4A11

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kendal mccarthy
Posts: 109
Joined: Wed Nov 14, 2018 12:22 am

4A11

Postby kendal mccarthy » Sun Feb 09, 2020 10:51 pm

"A calorimeter was calibrated with an electric heater, which supplied 22.5 kJ of energy as heat to the calorimeter and increased the temperature of the calorimeter and its water bath from 22.45°C to 23.97°C. What is the heat capacity of the calorimeter?"

I set up this problem doing
qcal=Ccal*deltaT

25kJ=Ccal(23.97-22.45)

Ccal=25kJ/(23.97-22.45)

I solved and got 16.4 but the answer key says 14.8. Am I doing something wrong?

Niharika 1H
Posts: 50
Joined: Thu Jul 25, 2019 12:16 am

Re: 4A11

Postby Niharika 1H » Sun Feb 09, 2020 10:53 pm

The q value given is 22.5, not 25. So:
Ccal=22.5kJ/(23.97-22.45) + 14.8026

kendal mccarthy
Posts: 109
Joined: Wed Nov 14, 2018 12:22 am

Re: 4A11

Postby kendal mccarthy » Sun Feb 09, 2020 10:54 pm

haha omg I just wrote down 25 instead of 22.5 ok that makes sense now

Juliet Stephenson 4E
Posts: 100
Joined: Wed Sep 18, 2019 12:21 am

Re: 4A11

Postby Juliet Stephenson 4E » Sun Feb 09, 2020 10:54 pm

It looks like you're using 25 kJ as the value for qcal but it should be 22.5 kJ.


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