Chemical reaction that can produce lots of heat from 2-3 simple liquid ingredients?
The first thing that comes to mind, if you have access to a stock-room, is mixing two solutions: (1) NaOH; (2) HCl. This can release a lot of heat if your solutions are concentrated enough, and it forms salt-water if your NaOH and HCl are of the same molarity. 

Edit: I'm on my lunch break, so I did some of the math... 

The chemical reaction of interest in this case is:

$$\ce{OH- + H+ -&gt; H2O}$$

The standard enthalpies of formation ($\Delta H_\mathrm f^\circ$) of these species are: 

$$\begin{array}{lr}
\hline
\text{Species} &amp; ΔH_\mathrm f^\circ/\pu{kJ mol-1} \\
\hline
\ce{OH-} &amp; -229.99 \\
\ce{H+}  &amp; 0.00       \\
\ce{H2O} &amp; -288.83 \\
\hline
\end{array}$$

Thus, the change in enthalpy for the reaction is: 

$$\Delta H^\circ = -288.83\ \mathrm{kJ/mol} - (-229.99\ \mathrm{kJ/mol}+0\ \mathrm{kJ/mol}) = -58.84\ \mathrm{kJ/mol}$$

Therefore, for $1\ \mathrm{mol}$ of $\ce{NaOH}$ + $1\ \mathrm{mol}$ of $\ce{HCl}$, you get $58.84\ \mathrm{kJ}$ of heat. Say you want to release enough heat to get the net solution up to $100\ \mathrm{^\circ C}$. 

Water has a heat capacity of $4.18\ \mathrm{J/(g\ ^\circ C)}$. Say you have $1\ \mathrm L$ of $\ce{NaOH}$ + $1\ \mathrm L$ of $\ce{HCl}$, you'll need enough heat to raise the temperature of $2\ \mathrm L$ of water to $100\ \mathrm{^\circ C}$. I'll assume the water starts off at $25\ \mathrm{^\circ C}$, so you have $2\,000\ \mathrm g$ and $75\ \mathrm{^\circ C}$ to go. 

$$4.18\ \mathrm{J/(g\ ^\circ C)} \cdot 2\,000\ \mathrm g \cdot 75\ \mathrm{^\circ C} = 62\,700\ \mathrm J = 62.7\ \mathrm{kJ}$$

How many moles of $\ce{NaOH}$ + $\ce{HCl}$ do you need for that much heat? 

$$\frac{q}{\Delta H^\circ} = \frac{62.7\ \mathrm{kJ}}{58.84\ \mathrm{kJ/mol}} = 1.065\ \mathrm{mol}$$

That would mean that you can mix $1\ \mathrm L$ of $1.065\ \mathrm M$ $\ce{NaOH}$ + $1\ \mathrm L$ of $1.065\ \mathrm M$ of $\ce{HCl}$, and would theoretically expect to get a temperature close to $100\ \mathrm{^\circ C}$. 

This might be off a bit because I've made some assumptions:


$\Delta H^\circ$ is constant with respect to temperature from $25\ \mathrm{^\circ C}$ to $100\ \mathrm{^\circ C}$. This may not be true. 
the dissolved salts in water dont significantly affect its heat capacity 
You have $2\,000\ \mathrm g$ of water in $1.065\ \mathrm M$ $\ce{NaOH}$ + $1.065\ \mathrm M$ $\ce{HCl}$ 


I think this could get you close though? I'm a bit surprised that the molarities aren't higher... It's a starting point at least.

Disclaimer of course: be careful with the $\ce{NaOH}$ + $\ce{HCl}$ solutions, they can be dangerous. Use proper chemistry hygeine protocols. The mixture should be benign, but you should confirm this with pH paper.