1. Cooling Curve Calculations

2. When dealing with a cooling curve…
draw the curve backwards
∆Hcondensation = - ∆Hvaporization and ∆Hsolidification = - ∆Hfusion
Specific heat values stay the same
C ice = 2.03 J/gºC
C water = J/gºC
C steam = 2.01 J/gºC
Hfus = 6.01 kJ/mol
Hvap = 40.7 kJ/mol

3. Example
How much heat is released when 10.0 g of steam at 125.0oC is cooled completely into ice at 0.0oC?

4. Example
How much heat is released when 10.0 g of steam at 125.0oC is cooled completely into ice at 0.0oC?
125.0-
∆T = – 125.0
= –25.0 1
100.0- 2
∆T = 0.0 – 100.0
= –100.0 3
0.0- 4

5. = -5030 J 1 KJ = -0.503kJ 1) q = 10.0 g x 2.01 J/goC x (-25.0oC ) x
Csteam
= J
1 KJ
= kJ
1) q = 10.0 g x 2.01 J/goC x (-25.0oC ) x
1000 J
∆T = -25.0
∆T =

6. 1 mol H2O = -22.6 kJ 2) 10.0 g H2O -40.7 kJ x x 18.02 g H2O 1 mol H2O
-∆Hvap
∆T =

7. 1 KJ 3) q = 10.0 g x 4.184 J/goC x -100.0oC ) = -4180 J = - 4.18kJ x
Cwater
1 KJ
3) q = 10.0 g x J/goC x oC )
= J
= kJ x
1000 J
∆T = -25.0
∆T =

8. 1 mol H2O = -3.34 kJ 4) 10.0 g H2O -6.01 kJ x x 18.02 g H2O 1 mol H2O
∆T =
-∆Hfus

9. kJ
Total kJ
- 4.18kJ + kJ kJ
= kJ