Calorimetry Questions

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Karen Ung 2H
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am
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Calorimetry Questions

Postby Karen Ung 2H » Sun Jan 21, 2018 5:36 pm

Do we have to add the heat of the calorimeter with the heat of the reaction to calculate the amount of heat released/absorbed from the reaction? Or is the heat absorbed by the calorimeter negligible?

vicenteruelos3
Posts: 42
Joined: Fri Sep 29, 2017 7:07 am

Re: Calorimetry Questions

Postby vicenteruelos3 » Sun Jan 21, 2018 6:29 pm

the heat absorbed by the calorimeter is equal to the heat released by the reaction

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

Re: Calorimetry Questions

Postby Aijun Zhang 1D » Sun Jan 21, 2018 6:31 pm

I think we should add it.
There is an example from my discussion.

20g of Cu at 45 degrees celsius is placed into 10g of water at 90 degree celsius. Find the final temperature.
If the question is set in calorimeter, then q(cu)+q(cal) = -q(H2O)
Because copper and calorimeter absorb heat, so the total heat is added. And it is equal to the heat released by water.


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