 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2019/2/c/solution

Use the ratio of the two measured coefficients so that the sample size cancels:

\frac{A _{2}}{A _{0}} = - \frac{T _{E}^{2}}{12} = - \frac{7500}{25} K^{2} = - 300 K^{2} .

Hence

T_{E} = 12 \cdot 300 K = 60 K .

The units of the

7500

J K

, so that its division by

T^{2}

has the same heat capacity units as the leading

25 J K^{- 1}

term.

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