Test #7


Moderators: Chem_Mod, Chem_Admin

Nha Dang 2I
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

Test #7

Postby Nha Dang 2I » Sun Feb 11, 2018 12:54 am

For #7, I understand that we have to include deltaH(fusion) into the calculations, but can someone explain where and how? And when calculating q(ice), would we be using the specific heat for ice in this case?

miznaakbar
Posts: 55
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 4 times

Re: Test #7

Postby miznaakbar » Sun Feb 11, 2018 1:45 am

You need to multiply delta H (fusion) by the moles of ice you have so that you account for the energy required to change phases from ice to water. You would then use the specific heat of liquid water in calculations of q, not ice, since you have already accounted for its phase change in your calculations.

Xin He 2L
Posts: 38
Joined: Fri Sep 29, 2017 7:05 am

Re: Test #7

Postby Xin He 2L » Sun Feb 11, 2018 1:47 am

I haven't done the question yet, but you put the enthalpy of fusion onto the ice side of the equation since it's the energy required to melt the ice. The equation should look like this,
Calculations for Ice Calculations for water
[Heat of Fusion of Ice+ n*C*delta T]= [n*C*deltaT]

miznaakbar
Posts: 55
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 4 times

Re: Test #7

Postby miznaakbar » Sun Feb 11, 2018 11:28 am

To specify my earlier response, the equation should look like this, where Temp final is unknown:

(melt ice) + (raise water at 0 degrees C to temp final) = - (decreasing tea water at 30 degrees C to temp final)

we want q(ice cube) = - q (tea water), so the negative sign is crucial:
[(Enthalpy of Fusion of Ice)(mols ice)(1000J/1kJ)] + [mass*C*delta T]= [- mass*C*deltaT]

You can convert the 250 mL of tea into 250g of tea water since you are given density, and then you use the specific heat capacity of water in both mCdeltaT equations (since you have already melted the ice, you will NOT use the specific heat of ice). Also, remember to convert the kJ in enthalpy to joules so that your terms will match up (the conversion factor is included in the above equation). Then, just rearrange the equation until you find Temp final, hope this helps!

Kayla Danesh 1F
Posts: 38
Joined: Thu Jul 13, 2017 3:00 am

Re: Test #7

Postby Kayla Danesh 1F » Sun Feb 11, 2018 8:15 pm

Does anyone have the correct final answer to this problem? I did the problem correctly but got the wrong final number

Xin He 2L
Posts: 38
Joined: Fri Sep 29, 2017 7:05 am

Re: Test #7

Postby Xin He 2L » Sun Feb 11, 2018 11:31 pm

The final answer I got was 15.9 degrees Celsius, and he should be posting the answers to the review guide soon

Mishta Stanislaus 1H
Posts: 45
Joined: Fri Sep 29, 2017 7:04 am

Re: Test #7

Postby Mishta Stanislaus 1H » Sun Feb 11, 2018 11:42 pm

The correct answer given during the review session was 16 degrees Celsius.

Lorie Seuylemezian-2K
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: Test #7

Postby Lorie Seuylemezian-2K » Mon Feb 12, 2018 9:56 pm

Can someone please show me their work with the numbers for this no matter what I do i keep getting the wrong answer?


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

Who is online

Users browsing this forum: No registered users and 8 guests