1 / 25

Chapter 10

Chapter 10. Causes of Change. Practice problem #1. Determine the energy, in Joules, needed to increase the temperature of 20.00 g of water from 300.0 K to 365.0 K. Ch 10 Section #1. Practice problem #2.

Download Presentation

Chapter 10

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 10 Causes of Change

  2. Practice problem #1 Determine the energy, in Joules, needed to increase the temperature of 20.00 g of water from 300.0 K to 365.0 K. Ch 10 Section #1

  3. Practice problem #2 Determine the energy, in Joules, released as 1.00 Kilograms of water cools from 72.0 oC to 30.0 oC. Ch 10 Section #1

  4. Practice problem #3 Determine the energy, in Joules, needed as 500.00 mL of water increases from 300.0 Kelvin to 370.0 Kelvin. (1.00 mL =1.00 gram for water) Ch 10 Section #1

  5. Practice problem #4 Determine the energy, in Joules, released as 1.000 Liter of water Cools from 350.0 Kelvin to 310.0 Kelvin Ch 10 Section #1

  6. Ch 10 Section #1 Practice problem #5 Determine the energy, in calories, needed as 2.00 x 104 milligrams of water cools from 85.0 oC to 5.00 oC.

  7. Practice problem # 6 Calculate the energy required as 2.50 x 103 grams of ice melts Ch 10 Section # 1

  8. Practice problem # 7 Calculate the energy required as 8.00 moles of ice melts Ch 10 Section # 1

  9. Practice problem # 8 Calculate the energy released as 4008.0 grams of water freezes. Ch 10 Section # 1

  10. Practice problem # 9 Calculate the energy required as 1.00 Kilomole of water freezes. Ch 10 Section # 1

  11. Practice problem # 10 Calculate the energy required as 5.00 x 103 grams of water boils (vaporizes). Ch 10 Section # 1

  12. Practice problem # 11 Calculate the energy required as 16.0 moles of water boils (vaporizes). Ch 10 Section # 1

  13. Practice problem # 12 Calculate the energy released as 8.016 x 103 grams of steam condenses. Ch 10 Section # 1

  14. Practice problem # 13 Calculate the energy released as 2.00 Kilomole of steam condenses. Ch 10 Section # 1

  15. Ch 10 Section #1 Practice problem # 14 Calculate the energy required as 180.0 g of Ice at 0.0 oC is heated to Steam at 100.0 oC. 0.0 oC Ice  0.0 oC Water Step #1 0.0 oC Water  100.0 oC Water Step #2 100.0 oC Water  100.0 oC Steam Step #3 Total Energy Required = Step #4

  16. Ch 10 Section #1 Practice problem # 15 Calculate the energy released as 5.00 moles of steam at 100.0 oC is cooled to 0.0 oC Ice. 100.0 oC Steam 100.0 oC Water Step #1 100.0 oC Water  0.0 oC Water Step #2 0.0 oC Water  0.0 oC Ice Step #3 Total Energy Released = Step #4

  17. Ch 10 Section #1 Bonus Practice problem: Extra Credit Calculate the total energy required as 20.0 moles of Ice at 248.0 K is heated to Steam at 423.0 K. For full credit you must show all six equations and answers.

  18. Bonus / Extra Credit x Ch 10 Section #1 Finish x Start

  19. Molar heat capacity The energy as heat needed to increase the temperature of 1.00 mole of a substance by 1.00 Kelvin (K). Ch 10 Section #1 Note: Values of C can be found in your textbook on page 343

  20. Practice problem # 16 Determine the energy as heat needed to increase the temperature of 20.0 moles of mercury by 15.0 K. The value of C for mercury is found on page 343. Ch 10 Section #1

  21. Practice problem # 17 Determine the energy as heat removed as 50.00 moles of Iron cools from 1500.0 oC to 150.0 oC. The value of C for Iron is found on page 343. Ch 10 Section #1

  22. Practice problem # 18 Determine the energy as heat needed as 3.60 x 102 grams of liquid water warms from 1.00 oC to 99.0 oC. The value of C for liquid water is found on page 343. Ch 10 Section #1

  23. Practice problem # 19 Calculate the molar heat capacity (C) of carbon, given that 126.0 Joules were needed to heat 24.00 grams of carbon by 2.000 x 102 K Ch 10 Section #1

  24. Ch 10 Section #1 Practice problem # 20 A 0.0700 mole of octane absorbs 3.50 x 103 J of energy. Calculate the temperature increase of octane if the molar heat capacity of octane is 254.0 J/K.mole

  25. Ch 10 Section #1 Practice problem # 21 Explain in your own words how you would convert the Specific Heat (Cp) of a material to the Molar Heat Capacity (C) of the same material.

More Related