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“The E n e r g y Associated with Phase Change” PROJECT

“The E n e r g y Associated with Phase Change” PROJECT. Ex: BORON. Properties of Elements. Melting Pt: 2079 o C (Chart) Heat of Fusion: 22.2 kJ/mol Change 2018 J/g Boiling Pt: 3675 o C ( Chart) Heat of Vaporization: 314 kJ/mol Change 28,545 J/g

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“The E n e r g y Associated with Phase Change” PROJECT

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  1. “The EnergyAssociated with Phase Change”PROJECT

  2. Ex: BORON Properties of Elements Melting Pt: 2079 oC(Chart) Heat of Fusion: 22.2 kJ/mol Change 2018 J/g Boiling Pt: 3675 oC (Chart) Heat of Vaporization: 314 kJ/mol Change 28,545 J/g Molar Mass: 11 g/mole (Per. Table) Specific Heat (Cp):1.03 J/g.C(Chart)

  3. Change Heat of Fusion and Heat of Vaporization to J/g Heat of fusion: 22.2 kJ/mol Change 2018 J/g Hfus Heat of vaporization : 314 kJ/mol Change 28545 J/g Hvap

  4. Purpose:Determine the energy (Q), in Joules, necessary to raise the temperature of 10.0 grams of the element from 10.0° C below the melting point to 10.0° C above the boiling point. Draw the phase change stair-step diagram. ·Label both axes with units and measurable quantity; ·Label the temperature axis with numbers. ·Also, label the melting point and boiling point. Write "A" at the initial temperature and a "B" at the final temperature on your plotted line.

  5. temperature added energy Heating curves and DH

  6. liquid solid gas Heating curves and DH temperature added energy

  7. liquid solid gas Heating curves and DH melting/ freezing pt temperature added energy

  8. liquid solid gas Heating curves and DH boiling/ cond. pt melting/ freezing pt temperature added energy

  9. liquid solid gas Heating curves and DH boiling/condensing occurring here melting/freezing occurring here boiling/ cond. pt melting/ freezing pt temperature added energy

  10. How is the total enthalpy change (DH) calculated for a substance whose temperature change includes a change in state?

  11. temperature Dt of solid absorbing energy added energy

  12. temperature Q = m x CsolidxDt added energy

  13. the energy absorbed as a solid melts becomes potential energy, so no Dt temperature Q= m x CsolidxDt added energy

  14. Q = mass x Hfus temperature Q= m x CsolidxDt added energy

  15. Dt of liquid absorbing energy Q = mass x DHfus temperature Q = m x CsolidxDt added energy

  16. Q = mass x DHfus Q = m x CliquidxDt temperature Q = m x CsolidxDt added energy

  17. the energy absorbed as a liquid boils becomes potential energy, so no Dt Q = mass X DHfus Q = m x CliquidxDt temperature Q = m x CsolidxDt added energy

  18. Q = mass X DHfus Q = m x CliquidxDt temperature Q = m x CsolidxDt added energy

  19. Dt of gas absorbing energy Q = mass X DHvap Q = mass X DHfus Q = m x CliquidxDt temperature Q = m x CsolidxDt added energy

  20. Q = m x CgasxDt Q = mass X DHvap Q = mass X DHfus Q = m x CliquidxDt temperature Q = m x CsolidxDt added energy

  21. Q = m x CgasxDt Q = mass X DHvap Q = mass X DHfus Q = m x CliquidxDt temperature Q = m x CsolidxDt added energy

  22. How many calculations for Q are required to determine the energy of this element according to the specifications listed in the Purpose? _______ Boron (J/g)

  23. How many calculations for Q are required to determine the energy of this elementaccording to the specifications listed in the Purpose?

  24. The DH of any substance being heated will be the sum of the DH of any Dt occurring plus DH of any phase change occurring Q = m x CgasxDt Q = mass X DHvap Q = mass X DHfus Q = m x CliquidxDt temperature Q = m x CsolidxDt added energy

  25. The DH of any substance being heated will be the sum of the DH of any Dt occurring plus DH of any phase change occurring Q = m x CgasxDt Q = mass x Hvap Q = mass x Hfus Q = m x CliquidxDt temperature Q = m x CsolidxDt added energy

  26. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? Q = m x CgasxDt Q = mass X DHvap Q = mass X DHfus Q = m x CliquidxDt temperature Q = m x CsolidxDt added energy

  27. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? 50 oC temperature 0 oC -20 oC added energy

  28. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? use the following values: Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC 50 oC temperature 0 oC -20 oC added energy

  29. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? use the following values: Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC 50 oC Q = m x CliquidxDt Q = mass X DHfus temperature 0 oC Q = m x CsolidxDt -20 oC added energy

  30. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? use the following values: Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC 50 oC Q3 = m x CliquidxDt Q = mass X DHfus temperature 0 oC Q1 = 10g x 2.1 J/goC x20oC -20 oC added energy

  31. DH2=10 g x 1mol/18g x 6.01kJ/mol EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? use the following values: Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC 50 oC Q3 = m x CliquidxDt Q2=10 g x 1mol/18g x 6.01 kJ/mol temperature 0 oC Q1 = 10g x 2.1 J/goC x 20oC -20 oC added energy

  32. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? use the following values: Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC Q3 = 10g x 4.186 J/goC x 50 oC 50 oC Q2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC Q1 = 10g x 2.1 J/goC x 20oC -20 oC added energy

  33. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? use the following values: Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC total DH = DH1 + DH2 + DH3 Q3 = 10g x 4.186 J/goC x 50 oC 50 oC Q2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC Q1 = 10g x 2.1 J/goC x 20oC -20 oC added energy

  34. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? use the following values: Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC total DH = DH1 + DH2 + DH3 = 420 J + 3340 J + 2093 J Q3 = 10g x 4.186 J/goC x 50 oC 50 oC Q2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC Q1 = 10g x 2.1 J/goC x 20oC -20 oC added energy

  35. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? use the following values: Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC total DH = DH1 + DH2 + DH3 5853 J = 420 J + 3340 J + 2093 J Q3 = 10g x 4.186 J/goC x 50 oC 50 oC Q2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC Q1 = 10g x 2.1 J/goC x 20oC -20 oC added energy

  36. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? It takes 5853 joules to heat up 10 grams of water from -20 oC to +50 oC. Q3 = 10g x 4.186 J/goC x 50 oC 50 oC Q2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC Q1 = 10g x 2.1 J/goC x 20 oC -20 oC added energy

  37. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? It takes 5853 joules to heat up 10 grams of water from -20 oC to +50 oC. Q3 = 10g x 4.186 J/goC x 50 oC 50 oC Q2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC Q1 = 10g x 2.1 J/goC x 20 oC -20 oC added energy

  38. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? It takes 5853 joules to heat up 10 grams of water from -20 oC to +50 oC. 50 oC temperature 0 oC -20 oC 5853 J added energy

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