8.19 part a

Moderators: Chem_Mod, Chem_Admin

Abbey Kerscher 3O
Posts: 24
Joined: Fri Jul 22, 2016 3:00 am

8.19 part a

Postby Abbey Kerscher 3O » Mon Jan 23, 2017 1:23 pm

How do you answer this I'm so confused.
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 degrees C to the boiling point of water, 100.0degrees C.

Andrew Nguyen 3G
Posts: 22
Joined: Wed Sep 21, 2016 2:59 pm

Re: 8.19 part a

Postby Andrew Nguyen 3G » Mon Jan 23, 2017 1:27 pm

I'm assuming you need to heat both the copper and the water to the same temperature. Because if they weren't, they'd start doing some 2nd Law of Thermodynamics stuff and the water might transfer its heat out to the copper and vice versa. We wouldn't want that to happen on paper, although it is what happens in real life but its better to pretend that both are being simultaneously heated to avoid complications.

Total Heat = heat for copper + heat for water
Total Heat = m c delta t + m c delta t
500. x c x 78 + 400. x c x 78

Look up whatever C is for copper (I think water is 4.18?) and add it all up.


Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

Who is online

Users browsing this forum: No registered users and 1 guest