1. **State the problem:** Calculate the heat capacity of the calorimeter when 15.00 mL of 3.00 M NaOH is mixed with 13.90 mL of 3.00 M HCl, causing a temperature rise from 22.00 °C to 30.40 °C. 2. **Identify the reaction and heat released:** The neutralization reaction releases heat according to the molar enthalpy of neutralization: $$\Delta H = -55.84\ \text{kJ/mol}$$ per mole of HCl. 3. **Calculate moles of HCl and NaOH:** $$\text{moles HCl} = 3.00\ \text{mol/L} \times 0.01390\ \text{L} = 0.0417\ \text{mol}$$ $$\text{moles NaOH} = 3.00\ \text{mol/L} \times 0.01500\ \text{L} = 0.0450\ \text{mol}$$ 4. **Determine limiting reactant:** HCl is limiting with 0.0417 mol. 5. **Calculate heat released by reaction:** $$q_{reaction} = 0.0417\ \text{mol} \times (-55.84\ \text{kJ/mol}) = -2.33\ \text{kJ} = -23300\ \text{J}$$ 6. **Calculate heat absorbed by solution:** Mass of solution = 28.90 g, specific heat capacity = 3.74 J/g°C, temperature change $$\Delta T = 30.40 - 22.00 = 8.40\ ^\circ C$$ $$q_{solution} = m \times c \times \Delta T = 28.90 \times 3.74 \times 8.40 = 908.3\ \text{J}$$ 7. **Calculate heat absorbed by calorimeter:** Heat released by reaction equals heat absorbed by solution plus calorimeter: $$q_{reaction} = q_{solution} + q_{calorimeter}$$ $$-23300 = 908.3 + q_{calorimeter}$$ $$q_{calorimeter} = -23300 - 908.3 = -24208.3\ \text{J}$$ Heat absorbed is positive, so: $$q_{calorimeter} = 24208.3\ \text{J}$$ 8. **Calculate heat capacity of calorimeter:** Heat capacity $$C = \frac{q_{calorimeter}}{\Delta T} = \frac{24208.3}{8.40} = 2881.5\ \text{J/}^\circ\text{C}$$ **Final answer:** $$\boxed{2882\ \text{J/}^\circ\text{C}}$$ (rounded to 4 significant figures)