4A.9 Help

Moderators: Chem_Mod, Chem_Admin

Andrew Pfeiffer 2E
Posts: 101
Joined: Sat Sep 28, 2019 12:16 am

4A.9 Help

Postby Andrew Pfeiffer 2E » Sun Feb 09, 2020 7:50 pm

I was working on the following homework problem 4A.9:

A piece of metal of mass 18.0 g at 100.0 C is placed in a vessel of negligible heat capacity but containing 50.2 g of water at 22.0 C. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings.

I ran into trouble setting up the equation. I know that we have to differentiate between Tfinal and T initial, but do we have to include initial and final masses or initial? And which heat capacity would we use: water's or copper's?

madawy
Posts: 81
Joined: Fri Aug 09, 2019 12:17 am

Re: 4A.9 Help

Postby madawy » Sun Feb 09, 2020 7:54 pm

should be something like this:
heat released = heat absorbed
20 (mass)*0.39 (specific heat of Cu) *(100-T) (change in temperature) = 50.7 (mass)* (specific heat of H2O) *(T-22)(change in temperature)
20*0.39*(100-T) = 50.7*4.18*(T-22)
= about 25

EthanPham_1G
Posts: 104
Joined: Sat Jul 20, 2019 12:17 am

Re: 4A.9 Help

Postby EthanPham_1G » Sun Feb 09, 2020 7:55 pm

Andrew Pfeiffer 2E wrote:I was working on the following homework problem 4A.9:

A piece of metal of mass 18.0 g at 100.0 C is placed in a vessel of negligible heat capacity but containing 50.2 g of water at 22.0 C. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings.

I ran into trouble setting up the equation. I know that we have to differentiate between Tfinal and T initial, but do we have to include initial and final masses or initial? And which heat capacity would we use: water's or copper's?


To solve for this problem, know that the heat lost from the metal is transferred to the water. Therefore, -q metal=q water.
Use the equation q=mCdelta T for both water and the metal (use their respective heat capacities) and set them equal and opposite to each other according to -q metal=q water. By doing this, you can solve for the final temperature.

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

Re: 4A.9 Help

Postby kendal mccarthy » Sun Feb 09, 2020 10:30 pm

I was having trouble with this one too. I think the Cu final should be the same as the Water final.
The setup should be:
20*0.39*(T-100) = 50.7*4.18*(T-22)

instead of 20*0.39*(100-T) = 50.7*4.18*(T-22)


Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest