# Homework Help: Thermodynamics Calorimetry Equilibrium Problem

1. Sep 7, 2011

### Calcfrenzy

My Physics III class is going through basic Thermodynamics and I have hit a bit of an impasse. This is relatively simple stuff, but for some reason, I cannot get this problem right. I've worked on it to the point of confusing myself pretty thoroughly. I would really appreciate some direction and some critiquing on what I am doing wrong.

1. The problem statement, all variables and given/known data
Water has a specific heat of 4.186 J/g · K and a heat of fusion of 334 J/g. Aluminum has a specific heat of 0.900 J/g · K

A 210-g aluminum calorimeter contains 624 g of water at 19° C. A 92-g piece of ice cooled to -19° C is placed in the calorimeter. (Assume that the specific heat of ice is always 2.02 J/g · K.)
(a) Find the final temperature of the system, assuming no heat is transferred to or from the system.

(b) A 206-g piece of ice at -19° C is added. How much ice remains in the system after the system reaches equilibrium?

2. Relevant equations

As far as I know, Q = mc$\Delta$T and Q = mL are the only equations needed for this problem (though that may very well be incorrect considering that I am failing to find the correct answer).

3. The attempt at a solution

I'll begin with part b) as that is the part I felt relatively confident about.

I set out by setting up the heat energies of each item in this system:

Qalu = mc$\Delta$T = 210(0.900)(0 - 19) = -3591 J

Qwater = mc$\Delta$T = 624(4.186)(0 - 19) = -49629 J

Qice1 = mc$\Delta$T = 206(2.02)[0 - (-19)] = 7906 J

Qice2 = mL = 334m

You'll notice that I assumed the equilibrium temperature is 0. I assumed that since according to the problem both ice and water remain in the system at equilibrium (apparently all the ice doesn't melt). I then set the expressions for the items that were losing heat (aluminum and water) equal to the items which absorbed (the ice). I also absolute valued the heat loss for the aluminum and water since I thought that was merely an expression of the fact that they were losing heat (is that where I went wrong?). Anyhow:

3591 J + 41676 J = 7906 J + 334m

334m = 37361 J

m = 136 g

Thus leaving 206g - 136g = 70g of ice unmelted in the system. This is incorrect. What am I doing wrong? I have exhausted all of my own ideas. Any help at all would be appreciated. As for problem a), even my professor wasn't exactly sure how to approach that one when I asked her, so I'll cross that bridge when I come to it.

Last edited: Sep 8, 2011
2. Sep 7, 2011

### RTW69

Make sure temperatures are in Kelvin not centigrade

3. Sep 8, 2011

### RTW69

The mass of water is 624 grams NOT 524 grams

What gets cooled = what gets heated

Your signs are incorrect for Q-alm and Q-water, they should be positive but you changed the signs to positive in the last step, not sure why.

4. Sep 8, 2011

### Calcfrenzy

Ah, thanks for catching the typo. I actually entered the un-typo-ed answer, however, and that didn't work. What exactly is the sign error I made? Could you expound upon that a little bit more? The negative of the energy lost by the water and aluminum should equal that gained by the ice -(Qwater + Qalu) = Qice. Is that incorrect?

Thanks!

5. Sep 8, 2011

### Calcfrenzy

Ah, I suppose I should have mentioned that I am working with an online homework system (Webassign).

6. Sep 8, 2011

### RTW69

OK I see what you did Qcold=-Qhot. Signs are OK

I think your answer is correct. 136 grams of ice are melted, 70 grams remain
I had a slightly different answer 135 grams melted, 71 grams remain but we are talking rounding differences.

Same approach for part "a" but all the ice will melt and water temp will be above 0 c, solve for final temperature

7. Sep 8, 2011

### Calcfrenzy

I broke part a) up into two parts like you said, went at it, and got an answer of 5.95oC which is correct! Part b) still evades me at the moment, but I'll work on it some more and let you know. Thank you for all your help and advice!