How does equalizing voltage for lead acid battery depend on the temperature?
Generally, the electrode reactions are both based on lead in different oxidation states. At the negative electrode, $\ce{Pb}$ is oxidized to $\ce{Pb^2+}$ during discharge.
$$\ce{Pb &lt;=&gt; Pb^2+ + 2 e-}$$
At the positive electrode, $\ce{Pb^4+}$ is reduced to $\ce{Pb^2+}$.
$$\ce{Pb^4+ + 2 e+ &lt;=&gt; Pb^2+}$$
For a classical lead–acid battery, the overall cell reaction is approximately
$$\ce{Pb + PbO2 + 2H+ + 2 HSO4- &lt;=&gt; 2 PbSO4 + 2 H2O}$$
As long as $\ce{Pb}$, $\ce{PbO2}$, and $\ce{PbSO4}$ are available at the electrodes, the equilibrium cell voltage depends only on the acid concentration (that's why equilibtrium cell voltage can be estimated based on measured acid density).
The dependence of the equilibrium voltage on concentration is given by the Nernst equation:
$$U_\text{cell}=\left(1.931+0.0592\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
This equation applies to one cell; a battery, however, has six cells in a row. Thus the total voltage is
$$U_\text{battery}=6\left(1.931+\frac{RT}{ZF}\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
where
$R$ is the gas constant,
$T$ is temperature,
$z$ is the number of electrons transferred in the cell reaction, and
$F$ is the Faraday constant.

We are only interested in the dependence of voltage $U$ on temperature $T$. All the other parameters may be taken as constants. Thus, our equation can be simplified to
$$U=\left(11.586+T\cdot k\right)\ \mathrm V$$
where $k$ is a constant.

The exact parameter values for a classical lead–acid battery might differ a bit from a modern battery with lead–carbon electrodes; however, we can calibrate our model using the data given by the manufacturer, i.e. $U=14.1\ \mathrm V$ at $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$
$$14.1\ \mathrm V=\left(11.586+298.15\ \mathrm K\cdot k\right)\ \mathrm V$$
and we find 
$$k=0.008432\ \mathrm{K^{-1}}$$
Therefore, our complete equation is
$$U=\left(11.586+T\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V$$
For a new temperature of $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$, we get an estimate of
$$\begin{align}U&amp;=\left(11.586+273.15\ \mathrm K\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V\\
&amp;\approx13.9\ \mathrm V\end{align}$$