 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 $F^{\prime }=0$. Thus the polarization-dependent entropy is $S=-a{P}^2/2$, givingThe heat absorbed on heating from the ordered to the disordered phase isThe same amount is released on cooling. If $\Delta S$ is defined in the opposite direction, the signed transition heat is correspondingly negative.