1. Section 7.3—Changes in State
What’s happening when a frozen ice pack melts?

2. Change in State
The energy being put into the system is used for breaking IMF’s, not increasing motion (temperature)
Breaking intermolecular forces requires energy
To melt or boil, intermolecular forces must be broken
A sample with solid & liquid will not rise above the melting point until all the solid is gone.
The same is true for a sample of liquid & gas

3. Melting
When going from a solid to a liquid, some of the intermolecular forces are broken
The Enthalpy of Fusion (Hfus) is the amount of energy needed to melt 1 gram of a substance
The enthalpy of fusion of water is cal/g or 334 J/g
All samples of a substance melt at the same temperature, but the more you have the longer it takes to melt (requires more energy).
Energy needed to melt 1 g
Energy needed to melt
Mass of the sample

4. Example
Example:
Find the enthalpy of fusion of a substance if it takes 5175 J to melt 10.5 g of the substance.

5. Example
Example:
Find the enthalpy of fusion of a substance if it takes 5175 J to melt 10.5 g of the substance.
H = enthalpy (energy)
m = mass of sample
Hfus = enthalpy of fusion
Hfus = 493 J/g

6. Vaporization
When going from a liquid to a gas, all of the rest of the intermolecular forces are broken
The Enthalpy of Vaporization (Hvap) is the amount of energy needed to boil 1 gram of a substance
The Hvap of water is cal/g or 2287 J/g
All samples of a substance boil at the same temperature, but the more you have the longer it takes to boil (requires more energy).
Energy needed to boil 1 g
Energy needed to boil
Mass of the sample

7. Example
Example:
If the enthalpy of vaporization of water is cal/g, how many calories are needed to boil 25.0 g of water?

8. Example
Example:
If the enthalpy of vaporization of water is cal/g, how many calories are needed to boil 25.0 g of water?
H = enthalpy (energy)
m = mass of sample
Hfus = enthalpy of fusion
DH = 1.37×104 cal

9. Changes in State go in Both Directions
Increasing molecular motion (temperature)
Changes in State go in Both Directions
Gas
Vaporizing or
Evaporating
Liquid
Melting
Condensing
Solid
Freezing

10. Going the other way
The energy needed to melt 1 gram (Hfus) is the same as the energy released when 1 gram freezes.
If it takes 547 J to melt a sample, then 547 J would be released when the sample freezes. DH will = -547 J
The energy needed to boil 1 gram (Hvap) is the same as the energy released when 1 gram is condensed.
If it takes 2798 J to boil a sample, then 2798 J will be released when a sample is condensed DH will = J

11. Example
Example:
How much energy is released with g of water is condensed? Hvap water = cal/g

12. Example
Example:
How much energy is released with g of water is condensed? Hvap water = cal/g
H = enthalpy (energy)
m = mass of sample
Hfus = enthalpy of fusion
Since we’re condensing, we need to “release” energy…DH will be negative!
DH = - 8.6×104 cal

13. Heating Curves
Heating curves show how the temperature changes as energy is added to the sample
Boiling & Condensing
Point
Melting & Freezing
Point

14. Going Up & Down
Moving up the curve requires energy, while moving down releases energy
+DH
-DH

15. States of Matter on the Curve
Liquid & gas
Energy added breaks remaining IMF’s
Liquid Only
Energy added increases temp
Gas Only Energy added increases temp
Solid Only Energy added increases temp
Solid & Liquid
Energy added breaks IMF’s

16. Different Heat Capacities
The solid, liquid and gas states absorb water differently—use the correct Cp!
Liquid Only
Cp = 1.00 cal/g°C
Gas Only Cp = 0.48 cal/g°C
Solid Only Cp = 0.51 cal/g°C

17. Changing States Liquid & gas Hvap = 547.2 cal/g Solid & Liquid
Hfus = cal/g

18. Adding steps together
If you want to heat ice at -25°C to water at 75°C, you’d have to first warm the ice to zero before it could melt.
Then you’d melt the ice
Then you’d warm that water from 0°C to your final 75°
You can calculate the enthalpy needed for each step and then add them together

19. Example Useful information: Cp ice = 0.51 cal/g°C Example:
Cp water = 1.00 cal/g°C
Cp steam = 0.48 cal/g°C
Hfus = cal/g
Hvap = cal/g
Example:
How many calories are needed to change 15.0 g of ice at -12.0°C to steam at 137.0°C?

20. Example Useful information: Cp ice = 0.51 cal/g°C Example:
Cp water = 1.00 cal/g°C
Cp steam = 0.48 cal/g°C
Hfus = cal/g
Hvap = cal/g
Example:
How many calories are needed to change 15.0 g of ice at -12.0°C to steam at 137.0°C?
Warm ice from -12.0°C to 0°C
91.8 cal
Melt ice
1213 cal
Warm water from 0°C to 100°C
1500 cal
Boil water
8208 cal
Warm steam from 100°C to 137°C
266 cal
Total energy = cal

21. How many needed to change 40.5 g of water at 25°C to steam at 142°C?
Let’s Practice
Useful information:
Cp ice = 0.51 cal/g°C
Cp water = 1.00 cal/g°C
Cp steam = 0.48 cal/g°C
Hfus = cal/g
Hvap = cal/g
Example:
How many needed to change 40.5 g of water at 25°C to steam at 142°C?

22. How many needed to change 40.5 g of water at 25°C to steam at 142°C?
Let’s Practice
Useful information:
Cp ice = 0.51 cal/g°C
Cp water = 1.00 cal/g°C
Cp steam = 0.48 cal/g°C
Hfus = cal/g
Hvap = cal/g
Example:
How many needed to change 40.5 g of water at 25°C to steam at 142°C?
Warm water from 25°C to 100°C
3038 cal
Boil water
22162 cal
Warm steam from 100°C to 142°C
816 cal
Total energy = cal