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### 8.25

Posted: **Sun Jan 21, 2018 5:38 pm**

by **Andrea Grigsby 1I**

for question 8.25, I understand everything up until the point where the answer key says Qreaction + Qcalorimeter =0, which then leads to the final answer. Can someone explain why the heat from the reaction + the heat of the calorimeter equals to 0?

Thanks :)

### Re: 8.25

Posted: **Sun Jan 21, 2018 5:54 pm**

by **Julian Krzysiak 2K**

The regular heat equation for any system/surroundings is Qsystem=-Qsurroundings, because the amount of energy must be conserved between the 2 environments, due to the Law of the Conservation of Energy. In this case, the reaction=system, and surroundings=calorimeter, so Qreaction= -Qcalorimeter.

Then we can just play with the equation, by moving the Qcalorimeter to the other side, which would change the sign to positive, and we would get Qreaction+Qcalorimeter=0

(eg. X=-Y -->X+Y=-Y+Y --> X+Y=0)

### Re: 8.25

Posted: **Sun Jan 21, 2018 11:49 pm**

by **Andrea Grigsby 1I**

oh yea, I completely forgot about that.

Thanks so much!

### Re: 8.25

Posted: **Wed Jan 24, 2018 4:23 pm**

by **Jana Sun 1I**

I had a question on another part of this problem. When we're using q in our calculations, why is our q positive instead of negative (because the question gives us a negative q)? Is it because for the subsequent experiment with HBr and KOH, the temperature rises?