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Katie Lam 1B
Posts: 52
Joined: Fri Sep 29, 2017 7:06 am


Postby Katie Lam 1B » Fri Jan 12, 2018 3:00 pm

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 =3.50 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?

The solutions manual doesn't mention anything about the volumes or molarities of the solutions. Do these not have any effect on the internal energy?

Thuy-Anh Bui 1I
Posts: 56
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 2 times

Re: 8.25

Postby Thuy-Anh Bui 1I » Fri Jan 12, 2018 3:24 pm

Since the question states that this is occurring in a constant-volume calorimeter, you don't need to worry about volumes. In addition, the heat capacity of a calorimeter is an extensive property so moles don't matter. Using q=CΔT rearranged to C=q/ΔT, you can calculate this heat capacity and then use it to calculate qcalorimeter and qreaction.

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