Achieve week 3 and 4 (Question 18)

[eopeyany 3D]

Hi, I was wondering if I could get any help in solving this question or getting started.
A constant‑volume calorimeter was calibrated by carrying out a reaction known to release 1.76 kJ of heat in 0.500 L of solution in the calorimeter (q=−1.76 kJ), resulting in a temperature rise of 3.48 ∘C. In a subsequent experiment, 250.0 mL of 0.10 M HClO2(aq) and 250.0 mL of 0.10 M NaOH(aq) were mixed in the same calorimeter and the temperature rose by 6.30 ∘C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

[Dayeon Kim 2D]

Hi,
First you need to find the heat capacity Ccal by using the Ccal = qcal/deltaT equation. To find qcal, you just do -q and q was given earlier in the problem. Then you divide the qcal by the given change in temperature. Now we find the heat capacity of the calorimeter released by the neutralization reaction which the equation is just qreaction + qcal = 0. Since qreaction = -qcal, you need to find -Ccal*deltaT. And in this case you use the next rise in temperature that they listed and multiply that by the Ccal that you found earlier to find you answer. (qreaction is equal to the change in internal energy).