Mixing 1 cubic meter of HCl and NaOH: Temperature rise of a metal container? 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 $Q$?
You have 1 mole/liter of each reactant.  How many liters of each solution do you have?  That gives you the number of moles of reactants. They will create way less than $2\cdot 10^6$ moles of water-you were counting the water that was already there.  Once you have the right number of moles of reactant, that is the number of moles of water produced.  Multiplying by your heat per mole gives $Q$.  Then divide by $mc_p$ to get $\Delta T$