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### 8.25

Posted: Fri Mar 09, 2018 11:44 pm
Problem is: 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 (q3.50 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 neutralization reaction?

What is this question asking for and what did they do conceptually? Why is the answer -Qcalorimeter? Also, why does the rest of the problem regarding HBr and KOH not matter?

### Re: 8.25

Posted: Sat Mar 10, 2018 9:09 am
The problem is asking you to calculate ΔU, the change in the internal energy. It is given that it is a constant-volume calorimeter, so you know that there is no ΔV therefore, w=0. That being said, we know that ΔU=q. Use the equation -qcal=CΔT to solve for C---you should get the value -478.1 J. We use -q because the calorimeter is said to be releasing heat in the given problem. Next, you use the equation ΔU=CΔT to get -1.2 kJ.