Moderators: Chem_Mod, Chem_Admin

Abigail Urbina 1K
Posts: 102
Joined: Thu Jul 27, 2017 3:01 am


Postby Abigail Urbina 1K » Sun Jan 21, 2018 2:52 pm

When calculating q for the calorimeter, why are you multiplying it by 2.49 K and not 7.32 K? How do you know which value to use?

Nehal Banik
Posts: 64
Joined: Thu Jul 13, 2017 3:00 am

Re: 8.25

Postby Nehal Banik » Sun Jan 21, 2018 3:14 pm

The calorimeter is being calibrated, therefore when it is calibrated the specific heat capacity of the calorimeter is being redefined, therefore you need to find the heat capacity of the calorimeter, then it states that there is a neutralization reaction and all that does it cause the temperature to increase again, but in this case you know the reaction is the exothermic because it is causing the temperature to go up, therefore the change in internal energy will be equal to the heat gained or lost because assuming there is no expansion or compression occurring. Then you multiply the specific heat capacity of the newly calibrated calorimeter and then multiply it by the change in the temperature due to the reaction and you find the change in heat that is caused by the reaction which should be -1.2 kj or something I'm not sure I don't have the exact answer in front of me. Hope that makes sense!

Return to “Concepts & Calculations Using First Law of Thermodynamics”

Who is online

Users browsing this forum: No registered users and 2 guests