Solution
= Solution
{c}
Use the ratio of the two measured coefficients so that the sample size cancels:
$$
\frac{A_2}{A_0}=-\frac{T_E^2}{12}=-\frac{7500}{25}\,\mathrm K^2=-300\,\mathrm K^2.
$$
Hence
$$
\boxed{T_E=\sqrt{12\cdot300}\,\mathrm K=60\,\mathrm K.}
$$
The units of the $7500$ coefficient are $\mathrm{J\,K}$, so that its division by $T^2$ has the same <heat capacity> units as the leading $25\,\mathrm{J\,K^{-1}}$ term.