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Rohan Chaudhari- 1K
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Postby Rohan Chaudhari- 1K » Mon Jan 15, 2018 1:38 pm

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

For this question, I used C=q/deltaT to figure out the heat capacity of the calorimeter, but am unsure on what to do next. What is the next step in this problem?

Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.25

Postby Yashaswi Dis 1K » Mon Jan 15, 2018 2:17 pm

-qcal = q (of the main reaction). You can find delta U from there since w=0 because there's constant volume (delta V is 0) so delta U is equal to q. There's a thread on this same question too which I answered in a little bit more detail and can check out here: viewtopic.php?f=75&t=25170

Hope that helps!


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