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General : Vaporization Question
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 Message 1 of 2 in Discussion 
From: MSN NicknameN_2006  (Original Message)Sent: 3/6/2008 8:38 PM
A 50.0 g piece of iron at 152 degrees C is dropped into a 20.0 g H2O(l) at 89 degrees C in an open, thermally insulated container.  How much water would you expect to vapourize, assuming no water splashes out?
 
Specific heats of iron and water are  0.45 and 4.21 J/g/C
and delta H of vap = 40.7kJ/mol H2O
 
 
Thanks..... I tired to being it but I don't know which to start from, or which equations to use.  I know ( at least I think so) the delta H condensation = Hliquid - Hvapour = -delta H vapourization, will be used to calculate the answer.


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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 3/6/2008 11:26 PM
Since some water is going to vaporize, we can assume that the final temperature of the iron and liquid water remaining is 100°C.
 
Here's an outline of the steps:

1)  First calculate the heat lost by the hot iron:

     Heat1  =  mass of iron  X  ΔT of the iron  X  0.444 J/g°C

2)  Some of this heat will be used to heat the water from 89°C to 100°C.  This heat is

     Heat2  =  mass of water  X  ΔT of the water  X  4.184 J/g°C

3)  The extra heat from the iron that will vaporize some of the water is

     Heat3  =  Heat1 �?Heat2

4)  So finally, the amount of water vaporized at 100°C is calculated from

     Heat3   =   mass of water vaporized  X  heat of vaporization of water
                =   mass in grams  X  2264 J/g
 

I used my book's values for the specific heats of iron and water, and the heat of vaporization of water in J/g.  They are a little different from yours but not too much. 

Steve