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Thermodynamics - Phase changes

  • Thread starter psychkub
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  • #1
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1. You have 0.9 L of water in an insulated (thermally isolated) cup at 20 C. You add 360 g of ice at −20 C in the cup to cool the water. Describe the final state of your system in terms of (i) number of moles of ice (ii) number of moles of water and (iii) the final temperature.



2.
Cp, ice=38.07J-1mole-1
Cp, liquid water=75.3JK-1mol-1
[tex]\Delta[/tex]Hmelting = 6.007kJmol-1




3.
First thing I did was find how many moles of water and ice I had (I'm aware I'm using rounded figures for the molar masses)

0.9L * 1000g/L * 1mol H20/18g/mol = 50mol H20
360g of ice * 1mol H20/18g/mol = 20mol ice

It's an isolated system so I know that the final temp of the water will equal that of the ice and hence I can say
-qwater = qice or
nCliquid(Tf-Ti) = nCice(Tf-Ti) + n[tex]\Delta[/tex]Hmelting

I did the calculations:
-(50)(75.3)(Tf-293) = (20)(38.07)(Tf-253) + (20)(6.007x103)

And I get Tf = 259.7K

Someone in class told me that Tf should be greater than 0, which if it is, that would make calculating the moles of water at the end soooo much easier, so either they are wrong, or I am doing something wrong.

Can anyone help me out?

Thanks!
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
859
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-(50)(75.3)(Tf-293) = (20)(38.07)(Tf-253) + (20)(6.007x103)
[/b]
This equation isn't right. There are three cases to consider. If there is so much ice that it doesn't even melt, then the [itex]\Delta H[/itex] term doesn't appear because no melting happens. If there is an amount of ice that makes some of it melt, but not all of it, you will have an equation that is similar to the one above, but the 'n' in front of [itex]\Delta H[/itex] will only be the amount that melts. Finally, all the ice could melt, and then the melted ice could heat up. In this case you don't have nCice(Tf-Ti), you have nCice(0-Ti) since the ice melts at 0C, and then after that it's heat capacity switches to the liquid heat capacity.

You need to figure out what case you're in first.
 
  • #3
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This equation isn't right. There are three cases to consider. If there is so much ice that it doesn't even melt, then the [itex]\Delta H[/itex] term doesn't appear because no melting happens. If there is an amount of ice that makes some of it melt, but not all of it, you will have an equation that is similar to the one above, but the 'n' in front of [itex]\Delta H[/itex] will only be the amount that melts. Finally, all the ice could melt, and then the melted ice could heat up. In this case you don't have nCice(Tf-Ti), you have nCice(0-Ti) since the ice melts at 0C, and then after that it's heat capacity switches to the liquid heat capacity.

You need to figure out what case you're in first.
So in order to figure out what case I'm in do I calculate the energy it takes to bring water from 20C to 0C, so

(50)(75.3)(0-20) = -75,300J

Then figure out the amount of energy it takes to bring the ice from -20C to 0, so

(20)(38.07)(0- (-20)) = 15,228

-75,300+15,228 = 60,072 left to melt the ice

So there is enough energy to bring the ice to 0C, now I need to figure out if there is enough energy to melt the ice. So,

n[tex]\Delta[/tex]H - (20)(6007)= 120,140J to melt all the ice.

So I there is not enough energy to melt all the ice (120140J-60,072 = 60,068J more needed to melt all the ice), so the final temp would be 0C since there is a mixture of both ice and water.

But now, how do I figure out how much ice I have left over?

do I just use algebra to solve for the number of moles left from 60,072J = n[tex]\Delta[/tex]fusion?
 
  • #4
LeonhardEuler
Gold Member
859
1
So in order to figure out what case I'm in do I calculate the energy it takes to bring water from 20C to 0C, so

(50)(75.3)(0-20) = -75,300J

Then figure out the amount of energy it takes to bring the ice from -20C to 0, so

(20)(38.07)(0- (-20)) = 15,228

-75,300+15,228 = 60,072 left to melt the ice

So there is enough energy to bring the ice to 0C, now I need to figure out if there is enough energy to melt the ice. So,

n[tex]\Delta[/tex]H - (20)(6007)= 120,140J to melt all the ice.

So I there is not enough energy to melt all the ice (120140J-60,072 = 60,068J more needed to melt all the ice), so the final temp would be 0C since there is a mixture of both ice and water.

But now, how do I figure out how much ice I have left over?

do I just use algebra to solve for the number of moles left from 60,072J = n[tex]\Delta[/tex]fusion?
Exactly! :smile:
 
  • #5
3
0
Thank you for your help LeonhardEuler!!!
 

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