1. Energy Requirements for changing state: In ice the water molecules are held together by strong intermolecular forces. The energy required to melt 1 gram of a substance is called the latent heat of fusion For ice it is 334 J/g The energy required to change 1 gram of a liquid to its vapor is called the latent heat of vaporization For steam it is 2259 J/g

2. It takes more energy to vaporize water than to melt it. This is because in melting you weaken the intermolecular forces. Here about 1/6 of the hydrogen bonds are broken. In vaporization you totally break them. All the hydrogen bonds are broken Fusion is when a solid melts to form a liquid Vaporization is when a liquid evaporates to form a gas.

3. A-B = Solid ice, temperature is increasing. Particles gain kinetic energy, vibration of particles increases. Heating and cooling curve for water heated at a constant rates. Ice

4. B-C = Solid starts to change state from solid to liquid. Temperature remains constant as energy is used to break inter- molecular bonds. H 2 O (s)  H 2 O (  ) energy required  334 J/g 0ºC

5. C-D = temperature starts to rise once all the solid has melted. Particles gain kinetic energy. Liquid water

6. D-E = Liquid starts to vaporize, turning from liquid to gas. The temperature remains constant as energy is used to break inter- molecular forces. H 2 O (  )  H 2 O (g) energy required  2259 J/g 100ºC

7. E-F = temperature starts to rise once all liquid is vaporized. Gas particles gain kinetic energy. steam

8. Calculating Energy Changes TemperatureTemperature Heat Added ( o C)  H f = 334 J/g  H v = 225 J/g 3. c l = 4.18 J/g o C 5. c v = 2.1 J/g o C 1. c s = 2.1 J/g o C Heating Diagram

9. Calculating Energy Changes Q (gained or lost) = m x L. H. (fusion/vaporization) Phase change: Temperature change: heat = mass specific (T f - T i ) heat Q (gained or lost) = m c  T

10. Problem How much energy is required to heat 25 g of liquid water from 25  C to 100  C and change it to steam?

11. Step 1: Calculate the energy needed to heat the water from 25  C to 100  C Q = 25g  4.184 J/g  C  75  C = Q = m  c   T

12. Step 2: Vaporization: Use the Latent Heat to calculate the energy required to vaporize 25g of water at 100  C.25g  1mol H 2 O / 18g mol -1 H 2 O = 1.4 mol H 2 O  vap H (H 2 O) = 1.4 mol H 2 O  40.6kJ/mol = 57 kJ Q = 25.0 g  2259 J/g =

13. Total energy change is:

14. 18.0 g x 2.06 J/g o C x 7 o C 18.0 g x 334 J/g 18.0 g x 4.184 J/g o C x 100 o C 18.0 g x 2259 J/g 18.0 g x 2.06 J/g o C x 25 o C = 260 J = 6012 J = 7531 J = 40668 J = 927 J = 55398 J Calculating Energy Changes Calculate the total amount of heat needed to change 1 mole of ice at -7 o C to steam at 125 o C. TT  phase TT TT

15. Calculating Energy Changes: Solid to liquid How much energy is required to melt 8.5 g of ice at 0  C? The molar heat of fusion for ice is 6.02 kJmol -1 Step 1: How many moles of ice do we have? n = m/Mn = 8.5g / 18gmol -1 = 0.47 mol H 2 O Step 2: Use the equivalence statement to work the energy (6.02 kJ is required for 1 mol H 2 O) kJ = 0.47 mol H 2 O  6.02 kJ / mol H 2 O = 2.8kJ

16. What is specific heat capacity? The amount of energy required to change the temperature of one gram of a substance by 1  C. Another name for specific heat is a calorie (1 calorie = 4.184 Joules) Specific heat capacity of liquid water (H 2 O (L) ) is 4.18 J g -1  C –1. Water (s) = 2.03 J g -1  C –1  0.5 cal/g to break up ice Water (g) = 2.0 J g -1  C –1 10  C11  C

17. Calculating the energy to increase the temperature of liquid water. Calculating specific heat using the equation: Q = ms (t f  t i ) orQ = energy (heat) required Q = ms  T ors = specific heat capacity Heat (H) = ms (t f  t i )m = mass of the sample  T = change in temperature in  C EXAMPLE: How much energy does it take to heat 10g of water from 50 to 100  C ? Specific heat capacity of water = 4.184 J g -1  C –1 Q = m  s   T Q = (10g)  (4.184 J g -1  C -1)  (50  C) = 2.1  10 3 J

18.  vap H (H 2 O) = 1.4 mol  40.6kJ/mol = 57 kJ

19. Heating Diagram