Say I (perfectly) mix $1\ \mathrm{m^3}$ of 1 M $\ce{HCl}$ with $1\ \mathrm{m^3}$ of 1 M $\ce{NaOH}$ at standard air pressure and room temperature within a metal vessel. Heat of neutralization for strong acids and bases is $-55.8\ \mathrm{kJ/mol}$ (is that actually just the bond enthalpy of an $\ce{H-OH}$ bond?).
So the $2\ \mathrm{m^3}$ of $\ce{H2O}$ amount to around $2 \cdot 10^6\ \mathrm{mol}$ of water ($M(\ce{H2O}) \overset{\wedge}{=} 20\ \mathrm{g\ mol^{-1}}$). Therefore, $\Delta_\mathrm n H$ is around $2 \cdot 10^6 \times 55.8\ \mathrm{kJ/mol}$, around $110 \cdot 10^6\ \mathrm{kJ}$.
I would now like to calculate the temperature rise.
The heat released is $Q = m c_p \Delta T$, where $m$ is the mass of the solution ($2000\ \mathrm{kg}$), $c_p$ is the specific heat capacity of the solution and $\Delta T$ ist the change in temperature. But that's exactly what I'm trying to find. So unless I know $Q$, I can't find $\Delta T$. Therfore, how do I find $Q$?