July 20, 2024, 09:58:07 AM
Forum Rules: Read This Before Posting

### Topic: supercooled water  (Read 6060 times)

0 Members and 1 Guest are viewing this topic.

#### plu

• Full Member
• Posts: 193
• Mole Snacks: +15/-7
• Gender:
##### supercooled water
« on: March 07, 2006, 08:23:49 PM »
Hello,

This is a question about supercooled water that I am having trouble on:

Data: Cp(ice) = 76.1 J/K mol    Cp(water) = 37.15 J/K mol    dHfus(H2O) = 6.01 kJ/mol

Consider 28.5 g of supercooled (liquid) water at -12.0 oC and 1.00 bar.  This metastable state suddenly freezes to ice at the same temperature and pressure.

(i) Treating the metastable state as an equilibirum state, calculate the heat released in this process.

This I calculated to be 1.58 mol x 6.01 kJ/mol = 9.51 kJ

(ii) What is the change in entropy of the system?

I am not sure here.  I figured that the energy released from the sudden fusion of the water would be applied to the newly-formed ice to heat it up to 0 oC .  Then, since dS = nCpln(T2/T1) at constant pressure, dS would equal 5.4 J/K mol...?

(iii) Assuming the temperature of the surroundings is -12 oC, what is the change in entropy of the surroundings?

Would this simply be dSsur = qp / T = -23 J/K mol ?  However, this wouldn't make sense since then dSsystem + dSsur < 0
« Last Edit: March 08, 2006, 10:53:49 PM by plu »

#### gregpawin

• Chemist
• Full Member
• Posts: 245
• Mole Snacks: +22/-5
• Gender:
• Ebichu chu chu chuses you!
##### Re:supercooled water
« Reply #1 on: March 09, 2006, 01:02:40 AM »
i think you use the equation you used for part 3 for part 2
I've got nothin'

#### plu

• Full Member
• Posts: 193
• Mole Snacks: +15/-7
• Gender:
##### Re:supercooled water
« Reply #2 on: March 09, 2006, 09:29:19 PM »
i think you use the equation you used for part 3 for part 2

I do not think this would work since temperature is not constant and thus only the inequality dS > qp/T is true