1. Heating / Cooling Curve Calculations
EQ: Why is an ideal heating curve not a straight line graph?

2. Ideal Heating/Cooling Curve
1) Heating a solid
2) Change from a solid to liquid
3) Heating a liquid
4) Change from a liquid to a gas
5) Heating a gas
gas
solid & liquid
liquid
liquid & gas
solid present

3. On the back of you notes write:

4. 1) Diagonal lines: KE changes Flat lines: Potential Energy changes
∆Hvap > ∆Hfus because it requires more energy

5. y= mx + b
q= mc∆T
The slope of the diagonal line is determined by the specific heat

6. Example
How much heat is needed to convert 50.0 grams of water from a temperature of 70.0 degrees to degree steam?
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

7. 1st: sketch part of the graph you need
110.0- 2
100.0-
∆T= 10.0C 1 3
70.0-
∆T= 30.0C
Heating: Draw graph going up
0.0-

8. 2nd: Perform calculation for each line segment
1) Diagonal (q = m c DT)
q = 50.0 g x J/goC x (30.0oC )
1 KJ
= 6.28 kJ
q = 6280 J x
1000 J

9. 1st: sketch part of the graph you need
110.0- 2
100.0-
∆T= 10.0C 1 3
70.0-
∆T= 30.0C
0.0-

10. 2) Flat line (Hvap = 40.7 kJ/mol)
1 mol H2O
40.7 kJ
= 114 kJ
50.0 g H2O x x
18.02 g H2O
1 mol H2O

11. 1st: sketch part of the graph you need
110.0- 2
100.0-
∆T= 10.0C 1 3
70.0-
∆T= 30.0C
0.0-

12. 3) Diagonal (q= mc∆T) q = (50.0 g)(2.01 J/goC)(10.0oC) q = 1010 J
1 KJ
= 1.01 kJ x
1000 J

13. 3rd : Add them up + +
1.01 kJ
= 121 kJ
6.28 kJ
114 kJ

14. Summary
Use H
(Flat lines)
Use q = m x c x DT (Diagonal lines)

15. When dealing with a cooling curve…
draw the curve backwards
∆H solidification & condensation are negative

16. Notebook Problems
Calculate the amount of heat needed to convert 10.0 grams of ice from a temperature of -23.0oC to water at 27.0oC.
Calculate the amount of heat released when 50.0 grams of steam at a temperature of 123.0oC cools into water at 77.0oC.

17. Answers
1) 4.94 kJ
2) 120. kJ