Textbook Question 4A.13

[kaela_losekoot_2A]

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution in the calorimeter (q = -23.50 kJ), resulting in a temperature rise of 7.32 C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

I understand how to calculate heat capacity, and I know multiplying that heat capacity by the change in temperature gives the correct answer. However, the equation for q is q=nCdeltaT, so I don't understand why we don't consider moles in the calculation?

Also, if we do include moles would the moles be the combined moles of the two reactants?
Thank you

[audrey 1A]

The question has a constant v so you don't need to worry about n. I hope that helps!