To expand on Steve Huntsman's comment, the entropy follows from Onsager's result for the free energy per site, $F=$
$$
-\beta^{-1}\left[\ln 2+ \frac{1}{2}\frac{1}{(2\pi)^2}\int_0^{2\pi}d\theta_1\int_0^{2\pi}d\theta_2 \ln(\cosh2\beta E_1\cosh2\beta E_2
-\sinh2\beta E_1\cos\theta_1-\sinh2\beta E_2\cos\theta_2)\right],
$$
and the thermodynamic relation,
$$
S=-\frac{\partial F}{\partial T},
$$
for the entropy per site. Here $\beta=1/(k_BT)$ and $E_1$ and $E_2$ are the horizontal and vertical interaction strengths. If you set both interaction strengths equal to 1 and use units where Boltzmann's constant equals 1, then the critical temperature is $2/\ln(\sqrt2 + 1)\approx2.269$. If you plot $S$, you should find that it interpolates between 0 at low temperature and $\ln2$ at high temperature, as expected. At the critical temperature, the graph has infinite slope.