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Energy Requirements for changing state:

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

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Energy Requirements for changing state:

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  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. Heating and cooling curve for water heated at a constant rates. A-B = Solid ice, temperature is increasing. Particles gain kinetic energy, vibration of particles increases. 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. H2O (s) H2O () 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. H2O () H2O (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. Heating Diagram Calculating Energy Changes Temperature 2. DHf = 334 J/g 4. DHv= 225 J/g 3. cl = 4.18 J/g • oC 5. cv = 2.1 J/g • oC 1. cs = 2.1 J/g • oC (oC) Heat Added

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

  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 = m  c T Q = 25g  4.184 J/g C 75 C =

  12. Step 2: Vaporization: Use the Latent Heat to calculate the energy required to vaporize 25g of water at 100C Q = 25.0 g  2259 J/g = .25g  1mol H2O / 18g mol-1 H2O = 1.4 mol H2O vap H (H2O) = 1.4 mol H2O  40.6kJ/mol = 57 kJ

  13. Total energy change is:

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

  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/M n = 8.5g / 18gmol-1 = 0.47 mol H2O Step 2: Use the equivalence statement to work the energy (6.02 kJ is required for 1 mol H2O) kJ = 0.47 mol H2O  6.02 kJ / mol H2O = 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 . 10 C 11 C Another name for specific heat is a calorie (1 calorie = 4.184 Joules) Specific heat capacity of liquid water (H2O (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

  17. Calculating the energy to increase the temperature of liquid water. Calculating specific heat using the equation: Q = ms (tf ti) or Q = energy (heat) required Q = ms T or s = specific heat capacity Heat (H) = ms (tf ti) 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 = 4.184J 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 (H2O) = 1.4 mol  40.6kJ/mol = 57 kJ

  19. Heating Diagram

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