Psichyk
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Need help with thermal equilibrium
Hi guys, I need some advise with an exercise about thermal equilibrium. I hope you can help me. It says:
"50 g of ice at 263 K are mixed adiabatically and at a constant pressure with 50 g of liquid water at 303 K.
a) What is the final temperature of the mixture?
b) Will be ice present in the equilibrium? How much??"
Specific heat of liquid water: 1 cal/g
Specific heat of ice: 0.5 cal/g
Enthalpy of fusion: 80 cal/g
Thanks.
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Magpie
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What about this don't you understand? Have you tried to work this problem? If so, show us how you did it.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Psichyk
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Well, I indeed tried to solve that problem, but I'm sure I didn't do it correctly. I've tried this:
Symbols used:
Q-> Heat
m-> Mass
T-> Unknown temperature
t-> Temperature
cp -> Specific heat
Since the process is adiabatic, QAbsorbed = -QReleased
Q = m * cp * Δt or Qt1t2 = m * cp * (t2 -
t1)
QAbsorbed = QIce 263273.15 + ΔHfus + QWater 273.15T
QReleased = QWater 303T
Replacing symbols:
50 * 0.5 * (273.15-263) + 80 * 50 + 50 * 1 * (T - 273.15) = -50 * 1 * (T - 303)
So:
T = 245.538 K, which is impossible.
I have no clue about how to solve this, so I appreciate your help, guys.
Thanks.
[Edited on 11-10-2016 by Psichyk]
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Magpie
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calories needed to bring water to the fp: (303°-273°)(50g)(1 cal/°-g) = 1500 cal
calories needed to to bring ice to the fp: (273°-263°)(50g)(0.5 cal/°-g) = 250
This should get you started.
The single most important condition for a successful synthesis is good mixing - Nicodem
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DraconicAcid
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When using a symbol for heat, it MUST be a lower case q. Capital letters are for state functions like enthalpy, entropy, and free energy. Heat is a
path function, and must be lower case.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Magpie
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Quote: Originally posted by Magpie  | calories needed to bring water to the fp: (303°-273°)(50g)(1 cal/°-g) = 1500 cal
calories needed to to bring ice to the fp: (273°-263°)(50g)(0.5 cal/°-g) = 250
This should get you started. |
to continue:
calories available to melt ice = 1500-250 = 1250 cal
ice melted = 1250 cal/(80 cal/g) = 15.6g
temp = fp = 273°K
[Edited on 12-10-2016 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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Psichyk
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I'm sorry for not replying before, I've had exams this week, but thank you so much for the help, guys.
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DraconicAcid
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I remember putting a question like that on a midterm for a second-year P-chem course. Some students thought the temperature of the water went up.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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