[BrandonTex]

I need to find the heat capacity of calorimeter. any help or answers on how to find this are much appreciated!

I have:
ΔH° = 64163 J/mol
n = 0.05 mol
change in temperature for the acid and the calorimeter = 5.97°C
change in temperature of the base = 6.31°C
volume of both base and acid = 50 mL
heat capacity of water = 4.16 J/mL °C

PLEASE I NEED this answered as soon as possible and appreciate all the help I can get!

Thank you.

[Borek]

Please read forum rules.

You have to write heat balance equation for your system.

[BrandonTex]

Here is the heat balance for my system:

- qrxn = - nΔH°= CcalΔTcal + CvVacidΔTacid + CvVbaseΔTbase


wat do i need to do now?

i got an answer of 109.6 and the girl im comparing my answer to, she got only 34.5

can u or someone tell which one is correct and how to go about getting this right answer?

Thank you 

[Borek]

change in temperature for the acid and the calorimeter = 5.97°C
change in temperature of the base = 6.31°C

Please elaborate. Description of the experiment/procedure won't hurt, right now it is hard to tell what is wrong. Most likely sign is reversed, but which and why - no idea.

[BrandonTex]

Ive given everything I have as the question states. I just need to figure out how to go about calculating the calorimeter or if u could tell me wat you would get from your own calculations for the calorimeter according to the variables ive listed above.

Thank you 

[Arkcon]

What is the formula you're using, from your text?  And what values, are you plugging in, and where.  There's a good chance, by the time you're done typing that, that you'll have solved the problem yourself, but still, this is work we all want to see.

[BrandonTex]

i used a formula I derived myself from the equation i showed above:

Original:- qrxn = - nΔH°= CcalΔTcal + CvVacidΔTacid + CvVbaseΔTbase

Derived: - nΔH° = (Ccal + CvVacid)(Tf – Ti) + CvVbase(Tf-Ti)

I need to plug them into the original equation above (derived a eq. only because I got a larger number than friend):
ΔH° = 64163 J/mol
n = 0.05 mol
change in temperature for the acid and the calorimeter = 5.97°C
change in temperature of the base = 6.31°C
volume of both base and acid = 50 mL
heat capacity of water = 4.16 J/mL °C

Using derived: I got 109.6
Compared to a friend: She got 34.5

I need you or someone to calculate it for me and tell me if im right or wrong.

Thank you, that’s all the info I have. 

[Borek]

Question - as listed twice - has no solution. We don't know why some acid and some base have changed their temperatures. I can try to guess, but I will be never sure. At the same time if you are not able to supply this information it may mean you don't understand what is going on and what is important in the question. That never helps in solving.