= Solution
Interpret the stated transition as an upper critical miscibility <temperature>: cooling increases the effective $\chi$ through $\chi_c$. Above $T_c$, the homogeneous mixing <free energy> is convex throughout $0<\phi<1$, with no two-phase common tangent. Below $T_c$, it acquires a region with $f''<0$, bordered by the <spinodal>. A lower straight line touches the curve at two binodal compositions $\phi_-$ and $\phi_+$; this <common-tangent construction> gives the equilibrium polymer-poor and polymer-rich phases.
The precise raw curve need not have two equal minima: subtracting an <affine function> of $\phi$ changes reference chemical potentials without changing <curvature> or coexistence compositions. The free-energy panels above subtract the tangent at $\phi_c$ from the upper-temperature curve and the common tangent from the lower-temperature curve. This makes the common tangent and the convex-versus-nonconvex distinction visible.
If the mean initial composition $\phi_0$ lies inside the low-temperature spinodal region, an arbitrarily small separation into equal-volume regions with compositions $\phi_0\pm\delta\phi$ changes the mean free <energy> by
$$
\frac{f(\phi_0+\delta\phi)+f(\phi_0-\delta\phi)}2-f(\phi_0)
=\frac12f''(\phi_0)(\delta\phi)^2+O((\delta\phi)^4)<0.
$$
Small fluctuations therefore grow spontaneously: this is <spinodal decomposition>. Interfacial costs limit very short-wavelength fluctuations, and the polymer-poor and polymer-rich domains subsequently coarsen towards their binodal compositions.
If $\phi_0$ lies just outside a spinodal branch but still inside the binodal, then $f''(\phi_0)>0$. Small fluctuations increase <free energy>, so the uniform mixture is <metastable>. Separation requires a sufficiently large nucleus to overcome the interfacial barrier, followed by <nucleation> and growth. There is no guarantee that this happens rapidly. If the initial composition is outside the binodal as well, it remains a stable uniform phase after cooling.
For a mean composition in the coexistence interval, conservation of <polymer> fixes the volume fraction $\alpha$ of the rich phase by the <lever rule>:
$$
\boxed{\alpha=\frac{\phi_0-\phi_-}{\phi_+-\phi_-},\qquad
1-\alpha=\frac{\phi_+-\phi_0}{\phi_+-\phi_-}.}
$$
The temperatures alone do not determine which mechanism occurs without the initial composition. Numerical free-energy curves additionally require $\chi(T)$; the plotted example uses $\chi(T)=\chi_cT_c/T$ with absolute temperatures $T_c=298.15\,\mathrm K$, $T_1=303.15\,\mathrm K$ and $T_2=293.15\,\mathrm K$.
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