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

Commit id: 8959870e009857bd3ddfd3fbaf1c5e0351ddb234

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\label{eq:NuclearCuContribution}  c = \alpha \left(\frac{H}{T}\right)^2 \mathrm{\ [J/K]}  \end{equation}  where $\alpha = 1.6035 \cdot 10^{-9}$ 10^{-9}$.