\left(\frac{2\delta}{r_{+}}+r_{+}^2F^{'}(r_{+})\right)\,
.$$ To get the explicit form of the Hawking temperature in terms of black hole horizon, we have to give the expression of $F^{'}$. Taking derivative on both sides of Eq. (\[Fequ\]) with respect to $r$, one has $$F^{'}(r_{+})=-\frac{(m-1)M r_{+} + m Q}{r_{+}^{m+1}\sum_{i=1}^{s}
i~\widehat{\alpha}_{i}\left(\frac{\delta}{r_{+}^2}\right)^{i-1}}\, .$$ On the other hand, from the equation (\[Fequ\]), we have $$\label{massparameter} M=-\frac{Q}{r_{+}}+
r_{+}^{m-1}\sum_{i=0}^{s}\widehat{\alpha}_{i}\left(\frac{\delta}{r_{+}^2}\right)^i\,
.$$ Thus we can express $F^{'}(r_+)$ as $$F^{'}(r_{+})=-\frac{Q+(m-1)r_{+}^m\sum_{i=0}^{s}
\widehat{\alpha}_{i}\left(\frac{\delta}{r_{+}^2}\right)^{i}}{r_{+}^{m+1}\sum_{i=1}^{s}
i~\widehat{\alpha}_{i}\left(\frac{\delta}{r_{+}^2}\right)^{i-1}}\, ,$$ and the Hawking temperature has the form $$\mathfrak{T}=\frac{1}{4\pi r_{+}\sum_{i=1}^{s}
i~\widehat{\alpha}_{i}\left(\frac{\delta}{r_{+}^2}\right)^{i-1}}
\left(\sum_{i=0}^{s}(m-2i-1)
\widehat{\alpha}_{i}\delta\left(\frac{\delta}{r_{+}^2}\right)^{i-1}
+\frac{Q}{r_{+}^{m-2}}\right)\,
.$$ Clearly, if we choose a suitable $Q$, the Hawking temperature may vanish. This case corresponds to the “extremal" black holes with vanishing Hawking temperature. For example, when $m=4$, choosing $$Q=-3\widehat{\alpha}_0 r_{+}^4-\widehat{\alpha}_1
r_{+}^2\delta+\widehat{\alpha}_2\delta^2$$ will lead to a vanishing temperature.

To get the mass of the black hole, we expand the metric $g_{00}$ in the large $r$ limit, subtract the corresponding one for a suitable reference background solution $F_b$, and then read off the mass with the coefficient in front of some power of the radial coordinate $r$. Here we choose the vacuum AdS solutions with vanishing $M$ and $Q$ as the reference background, i.e., $$\sum_{i=0}^{s}\widehat{\alpha}_i F_b^i = 0 \, .$$ For large $r$, we can expand $F$ 