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Chemistry Two

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Chemistry II

A 20.0 g sample of ice at -5 degrees C is added to 50.0 g of water at 40 degrees C. a 20.0 g sample of ice at -5 degrees C is added to 50.0 g of water at 40 degrees C. What is the final temperature of the system?

S (of ice)= 2.0 J/gram celcius
S (of water) = 4.18 J/gram celcius
Change in enthalpy of fusion is 6.01 Jk/mole
There are 4 processes in this system :
1. Sample of ice at -5 degrees C raise to ice at 0 degrees C (before it begin to melt)
2. Water at 0 degree C (right after melting from ice)
3. Water at 0 degree C raise to water at x degree C (after mixed with water at 40 degree C)
4. Water at 40 degree C drop to water at x degree C (after mixed with ice).

We have to calculate process 3 and 4 to find out the x, which is the final temperature reached by the system.

Process 3
Q = m*S*delta Temperature
Q = 20 grams (since it WAS ice) * 4.18 J/gram Celsius (because NOW it is water) * [x - To] (because x will be higher than To, which is 0 degree C)
Q = 20 grams * 4.18 J/gram Celsius * (x-0)
Q = 83.6 (x-0) J/ Celsius
Q = 83.6 x - 0

Process 4
Q = m*S*delta Temperature
Q = 50 grams * 4.18 J/gram Celsius * [T1-x] (because x will be lower than T1, which is 40 degrees C)
Q = 209 (40-x)
Q = 8360 - 209x

Then we do the elimination, and you'll find x, which is the final temperature
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