8.41

Moderators: Chem_Mod, Chem_Admin

Aya Shokair- Dis 2H
Posts: 58
Joined: Fri Sep 29, 2017 7:07 am

8.41

Postby Aya Shokair- Dis 2H » Thu Jan 18, 2018 9:25 pm

A 50.0-g ice cube at 0.0 C is added to a glass containing
400.0 g of water at 45.0 C. What is the fi nal temperature of
the system (see Tables 8.2 and 8.3)? Assume that no heat is lost
to the surroundings.

I understand that q(ice)=-q(water)
So i imagined that we would solve it like this
m(ice) x C(s,ice) x change in temp (ice) = - m(water) x C(s, water) x change in temp (water)

But in the solutions they did something different, and I'm confused.

Lucian1F
Posts: 87
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

Re: 8.41

Postby Lucian1F » Thu Jan 18, 2018 9:29 pm

The temperature of the water is higher than 0 degrees Celsius so you need to take the amount of energy it takes to melt the given amount of ice first and consider that in your calculations.

Payton Schwesinger 1J
Posts: 45
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.41

Postby Payton Schwesinger 1J » Fri Jan 19, 2018 2:45 pm

On the left side of your equation you have to also account for the amount of heat that will be used to melt the ice into water before the temperature equalizes.

Jessica Jones 2B
Posts: 80
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.41

Postby Jessica Jones 2B » Fri Jan 19, 2018 3:05 pm

You will need to account for the phase change from ice to water, deltaHfusion.
So your q-ice would include both the mCdeltaT and deltaHfusion.


Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 2 guests