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anonymous

  • one year ago

Need help and an explanation on how to do this, please. When 0.7521g of benzoic acid was burned in a calorimeter containing 1,000g of water, a temperature rise of 3.60 degrees C was observed. What is the heat capacity of the bomb calorimeter, excluding the water? The heat of combustion of benzoic acid is -26.42 kJ/g.

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  1. anonymous
    • one year ago
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    Is there an equation for you to plug these numbers into?

  2. anonymous
    • one year ago
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    \[Q=(m)(c)(\Delta)(T)\]

  3. anonymous
    • one year ago
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    I'm not too sure which one to use.

  4. anonymous
    • one year ago
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    How come we use that equation even though there isn't really a change in temperature?

  5. anonymous
    • one year ago
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    It says the temperature increased 3.6 degrees C so that is the temperature change

  6. anonymous
    • one year ago
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    Ah, I see. But we don't have to calculate a temperature change in this case, right?

  7. anonymous
    • one year ago
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    Nope! its already done for you! :)

  8. anonymous
    • one year ago
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    The heat capacity would be "c" in the equation?

  9. anonymous
    • one year ago
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    yes!

  10. anonymous
    • one year ago
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    so what i'm not really understanding is: they've given the mass (0.7521g of benzoic acid) and they've given the temperature (3.60 degrees C).. where/how do we use the 1,000g of water and the heat of combustion of benzoic acid in the question?

  11. anonymous
    • one year ago
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    If i'm correct, the water is just something thrown in there to mess you up, but Q is the -26.42

  12. anonymous
    • one year ago
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    So this would the equation -26.42= (0.7521)(c)(3.6)

  13. anonymous
    • one year ago
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    C is the heat capacity

  14. anonymous
    • one year ago
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    would the answer be -9.76J?

  15. anonymous
    • one year ago
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    Thats what I got!

  16. anonymous
    • one year ago
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    I did find an explanation to this problem on yahoo answers (and i'm not doubting your way) but they got another answer and used steps i don't understand :(

  17. anonymous
    • one year ago
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    would you like the link?

  18. anonymous
    • one year ago
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    Sure!

  19. anonymous
    • one year ago
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    https://answers.yahoo.com/question/index?qid=20120424181717AAxs958

  20. anonymous
    • one year ago
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    Woah. Thats a lot of work. I checked the guys bio and he is a chemistry professor so i would probably go with his answer, but I assumed that it would be the explanation I gave you. I don't understand any of that, but then again he is a professor. You could write both solutions down and then ask your teacher which one is right. If your class has been doing Specific Heat and Heat Capacity equations then I would use mine, but if that is not the case then the other guy is probably correct with is crazy math.

  21. anonymous
    • one year ago
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    ha! That's how i felt :( I'll definitely ask on both because not knowing the way he did on yahoo answers means we didn't go over it in class but yours made sense as well.. even though now i'm double confused lol but thank you for your help!

  22. anonymous
    • one year ago
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    i found 1 more explanation actually!

  23. anonymous
    • one year ago
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    this person did it with less math

  24. anonymous
    • one year ago
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    Q(comb) = (0.7521g)((-26.42KJ/g) = -19.87KJ Q(H20) = (3.6C)(1000g)[4.184J/(g-C)] = 15.062KJ Q(cal) = 19.87KJ - 15.062KJ = 4.81KJ SpH = 4.81KJ/3.6C = 1.34 KJ/C

  25. anonymous
    • one year ago
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    That looks less confusing but thats above what I know

  26. anonymous
    • one year ago
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    Alright, great! Thank you.

  27. anonymous
    • one year ago
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    No problem!

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