1 / 21

Homework:

Homework:. Page 498, problems 12-13 Page 500, questions 15-19. 12. If 335g of water at 65.5 o C loses 9750J of heat, what is the final temperature of the water?. 12. If 335g of water at 65.5 o C loses 9750J of heat, what is the final temperature of the water? q = cm∆T.

harken
Download Presentation

Homework:

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. Homework: Page 498, problems 12-13 Page 500, questions 15-19

  2. 12. If 335g of water at 65.5oC loses 9750J of heat, what is the final temperature of the water?

  3. 12. If 335g of water at 65.5oC loses 9750J of heat, what is the final temperature of the water? q = cm∆T

  4. 12. If 335g of water at 65.5oC loses 9750J of heat, what is the final temperature of the water? q = cm∆T ∆T = q / cm

  5. 12. If 335g of water at 65.5oC loses 9750J of heat, what is the final temperature of the water? q = cm∆T ∆T = q / cm = 9750J / (4.184J/go)(335g)

  6. 12. If 335g of water at 65.5oC loses 9750J of heat, what is the final temperature of the water? q = cm∆T ∆T = q / cm = 9750J / (4.184J/go)(335g) = 7oC

  7. 12. If 335g of water at 65.5oC loses 9750J of heat, what is the final temperature of the water? q = cm∆T ∆T = q / cm = 9750J / (4.184J/go)(335g) = 7oC 65.5 – 7 = 58.5oC

  8. The temperature of a sample of water increases from 20.0oC to 46.6oC as it absorbs 5650J of heat. What is the mass of the sample?

  9. The temperature of a sample of water increases from 20.0oC to 46.6oC as it absorbs 5650J of heat. What is the mass of the sample? • q = cm∆T

  10. The temperature of a sample of water increases from 20.0oC to 46.6oC as it absorbs 5650J of heat. What is the mass of the sample? • q = cm∆T • m = q / c∆T

  11. The temperature of a sample of water increases from 20.0oC to 46.6oC as it absorbs 5650J of heat. What is the mass of the sample? • q = cm∆T • m = q / c∆T • = 5650J / (4.184J/go)(26.6o)

  12. The temperature of a sample of water increases from 20.0oC to 46.6oC as it absorbs 5650J of heat. What is the mass of the sample? • q = cm∆T • m = q / c∆T • = 5650J / (4.184J/go)(26.6o) • = 50.8g

  13. Why does ∆H for an exothermic reaction have a negative value?

  14. Why does ∆H for an exothermic reaction have a negative value? • The reaction gives off heat, so the products have less heat than the reactants did. The change is negative.

  15. Why is a measured amount of water an essential part of a calorimeter?

  16. Why is a measured amount of water an essential part of a calorimeter? • q = cm∆T • c = 4.184 J/go

  17. Explain the meaning of ∆Hrxn. Why is ∆Hrxn sometimes positive and sometimes negative?

  18. Explain the meaning of ∆Hrxn. Why is ∆Hrxn sometimes positive and sometimes negative? • Positive for endothermic reactions • Negative for exothermic reactions.

  19. Could another liquid be used just as effectively as water in a calorimeter? Why or why not?

  20. Could another liquid be used just as effectively as water in a calorimeter? Why or why not? • q = cm∆T • c = _____ J/go

  21. Explain how you would design and experiment to determine the specific heat of a 45g piece of metal.

More Related