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State Changes

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alamea
10 diapositives · 328 Views · Jul 10, 2014

State Changes. Heat of Fusion Heat energy required to convert a solid at its melting point to a liquid Value for water: 333 J/g Heat of Vaporization Heat energy required to convert a liquid at its boiling point to a solid Value for water: 2260 J/g. Vaporization

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State Changes
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  1. State Changes Heat of Fusion Heat energy required to convert a solid at its melting point to a liquid Value for water: 333 J/g Heat of Vaporization Heat energy required to convert a liquid at its boiling point to a solid Value for water: 2260 J/g

  2. Vaporization endothermic process (H positive) Condensation exothermic process (H negative) Melting endothermic process (H positive) Freezing exothermic process (H negative)

  3. Finding the Energy to Change the State of a Given Mass of a Substance -Pause the video and solve the problems below. J = (mass)(heat of vaporization or heat of fusion) Find the heat energy needed to change 50.0 g of water from liquid to gas. How much heat energy will be released when 50.0 g of water freeze (change from liquid to solid)?

  4. Finding the Energy to Change the State of a Given Mass of a Substance -Pause the video and solve the problems below. J = (mass)(heat of vaporization or heat of fusion) Find the heat energy needed to change 50.0 g of water from liquid to gas. (2260J/g)(50.0g) = 1.13 x 106 J How much heat energy will be released when 50.0 g of water freeze (change from liquid to solid)? (-333J/g)(50.0g) = -1.67 x 105 J

  5. Graphing Temperature v. Time During a State Change 120 Temperature (C) -10 Time (min)

  6. Why do the plateaus occur in this graph even though heat was steadily added over time?

  7. Cpgas = 1.86 J/gK Heat of vaporization = 2260J/g Specific heat of water = 4.184J/gK heat of fusion = 333 J/g Specific heat of ice = 2.06 J/gK

  8. Find the energy needed to convert 500.g of ice at -50.0 C to steam at 200. C. Pause the video while you solve the problem. • Solve this in 5 steps by finding energy needed to: • Heat ice from -50.0 to 0 using specific heat of ice • Melt ice using heat of fusion of ice • Heat water from 0 to 100 using specific heat of water • Boil water using heat of vaporization of water • Heat steam from 100 to 200. using specific heat of steam • Then add all of these q values together to find the total amount of energy needed.

  9. Find the energy needed to convert 500.g of ice at -50.0C to steam at 200. C. Pause the video while you solve the problem. • Solve this in 5 steps by finding energy needed to: • Heat ice from -50.0 to 0 using specific heat of ice • q = (2.06J/gK)(50.0g)(0 C - -50.0C) = 5.15 x 104J • Melt ice using heat of fusion of ice • q = (333J/g)(50.0g) = 1.67 x 105J • Heat water from 0 to 100 using specific heat of water • q = (4.184J/gK)(50.0g)(100 C - 0C) = 2.09 x 105J • Boil water using heat of vaporization of water • q = (2260J/g)(50.0g)= 1.13 x 106J • Heat steam from 100 to 200. using specific heat of steam • q = (1.86J/gK)(50.0g)(200. C – 100.0C) = 9.30 x 104J • Then add all of these q values together to find the total amount of energy needed. 1650kJ

  10. Let’s try something harder!!! Pause the video while you solve the problem. Suppose you pour 250. mL of tea (density 1g/mL) at 18.2C into a glass with 5 ice cubes (each weighing 15g). What quantity of ice will melt and what mass will remain? (Hint: the tea will cool until it is at thermal equilibrium with the ice at 0 C.) Assume the specific heat of tea is 4.2J/gK.