8.19

Moderators: Chem_Mod, Chem_Admin

Maeve Gallagher 1J
Posts: 56
Joined: Fri Sep 29, 2017 7:07 am

8.19

Postby Maeve Gallagher 1J » Tue Jan 16, 2018 5:18 pm

Calculate the heat that must be supplied to a 500.0-g copper kettle containing 400.0 g of water to raise its temperature from 22.0 'C to the boiling point of water, 100.0 'C.

Why does the solutions manual have you add the q of the copper and the q of the water instead of setting q copper = -q water as it does for other similar problems?

Rana YT 2L
Posts: 49
Joined: Thu Jul 27, 2017 3:01 am

Re: 8.19

Postby Rana YT 2L » Tue Jan 16, 2018 7:56 pm

the q values for the water and the copper are added in the solution manual because no heat is technically being lost. Instead, heat is simply being transferred into the kettle and then into the water. Because of this, there is no loss in heat in either kettle nor the copper, meaning that no negative is necessary

Wenting Hu 2H
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am

Re: 8.19

Postby Wenting Hu 2H » Thu Jan 18, 2018 11:06 am

For q(copper)= - q(water), does it matter where the negative sign is? I learned today in discussion that it did but then it doesn't make sense to me. Isn't the water gaining the heat that the copper lost? So shouldn't it be -q(copper)=q(water)?


Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 0 guests