Sciencemadness Discussion Board

Need help with thermal equilibrium

Psichyk - 10-10-2016 at 15:26

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.


Magpie - 10-10-2016 at 15:56

What about this don't you understand? Have you tried to work this problem? If so, show us how you did it.

Psichyk - 10-10-2016 at 16:36

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]

Magpie - 10-10-2016 at 19:33

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.

DraconicAcid - 10-10-2016 at 22:04

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.

Magpie - 11-10-2016 at 20:37

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]

Psichyk - 14-10-2016 at 14:56

I'm sorry for not replying before, I've had exams this week, but thank you so much for the help, guys.

DraconicAcid - 14-10-2016 at 16:36

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.