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Assume an asymptotically flat Reissner-Nordstrom spacetime, with $M>|Q|$, time normalized at infinity, and $G=\hslash =c_{\mathrm{light}}=k_B=1$. PutNear $r_{+}$, the Euclidean radial-time metric is $f^{\prime }(r_{+})(r-r_{+})d{\tau }^2+d{r}^2/[f^{\prime }(r_{+})(r-r_{+})]$. With $\rho =2\sqrt{(r-r_{+})/f^{\prime }(r_{+})}$, it becomesThe Euclidean black-hole regularity condition removes a conical defect only if $\tau$ has period $4\pi /f^{\prime }(r_{+})$. Therefore the Reissner-Nordstrom Hawking temperature isEquivalently 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 $y=r_{+}/q\ge 1$. The horizon equation givesThe mass is increasing with $y>1$, andThus the fixed-charge Reissner-Nordstrom temperature maximum occurs atThe 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.

Under the printed thermal-emission cutoff, charged emission is absent throughout precisely when the entire mass path remains at or below the threshold: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: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 $T\sim \sqrt{M-q}/(\sqrt{2}\pi q^{3/2})$, 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.