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Finding Specific Heat of 2 substances mixed

  1. Jan 8, 2005 #1
    In an experiment, .02 kg of substance is heated in a bath of boiling water until its temperature is apporoximately 100*C. The substance is the placed in an insulated container that is holding .10 kg of water at 25*C. The final equilibrium temperature of the mixture is 32*C. The specific heat of substance, in kilojoules kilogram . degree Celsius, is.

    I tried making the change of heat of the substance equal to change of heat of wate, but that didnt give me the right awnser, I know this because it is a multiple choice awnser. I would really appreciate help on this. Thxs before hand.
  2. jcsd
  3. Jan 8, 2005 #2
    Oops, that title is so wrong, suppose to be, Finding Specific Heat of Substance. What was I thinking. :mad:
  4. Jan 8, 2005 #3
    That should work... what are the answers available?
  5. Jan 8, 2005 #4
    A. 2.16
    B. 2.095
    C. 2.57
    D. 4.19
    E. 1.0
  6. Jan 8, 2005 #5
    I got 2.16...
  7. Jan 8, 2005 #6
    I did:
    (.02 kg)(68*C)c=(.1 kg)(1)(7*C)

    and got c= .5147
  8. Jan 8, 2005 #7
    How did your get 2.16?
  9. Jan 8, 2005 #8
    you got the answer in calories/(K*kg) multiply it by the specific heat of water in KJ
  10. Jan 8, 2005 #9
    I dont understand.
  11. Jan 8, 2005 #10


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    My answer is "A".Post your work.

  12. Jan 8, 2005 #11
    I did.....
  13. Jan 8, 2005 #12


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    That "1" in the formula for heat absorbed by water.It's 1 calory.It should be in [itex] KJ Kg^{-1}K^{-1} [/itex]

  14. Jan 8, 2005 #13
    For the specific heat of water, use 4.19kJ/(K*kg) instead of 1 calorie/(K*kg)
  15. Jan 8, 2005 #14
    So its

    (.02 kg)(68*C)c=(.1)(7*C)(1 ?)

    what is the specific heat measured in, isn't cal/g*C?
  16. Jan 8, 2005 #15
    If you actually read between the line... you would notice that we've actually told you how to do it...

    Dextercioby, should we consider helping someone this rude?
  17. Jan 8, 2005 #16


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    Okay,u did it fine,but put wrong units.If u put instead of calories [itex] KJ Kg^{-1}K^{-1} [/itex],your answer will coincide with the one at point "a".

  18. Jan 8, 2005 #17
    apchem, I posted that before, because, dex said i got A, and told me to post my work. I said I did post it, and I didnt really care what the letter awnser was.

    Anyways thxs for the help, didnt mean to be rude.
  19. Jan 8, 2005 #18
    Use the 4.19 in the place of 1.... since that is the correct unit not 1 calorie/(K*kg)
  20. Jan 8, 2005 #19


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    Yes,it's just that u need to express it in KJ/Kg*C (or Kelvin,they are the same thing,just scaling is different).

  21. Jan 8, 2005 #20
    I c how, 1 cal/g*C= 4.184 J/kg*C, correct?

    Oops, 4.184 is already in KJ, I just noticed, that 1 cal/g*C=4184 J/kg*C. Thats whats started this whole question.

    Well thxs for all the help, and again I didn't mean to be rude. :smile:
    Last edited: Jan 8, 2005
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