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Calcfrenzy
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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.
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?
As far as I know, Q = mc[itex]\Delta[/itex]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).
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[itex]\Delta[/itex]T = 210(0.900)(0 - 19) = -3591 J
Qwater = mc[itex]\Delta[/itex]T = 624(4.186)(0 - 19) = -49629 J
Qice1 = mc[itex]\Delta[/itex]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.
Thank you in advance!
Homework Statement
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?
Homework Equations
As far as I know, Q = mc[itex]\Delta[/itex]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).
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[itex]\Delta[/itex]T = 210(0.900)(0 - 19) = -3591 J
Qwater = mc[itex]\Delta[/itex]T = 624(4.186)(0 - 19) = -49629 J
Qice1 = mc[itex]\Delta[/itex]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.
Thank you in advance!
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