# Calculate Heat capacity with temperature and total energy

I have a molecular Dynamics simulation and want to calculate the heat capacity with constant volume of my material.

As output parameters I can get temperature, potential energy, kinetic energy, total energy, enthalpy, pressure, volume and density.

I once saw a calculation only with temperature and total energy. But it like a year ago and I cannot find it anymore. (But I guess this would be possible).

I get my data as a huge table with the value for each time step, so I can take the average for the last x timesteps in Matlab or use it in other ways.

My simulation is already on a constant volume, so my output values are under constant volume.

EDIT: I forgot 2 important information. I can calculate the mass of my material per hand and I raised my temperature from 300K to 320K in 0.2 Nanoseconds (I heat up around 1000 atoms. That should explain the fast heating)

Any suggestions?

Heat capacity is related to fluctuations of energy: $\overline{(\Delta E)^2} = k_B T^2 C_{V,N}$. Dispertion of $E$ can be found from numerical data.

• Is kb the Boltzmann constant? Therefor it is 1.380e-23 and when I divide through it to get my cv I will end up with a really large number. My cv is now like 4.7e+19 which is definetly wrong. Did I understood something wrong? – ChrizZly Dec 12 '17 at 11:09
• I also forgot that I have a solid metal and not a gas, so I cannot even use the Boltzman constant. – ChrizZly Dec 12 '17 at 12:27
• Yes, $k_B$ is Boltzmann constant. This constant is needed no matter gas or solid metal. Dispertion of the energy measured in appropriate units is really small, division by $k_B$ must not be a problem. In what units do you measure energy? – Gec Dec 12 '17 at 16:01

If the temperature at any point is changed, the local gradient heat flow is

$$\frac{\partial T}{\partial t} = -\frac{1}{\rho C_p} \frac{\partial Q}{\partial x}$$

We will use this equation in a moment - the heat energy per unit area is,

$$Q = -k \nabla T$$

we can create a gradient such that

$$\nabla Q = \frac{\partial Q}{\partial x} = -k R T$$

Then we can retrieve the definition of the heat flow equation in terms of the Ricci curvature again $$R$$, keep in mind, $$k$$ is the thermal conductivity. In the case above, the curvature $$R$$ has replaced the definition of the gradient $$\nabla$$. If the temperature at any point changed, the local gradient heat flow is, after we multiply through by $$-\frac{1}{\rho C_p}$$ we get,

$$-\frac{1}{\rho C_p}\nabla Q = -\frac{1}{\rho C_p}\frac{\partial Q}{\partial x} = \frac{k}{\rho C_p} R T = \alpha R T = \frac{\partial T}{\partial t}$$

Went off on a bit of a tangent, but if the second equation describes Fouriers relation as heat energy flux through an area, then the volume is found simply as

$$\mathbf{Q} = -k \nabla^2 T$$

defining the heat-energy density.