# Calorimetry Please HELP Test On Wednesday

1. Nov 5, 2007

### BuBbLeS01

Calorimetry....Please HELP!!! Test On Wednesday!!

1. The problem statement, all variables and given/known data
500g of soda in a 90g glass at 20 deg C, an ice cube of mass 100g is added. Will all of the ice melt? What is Tf?
Tice = -20 deg C

2. Relevant equations
calorimeter - glass
What loses heat - glass and soda
What gains heat - ice
Mh20 = Mice

3. The attempt at a solution
Okay so I wrote down then different Q's that I need to calculate to figure out if all the ice melts...
Qglass = Mg*Cg*CH Temp (What temps do I use?)
Qh20 = Mh20 * Ch20*CH Temp (What temps do I use?)
Qice to 0 deg C = Mice*Cice*(0- -20)
Qheat fusion at 0 deg C = Mice*Lfice
Qh20 to Tf = Mh20*Ch20*(Tf-0)

Then I have to separate the Q's into the energy required to Tf and the energy required to melt the ice?? This is where I am confused

2. Nov 5, 2007

### Antineutron

What does the sum of all the Qs equal to? What temperature did the soda and the glass start out as? What does change in temperature mean?

3. Nov 5, 2007

### BuBbLeS01

Sum of all Q's = 0.
Soda and glass were at a temperature of 20 deg C initially so do I just put Tf - 20? Since the Tf is unknown and the same for both?

4. Nov 5, 2007

yes.

5. Nov 6, 2007

### BuBbLeS01

Now this is where I get confused...separating the Q's to get the heats to cool the soda and to melt the ice...

Qglass = Mg*Cg*(Tf-20)
Qh20 = Mh20 * Ch20*(Tf-20)
Qice to 0 deg C = Mice*Cice*(0- -20)
Qheat fusion at 0 deg C = Mice*Lfice
Qh20 to Tf = Mh20*Ch20*(Tf-0)

So to melt the ice I would use...Qice to 0 deg C + Qheat fusion at 0 deg C + Qh20 to Tf

And the other would be Qglass + Qh20?? I don't understand this one.

6. Nov 6, 2007

### Antineutron

Although the sum of all the Q's are 0, make sure you put the correct sign for each Q. They are not all positive.

7. Nov 6, 2007

### BuBbLeS01

-Qglass = Mg*Cg*(Tf-20)
-Qh20 = Mh20 * Ch20*(Tf-20)
Qice to 0 deg C = Mice*Cice*(0- -20)
Qheat fusion at 0 deg C = Mice*Lfice
Qh20 to Tf = Mh20*Ch20*(Tf-0)

If it loses heat it is negative right? Now what do I do?

8. Nov 6, 2007

### learningphysics

The water/ice will not change temperature until all of it melts...

So the rest of the materials will keep dropping in temperature until the ice melts...

are you using glass to refer to glass+soda?

seems to me like you only need these:
Qglass = Mg*Cg*(Tf-20)
Qh20 to Tf = Mh20*Ch20*(Tf-0)
Qheat fusion at 0 deg C = Mice*Lfice

the glass+soda goes from 20 to the Tf... the ice melts at 0 degrees... the water it melts into goes from 0 to Tf...

these 3 add to zero:
Mg*Cg*(Tf-20) + Mh20*Ch20*(Tf-0) + Mice*Lfice = 0

Tf can't be a negative number... if the solution to the above turns to be Tf<0 that means that the ice doesn't all melt and Tf = 0.

or another way to check if all the ice melts... get Mice*Lfice. Now also get:
-Mg*Cg*(0-20)... that's the heat lost by the soda+glass in coming down to 0 degrees... if the heat lost is less than Mice*Lfice... then all the ice doesn't melt and the final temperature is 0.

9. Nov 6, 2007

### BuBbLeS01

No I was doing the Qglass and Qwater seperate.

All I know is that we did this problem in class a week ago and thought I understood it and now I am trying to do it and I can't and I am very confused and its on our test tomorrow :(.

In class we used...
Qice = (Mice * CH T) * (Cice + Mice * Lf) = 37,420 J

Qcool = (Msoda * Cwater * CH Tsoda) + (Mglass * Cglass * CH Tglass) = -42,995 J

I don't know how to pick what Q's go into Qice and Qcool? And why I came up With 5 Q's and we only are using 4?

10. Nov 6, 2007

### learningphysics

Can you post Cice, Lf, Cwater and Cglass if you have them?

11. Nov 6, 2007

### BuBbLeS01

Yes....
Cice = 2.06 J/g Deg C
Lf = 333 J/g
Cwater = 4.18 J/g Deg C
Cglass = 0.664 J/g Deg C

12. Nov 6, 2007

### learningphysics

Well... I'm confused by what you were doing in class... the ice won't change temperature unless it goes below 0 right? from 0 onwards the ice is in its melting phase... ?

here's how I'd do the problem... don't know if it's completely right...

msoda*csoda*(Tf-20) + mglass*cglass*(Tf-20) + 333*100 + 100*cwater*(Tf-0) = 0

500*4.18*(Tf-20) + 90*0.664*(Tf-20) + 33300 + 100*4.18*Tf = 0

so I'm using csoda = cwater = 4.18

2567.76 Tf = 9695.2

Tf = 3.7757 degrees Celsius

is this what you got in class?

13. Nov 6, 2007

### BuBbLeS01

We got 2.17 deg celsius

14. Nov 6, 2007

### BuBbLeS01

We got 2567.76 but instead of 9695.2 we got 5575.2

15. Nov 6, 2007

### BuBbLeS01

how did u figure out if the ice melted?

16. Nov 6, 2007

### learningphysics

Oh... I didn't notice that Tice = -20C ok... now I see...

msoda*csoda*(Tf-20) + mglass*cglass*(Tf-20) + 100*cice(0-(-20)) + 333*100 + 100*cwater*(Tf-0) = 0

500*4.18*(Tf-20) + 90*0.664*(Tf-20) + 100*2.06*20 + 33300 + 100*4.18*Tf = 0

2567.76 Tf = 5575.2

Tf = 2.17 degrees celsius

all the ice melts.

17. Nov 6, 2007

### BuBbLeS01

Oh good you got the same answer! So in class she told us in order to determine whether the all the ice melts or not you have to find the amount of heat it takes to melt the ice and the amount of heat it takes to cool the glass. So we got...
Qice = 37420 J
Qcool = 42995 J
Do you know how we got those? If so can you explain it to me?

18. Nov 6, 2007

### learningphysics

If Tf comes out >=0 ... that means all the ice melted...

If Tf comes out <0... then the ice didn't melt... (reason is we have an internal contradiction... assuming the ice melts leads to Tf<0 which implies the ice doesn't melt... that means our initial assumption is wrong... and Tf may be wrong too).

If Tf comes out <0 then:

1) we should see ho much energy is obtained from the temperature of the soda and glass going to 0 degrees... see how much energy is required to get the ice up to 0 degrees. if the energy from the soda and glass is < thant the temperature required to get the ice up to 0 degrees... that means that the ice doesn't melt at all...

2) if the ice doesn't melt at all... then we can find Tf again, but without using the heat of fusion or the melted water...

3) if there is enough energy from the soda+glass to get the ice up to 0... then Tf is 0. then we can see the difference in energy and see how much of the ice melts...

19. Nov 6, 2007

### learningphysics

yes, that method also makes sense... I think that if Tf comes out greater than or equal to 0... then everything is good and the ice has melted... but I might have missed something...

Qice is the energy the ice gains in going from -20 to 0... and then melting (heat of fusion)

So that comes out to: 100*2.06*(0-(-20)) + 333*100 = 37420 J

Qcool is the energy the glass and soda lose in going from 20C to 0...

So that comes out to (since it's energy lost I take initial temp.-final temp instead of final temp-initial temp.): 500*(20-0)*4.18 + 90*(20-0)*0.664 = 42 995.2 J

Last edited: Nov 6, 2007
20. Nov 6, 2007

### BuBbLeS01

Thank you so much for helping me!!!

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