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Calculate Heat capacity with temperature and total energy

Physics Asked by ChrizZly on March 29, 2021

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?

2 Answers

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.

Answered by Gec on March 29, 2021

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.

Answered by Gareth Meredith on March 29, 2021

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