 Solution

For electrode area $A$, the change in spontaneous polarization density supplies a charge scale $\Delta Q=pA\Delta T$, while the film's capacitance is $C=\epsilon A/d$. The voltage magnitude is therefore $|\Delta V|=|p|d|\Delta T|/\epsilon$. More explicitly, for an initially compensated film with no change in free electrode charge, $\Delta D=\epsilon \Delta E+p\Delta T=0$. Taking $z$ from the lower electrode to the upper one gives $\Delta V_{\mathrm{upper}-\mathrm{lower}}=-d\Delta E=pd\Delta T/\epsilon$.The measurable magnitude is $0.339\mathrm{V}$, with polarity reversed by reversing the poling direction or the temperature change. This is an ideal open-circuit estimate: leakage, the measuring instrument's input capacitance and finite charge-relaxation time can reduce or reshape the observed voltage.