative temperature still be consistent with a positive $C_V$, as the temperature is increasing past $\infty$ to $-\infty$ and beyond? (Or, less weirdly, $\beta\propto 1/T$ is constantly decreasing as energy is added, which is exactly what we expect from more conventional systems and a positive $C_V$?) – Chemomechanics Dec 6 '17 at 23:54
• Indeed @Chemomechanics the thermodynamic beta is decreasing past zero. – Pieter Dec 7 '17 at 1:11
• You've got a sign error somewhere, because your last line implies $C_V < 0$. – Chris Dec 8 '17 at 0:34
• Indeed. I will edit it now. – David Sousa Dec 8 '17 at 2:13