Nathanael A. Fortune edited section_Application_to_Specific_Heat__.tex  over 8 years ago

Commit id: 2d13f4dc18db97f4ef907196bb06803e07fed946

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In our Jan 2015 heat capacity run, we used a 0.40 mg sample of $\kappa$-(BEDT-TTF)$_2$Cu(NCS)$_2$. In our July 2015 run, we used a 0.50 mg sample. Assuming a molar mass of $832.98 \mathrm{g/mole}$ and 1 Cu atom per mole, the nuclear contribution to the measured heat capacity of $\kappa$-(BEDT-TTF)$_2$Cu(NCS)$_2$ is predicted to be  \begin{equation}  \label{eq:NuclearCuContribution}  c\mathrm{\ [J/K]}  = \alpha \left(\frac{H}{T}\right)^2 \mathrm{\ [J/K]}  \end{equation}  where $\alpha = 1.6035 \cdot 10^{-9}$