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What happens when we add energy to a solid at constant pressure

What happens when we add energy to a solid at constant pressure. gas. What are the energy changes that occur?. D H = D H fus x # mols. DH = m x C gas x D t. D H = D H vap x # mols. DH = m x C solid x D t. DH = m x C liquid x D t. What would these changes look like

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What happens when we add energy to a solid at constant pressure

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  1. What happens when we add energy to a solid at constant pressure gas

  2. What are the energy changes that occur? DH = DHfusx # mols DH = m xCgasxDt DH = DHvapx # mols DH = m xCsolidxDt DH = m xCliquidxDt

  3. What would these changes look like on a temperature versus energy graph? DH = DHfusx # mols DH = m xCgasxDt DH = DHvapx # mols DH = m xCsolidxDt DH = m xCliquidxDt

  4. temperature added energy Heating curves and DH

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

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

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

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

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

  10. temperature Dt of solid absorbing energy added energy

  11. temperature DH = m x CsolidxDt added energy

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

  13. DH = DHfusx # mols temperature DH = m x CsolidxDt added energy

  14. Dt of liquid absorbing energy DH = DHfusx # mols temperature DH = m x CsolidxDt added energy

  15. DH = DHfusx # mols DH = m x CliquidxDt temperature DH = m x CsolidxDt added energy

  16. the energy absorbed as a liquid boils becomes potential energy, so no Dt DH = DHfusx # mols DH = m x CliquidxDt temperature DH = m x CsolidxDt added energy

  17. DH = DHvapx # mols DH = DHfusx # mols DH = m x CliquidxDt temperature DH = m x CsolidxDt added energy

  18. Dt of gas absorbing energy DH = DHvapx # mols DH = DHfusx # mols DH = m x CliquidxDt temperature DH = m x CsolidxDt added energy

  19. DH = m x CgasxDt DH = DHvapx # mols DH = DHfusx # mols DH = m x CliquidxDt temperature DH = m x CsolidxDt added energy

  20. DH = m x CgasxDt DH = DHvapx # mols DH = DHfusx # mols DH = m x CliquidxDt temperature DH = m x CsolidxDt added energy

  21. 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 DH = m x CgasxDt DH = DHvapx # mols DH = DHfusx # mols DH = m x CliquidxDt temperature DH = m x CsolidxDt added energy

  22. 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 DH = m x CgasxDt DH = DHvapx # mols DH = DHfusx # mols DH = m x CliquidxDt temperature DH = m x CsolidxDt added energy

  23. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC? DH = m x CgasxDt DH = DHvapx # mols DH = DHfusx # mols DH = m x CliquidxDt temperature DH = m x CsolidxDt added energy

  24. 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

  25. 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

  26. 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 DH3 = m x CliquidxDt DH2 = DHfusx # mols temperature 0 oC DH1 = m x CsolidxDt -20 oC added energy

  27. 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 DH3 = m x CliquidxDt DH2 = DHfusx # mols temperature 0 oC DH1 = 10g x 2.1 J/goC x20oC -20 oC added energy

  28. 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 DH2=10 g x 1mol/18g x 6.01 kJ/mol DH3 = m x CliquidxDt temperature 0 oC DH1 = 10g x 2.1 J/goC x 20oC -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 DH3 = 10g x 4.186 J/goC x 50 oC 50 oC DH2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC DH1 = 10g x 2.1 J/goC x 20oC -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 total DH = DH1 + DH2 + DH3 DH3 = 10g x 4.186 J/goC x 50 oC 50 oC DH2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC DH1 = 10g x 2.1 J/goC x 20oC -20 oC added energy

  31. 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 DH3 = 10g x 4.186 J/goC x 50 oC 50 oC DH2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC DH1 = 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 total DH = DH1 + DH2 + DH3 5853 J = 420 J + 3340 J + 2093 J DH3 = 10g x 4.186 J/goC x 50 oC 50 oC DH2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC DH1 = 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? It takes 5853 joules to heat up 10 grams of water from -20 oC to +50 oC. DH3 = 10g x 4.186 J/goC x 50 oC 50 oC DH2=10 g x1mol/18g x 6.01 kJ/mol temperature 0 oC DH1 = 10g x 2.1 J/goC x 20 oC -20 oC added energy

  34. Example II: Ice at 0oC is placed in a Styrofoam cup containing 0.32 kg of lemonade at 27oC. The specific heat capacity of lemonade is virtually the same as that of water. After the ice and lemonade reach an equilibrium temperature, some ice still remains. Assume that the cup absorbs a negligible amount of heat. miceg X 1mol/18g X 6010J/mole = 320g X 4.18J/g.CoX (27oC -0oC) mice = 108.1g

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