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Liquids and Solids

Liquids and Solids. …if it’s not a gas…. Well, duh. Ingredients: Water. Physical Parameters of Water. Formula FM Shape Polar? Density . Physical Parameters of Water. Formula H 2 O FM 18.02 g/mol Shape bent, 104.5 o Polar? Yes Density 1.00 g/ml.

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Liquids and Solids

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  1. Liquids and Solids …if it’s not a gas…

  2. Well, duh. Ingredients: Water

  3. Physical Parameters of Water • Formula • FM • Shape • Polar? • Density

  4. Physical Parameters of Water • Formula H2O • FM 18.02 g/mol • Shape bent, 104.5o • Polar? Yes • Density 1.00 g/ml

  5. Thermal characteristics of Water • MP • BP • C • Hfus • Hvap

  6. Thermal characteristics of Water • MP 0.0oC • BP 100.0oC • C 4.18 J/goC • Hfus6.0 kJ/mol • Hvap41 kJ/mol

  7. If you add heat to matter, it may… a) b) c) d)

  8. If you add heat to matter, it may… a) warm up. b) melt c) boil d) expand (tough to calculate, don’t bother)

  9. Let’s try to warm up a cup of cold coffee. Step 1: Add heat.

  10. Let’s try to warm up a cup of cold coffee. Step 1: Add heat. Well, that was easy.

  11. Let’s try to warm up a cup of cold coffee. What if you add half as much heat?

  12. Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) b) c)

  13. Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) Raise the temperature only half as much. b) c)

  14. Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) Raise the temperature only half as much. b) Use half as much coffee (and cup) c)

  15. Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) Raise the temperature only half as much. b) Use half as much coffee (and cup) c) Use a different substance

  16. The effect of heat, q! • When something warms up: The heat, q, depends on: • The mass of the sample (m) • The change in temperature (DT) • The nature of the sample (C)

  17. The effect of heat (q) • When something warms up: The heat, q, depends on: • The mass of the sample (m) • The change in temperature (DT) • The nature of the sample (C) C is the specific heat capacity for a given substance. Its units are (J/goC)

  18. If you add heat to a sample, it may… a) warm up. q=mCDT b) melt c) boil d) expand (tough to calculate, don’t bother)

  19. q=mCDT • q – heat, in Joules • m –mass, in grams • C –specific heat capacity, in J/goC • DT—change in temperature (Tfinal-Tinitial)

  20. Cwater=4.184 J/goC • Cethanol =2.4 J/goC • Cice =2.1 J/goC • CAl =.90 J/goC • CFe =.46 J/goC • Cglass =.50 J/goC • CAg =.24 J/goC

  21. How much heat? • How much heat does it take to raise 50.g water from 15oC to 80.oC? • q=mCDT

  22. How much heat? • How much heat does it take to raise 50.g water from 15oC to 80.oC? • q=mCDT = 50.g x 4.18 J/goC x (80.oC-15oC)

  23. How much heat? • How much heat does it take to raise 50.g water from 15oC to 80.oC? • q=mCDT = 50.g x 4.18 J/goC x (80.oC-15oC) = 50.g x 4.18 J/goC x (65oC)

  24. How much heat? • How much heat does it take to raise 50.g water from 15oC to 80.oC? • q=mCDT = 50.g x 4.18 J/goC x (80.oC-15oC) = 50.g x 4.18 J/goC x (65oC) =14000 J (14 kJ)

  25. What is the change in temperature? • If you add 1550 J to 12 g water, how much will it heat up? • DT =q/mC

  26. What is the change in temperature? • If you add 1550 J to 12 g water, how much will it heat up? • DT =q/mC= 1550 J / (12 g x 4.18 J/goC )

  27. What is the change in temperature? • If you add 1550 J to 12 g water, how much will it heat up? • DT =q/mC= 1550 J / (12 g x 4.18 J/goC ) = 31oC

  28. What is the change in temperature? • If you add 1550 J to 12 g water, how much will it heat up? • DT =q/mC= 1550 J / (12 g x 4.18 J/goC ) = 31oC If the temperature starts at 25oC, it will heat up to …

  29. What is the change in temperature? • If you add 1550 J to 12 g water, how much will it heat up? • DT =q/mC= 1550 J / (12 g x 4.18 J/goC ) = 31oC If the temperature starts at 25oC, it will heat up to 56oC

  30. Calorimetry • --the measurement of heat.

  31. Calorimetry • --the measurement of heat. • If one thing gains heat…

  32. Calorimetry • --the measurement of heat. • If one thing gains heat… …something else lost it.

  33. If 75 g of a metal at 96oC is placed in 58 g of water at 21oC and the final temperature reaches 35oC, what is the specific heat capacity of the metal?

  34. Step 1 • How much heat did the water gain?

  35. Step 1 • How much heat did the water gain? q=mCDT Mass of water, in grams Specific heat of water, 4.18 J/goC Change in the temperature of water, in oC

  36. Step 2 • How much heat did the metal lose?

  37. Step 2 • How much heat did the metal lose? • Heat lost = - heat gained • qlost=-qgained

  38. Step 3 • What is the specific heat capacity of the metal?

  39. Step 3 • What is the specific heat capacity of the metal? C=q/mDT Heat lost by metal Mass of metal, in grams Change in the temperature of metal, in oC Specific heat of metal, in J/goC

  40. If 75 g of a metal at 96oC is placed in 58 g of water at 21oC and the final temperature reaches 35oC, what is the specific heat capacity of the metal?

  41. If 75 g of a metal at 96oC is placed in 58 g of water at 21oC and the final temperature reaches 35oC, what is the specific heat capacity of the metal? .74 J/goC

  42. Heats of fusion and vaporization • How much heat is required to melt 150 g of water (at its melting point)?

  43. Heats of fusion and vaporization • How much heat is required to melt 150 g of water (at its melting point)? • (Step 1: Convert to moles)

  44. Heats of fusion and vaporization • How much heat is required to melt 150 g of water (at its melting point)? • 150 g x 1mol/18.02 g =

  45. Heats of fusion and vaporization • How much heat is required to melt 150 g of water (at its melting point)? • 150 g x 1mol/18.02 g = 8.3 mol

  46. Heats of fusion and vaporization • How much heat is required to melt 150 g of water (at its melting point)? • 150 g x 1mol/18.02 g = 8.3 mol • (Step 2: Apply the heat of fusion)

  47. Heats of fusion and vaporization • How much heat is required to melt 150 g of water (at its melting point)? • 150 g x 1mol/18.02 g = 8.3 mol • Q=nHf=8.3 mol x 6.01 kJ/mol =

  48. Heats of fusion and vaporization • How much heat is required to melt 150 g of water (at its melting point)? • 150 g x 1mol/18.02 g = 8.3 mol • Q=nHf=8.3 mol x 6.01 kJ/mol = 50.kJ

  49. Heats of fusion and vaporization • How much heat is required to boil 250 g of water (at its boiling point)?

  50. Heats of fusion and vaporization • How much heat is required to boil 250 g of water (at its boiling point)? • (Step 1: Convert to moles)

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