Calorimetry measured the heat of a hot metal in water. The same coffee-cup calorimeter can measure the heat of a reaction, which then gives the $\Delta H$ of a thermochemical equation (Heat from Amounts of Reactant in reverse).
Key idea
Run the reaction in an insulated cup of solution and measure $\Delta T$. The solution absorbs $q_\text{soln} = mc\Delta T$; the reaction released the same amount: $q_\text{rxn} = -q_\text{soln}$. Then $$\Delta H = \frac{q_\text{rxn}}{\text{moles of reaction}},$$ using the moles of the limiting reactant (adjusted by its coefficient). At constant pressure, this heat is $\Delta H$.
Why it works
The cup is insulated, so no heat leaves: every joule the reaction gives off stays in the solution and shows up as a temperature rise. If the solution gains 2790 J, the reaction (the system) lost 2790 J, so $q_\text{rxn}$ is negative and the reaction is exothermic. Dividing by moles turns a result about this particular cup into a property of the reaction, per mole as written. (We treat dilute solutions as water: density 1.00 g/mL and $c = 4.184\ \text{J/(g·°C)}$.)
Worked example: Neutralization
$50.0\ \text{mL}$ of $1.00\ \text{M}$ HCl and $50.0\ \text{mL}$ of $1.00\ \text{M}$ NaOH, both at $22.00^\circ\text{C}$, are mixed. The temperature rises to $28.67^\circ\text{C}$. Find $\Delta H$ for $\mathrm{H^+} + \mathrm{OH^-} \to \mathrm{H_2O}$.
- Mass of solution: $100.0\ \text{mL} \times 1.00\ \text{g/mL} = 100.0\ \text{g}$; $\Delta T = 6.67^\circ\text{C}$.
- $q_\text{soln} = (100.0)(4.184)(6.67) = 2791\ \text{J}$, so $q_\text{rxn} = -2791\ \text{J}$.
- Moles reacting: $0.0500\ \text{L} \times 1.00\ \text{M} = 0.0500\ \text{mol}$ of each.
- $$\Delta H = -2791/0.0500 = -55{,}800\ \text{J/mol} = -55.8\ \text{kJ/mol}$$
Answer: $-55.8\ \text{kJ/mol}$
Worked example: A salt that cools
$5.00\ \text{g}$ of $\mathrm{NH_4NO_3}$ ($80.06\ \text{g/mol}$) dissolves in $50.0\ \text{g}$ of water; the temperature falls by $6.97^\circ\text{C}$. Find $\Delta H_\text{soln}$ (solution mass $55.0\ \text{g}$).
- $q_\text{soln} = (55.0)(4.184)(-6.97) = -1604\ \text{J}$: the solution lost heat.
- $q_\text{rxn} = +1604\ \text{J}$; moles: $5.00/80.06 = 0.06245$.
- $$\Delta H = 1604/0.06245 = +25{,}700\ \text{J/mol} = +25.7\ \text{kJ/mol}$$
Answer: $+25.7\ \text{kJ/mol}$
A coffee-cup calorimeter rests on a few assumptions: no heat escapes through the cup, the cup itself absorbs almost nothing, and a dilute solution has water's density and specific heat. Careful work corrects for the cup with a small calorimeter constant, found the same way as for a bomb calorimeter.
Common mistake
Reporting $q$ of the solution as $\Delta H$, without changing its sign or dividing by the moles of the limiting reactant. The solution warmed, so it gained heat; the reaction lost it. Check: a temperature rise means an exothermic reaction and a negative $\Delta H$; and $\Delta H$ is per mole, so it shouldn't change if you double the amounts.