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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.**

C

C

[tex]\Delta[/tex]H

C

_{p, ice}=38.07J^{-1}mole^{-1}C

_{p, liquid water}=75.3JK^{-1}mol^{-1}[tex]\Delta[/tex]H

_{melting}= 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

-q

nC

I did the calculations:

-(50)(75.3)(T

And I get T

Someone in class told me that T

Can anyone help me out?

Thanks!

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

-q

_{water}= q_{ice}ornC

_{liquid}(T_{f}-T_{i}) = nC_{ice}(T_{f}-T_{i}) + n[tex]\Delta[/tex]H_{melting}I did the calculations:

-(50)(75.3)(T

_{f}-293) = (20)(38.07)(T_{f}-253) + (20)(6.007x10^{3})And I get T

_{f}= 259.7KSomeone in class told me that T

_{f}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!