= Solution
In equilibrium without <hysteresis>, the system selects the lower free-energy minimum. At $T_2$ the <order parameter> jumps between $0$ and $2b/(3c)$, which identifies a <first-order phase transition>. Differentiating the minimized <free energy> does not add a $dP/dT$ contribution because $F'=0$. Thus the polarization-dependent <entropy> is $S=-aP^2/2$, giving
$$
S_{\rm ordered}-S_{\rm disordered}=-\frac{2ab^2}{9c^2}.
$$
The heat absorbed on heating from the ordered to the disordered phase is
$$
\boxed{Q_l=\frac{2ab^2T_2}{9c^2}}.
$$
The same amount is released on cooling. If $\Delta S$ is defined in the opposite direction, the signed transition heat is correspondingly negative.
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