How explain non-linear heat transfer in graphite

Let's say we have an isolated system that contains two equally sized volumes of graphite that are in thermal contact with each other. At $t=0$ one graphite block has a temperature of 298 K, while the other block has a temperature of 498 K. Eventually a thermal equilibrium is reached between the two graphite blocks, and the equilibrium temperature is not simply the average temperature

$$\langle T \rangle \neq \frac{298 + 498}{2} = 398$$

but is instead 410 K. How explain this in therms of statistical thermodynamics?

Derivation of an expression for the heat capacity with constant volume

I have that the heat capacity for constant volume, $C_V$, is

$$C_V = \left( \frac{\partial U}{\partial T} \right)_V$$

and that the internal energy, $U$, is

$$U = kT^2 \left( \frac{\partial \ln Q}{\partial T} \right)_{N,V}$$

Assuming interacting and distinguishable particles (because of solid-state graphite) we have that the natural logarithm of the canonical partition function $Q$ is

$$\ln Q = \ln q^N = N\ln q_{tr} + N\ln q_{rot} + N\ln q_{vib} + N\ln q_{el}$$

where $q_{tr}$, $q_{rot}$, $q_{vib}$, and $q_{el}$ are the translational, rotational, vibrational, and electronic contributions to the molecular partition function $q$, and $N$ is the number of particles. We obtain models for these contributions based on the particle in a box, rigid rotor, and harmonic oscillator, and neglecting the electronic contribution. We then get

$$\ln Q = N \ln \frac{V\cdot e \left( \frac{2 \pi mkT}{h^2} \right)^{3/2}}{N} + N \ln \frac{T}{\theta_{rot}} + n \ln \frac{1}{1-e^{h\nu_0 /kT}}$$

This is the same as

$$\ln Q = N \ln V + N \ln e + \frac{3}{2}N \ln \left( \frac{2\pi mkT}{h^2} \right) + \frac{3}{2}N \ln T + N \ln T - N \ln \theta_{rot} - N \ln (1 - e^{-h\nu_0/kT})$$

To get an expression for the heat capacity, we need to take the partial time derivative of $\ln Q$. Only three terms depend on T and will be non-zero. Therefore

$$\frac{\partial \ln Q}{\partial T} = \frac{\partial}{\partial T} \frac{3}{2}N\ln T + \frac{\partial}{\partial T}N\ln T - \frac{\partial}{\partial T}N\ln (1-e^{-h\nu_0/kT})$$

which becomes

$$\frac{\partial \ln Q}{\partial T} = \frac{3N}{2T} + \frac{N}{T} + \frac{h\nu_0 e^{-h\nu_0/kT}}{kT^2(1-e^{-h\nu_0/kT})}$$

Plugging this into the expression for $U$ gives

$$U = \frac{3}{2}NkT + NkT + \frac{h\nu_0}{e^{h\nu_0/kT}}$$

Taking again the partial time derivative to get $C_V$, we finally get the following expression for the heat capacity

$$\underline{C_V = \frac{3}{2}Nk + Nk + \frac{h^2\nu_0^2 e^{h\nu_0/kT}}{kT^2(e^{h\nu_0/kT}-1)^2}}$$

However, I am not sure how to use this to explain the non-linear heat transfer observed for the isolated system described at the beginning of the question. I see that $C_V$ is not linear in $T$, but I don't know whether using the heat capacity is the correct way of explaining this in the first place.

I made this simple sketch to try and understand why the final temperature is closer to 498 K than to 298 K. I found the temperature in the middle of $C_v(298)$ and $C_V(498)$. Is this a good way of thinking about this? For $T \rightarrow \infty$, the heat capacity will also go to infinity, so I assumed this shape of the graph:

• "and the equilibrium temperature is not simply the average temperature" Is this based on observations? – Gert Sep 6 '16 at 21:35
• I don't know, the exercise text stated this. I suppose we can measure this. – Yoda Sep 7 '16 at 8:24

You can see from your final equation that the heat capacity is not necessarily constant, but, instead, can depend on temperature. So, if $$\int_{298}^{T_f}C_v(T)dT=\int_{T_f}^{498}C_v(T)dT$$ there is no guarantee that $T_f=398$ (unless $C_v$ is constant, independent of temperature).