= Solution
Assume an asymptotically flat <Reissner-Nordstrom spacetime>, with $M>|Q|$, time normalized at infinity, and $G=\hbar=c_{\rm light}=k_B=1$. Put
$$
q=|Q|,\qquad r_\pm=M\pm\sqrt{M^2-q^2},\qquad
f(r)=\frac{(r-r_+)(r-r_-)}{r^2}.
$$
Near $r_+$, the Euclidean radial-time metric is $f'(r_+)(r-r_+)\,d\tau^2+dr^2/[f'(r_+)(r-r_+)]$. With $\rho=2\sqrt{(r-r_+)/f'(r_+)}$, it becomes
$$
d\rho^2+\left(\frac{f'(r_+)}2\right)^2\rho^2d\tau^2.
$$
The <Euclidean black-hole regularity condition> removes a conical defect only if $\tau$ has period $4\pi/f'(r_+)$. Therefore the <Reissner-Nordstrom Hawking temperature> is
$$
\boxed{T=\frac{f'(r_+)}{4\pi}
=\frac{r_+-r_-}{4\pi r_+^2}
=\frac{\sqrt{M^2-q^2}}{2\pi(M+\sqrt{M^2-q^2})^2}.}
$$
Equivalently this is the <Reissner-Nordstrom horizon surface gravity> divided by $2\pi$. The derivation assumes a nondegenerate horizon; the value zero at $M=q$ is obtained by the nonextremal limit, not by imposing conical regularity directly on the degenerate geometry.
For fixed nonzero $q$, write $y=r_+/q\geq1$. The horizon equation gives
$$
\frac Mq=\frac12(y+y^{-1}),\qquad
qT=\frac1{4\pi}(y^{-1}-y^{-3}).
$$
The <mass> is increasing with $y>1$, and
$$
\frac{d(qT)}{dy}=\frac{3-y^2}{4\pi y^4}.
$$
Thus the <fixed-charge Reissner-Nordstrom temperature maximum> occurs at
$$
\boxed{r_+=\sqrt3\,q,\qquad
M_{\max}=\frac{2q}{\sqrt3},\qquad
T_{\max}=\frac1{6\sqrt3\,\pi q}.}
$$
The curve starts at zero at $M=q$, rises to this maximum, then falls as $1/(8\pi M)$ for large <mass>. During evaporation from $M_0\gg q$, the <mass> moves from right to left: the <temperature> initially increases, reaches its maximum, and then decreases towards zero.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-69-temperature.png]
{title=Reissner-Nordstrom temperature at fixed charge, its maximum and the evaporation direction}
Under the printed thermal-emission cutoff, charged emission is absent throughout precisely when the entire <mass> path remains at or below the threshold:
$$
\boxed{m\geq T_{\max}\quad\Longleftrightarrow\quad
m|Q|\geq\frac1{6\sqrt3\,\pi}.}
$$
Equality is allowed because emission is assumed to require $T>m$, not merely $T=m$. The initial large <mass> ensures that the path includes the maximum.
With charge then conserved, neutral <Hawking radiation> lowers the <mass> until the system approaches the <charge-preserving Reissner-Nordstrom evaporation endpoint>:
$$
\boxed{M_{\rm final}=|Q|,\qquad T_{\rm final}=0,\qquad
A_{\rm final}=4\pi Q^2.}
$$
The limiting state is an <extremal black hole>, rather than a neutral zero-mass endpoint. Exact attainment in finite time is not established: near extremality $T\sim\sqrt{M-q}/(\sqrt2\pi q^{3/2})$, and the ideal radiation rate tends to zero. The conclusion is within the assumed thermal cutoff model, which omits nonthermal charge creation and other quantum corrections. For $Q=0$, the Schwarzschild <temperature> instead grows without bound in the extrapolated model, so no finite $m$ satisfies the stated all-the-way <temperature> bound.
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