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Specific heat question please help =(

  1. Apr 12, 2005 #1
    how're all the fine scientists feeling today?

    well i jst had a question regarding my specific heat & Co. homework.

    ok here it is: "When a 25g block of metal alloy at 215 degrees celcius (i dont know how to make that round degree symbol on here) is dropped into 85g of water at 22 deg.C, the final temperature is 37 deg.C. What is the specific heat of the alloy?

    i worked it out 2 different times and got 2 different answers each time. :cry: :uhh:
    once i got 15 J/(g*degrees Celcius) and then the second time i got 1.2 J/(g*degrees Celcius)

    any help at all appreciated!!!!!! thanks in advance!!!!!
     
    Last edited: Apr 12, 2005
  2. jcsd
  3. Apr 12, 2005 #2

    dextercioby

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    Okay.I don't use sign conventions for given and received heat,so i'll simply write

    [tex] Q_{\mbox{given}}=m_{\mbox{alloy}}c_{\mbox{alloy}}(215-37) [/tex]

    [tex] Q_{\mbox{received}}=m_{\mbox{water}}c_{\mbox{water}}(37-22) [/tex]

    Set the 2 #-s equal and then solve for the unknown.

    Daniel.

    P.S.Pay attention with the units.I'd advise u to use SI-mKgs.
     
  4. Apr 12, 2005 #3
    wouldnt it be (37-215) because delta T is always t final minus t initial??

    and what did you mean by SI-mKgs ? :confused:
     
  5. Apr 12, 2005 #4

    dextercioby

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    Nope,i don't use convention signs.The heats are always positive and equal...(in this 2 body thermal interraction).

    SI-mKgs from Système International-metre,Kilogramme,sécond .

    Daniel.
     
  6. Apr 12, 2005 #5
    oh.

    well when heat is negative, it meant that its lost, right? thats what i was thought. let me get another crack at it with what you said. but i dont understand why u would have to set them equal together, since Q of water plus Q of metal alloy should equal 0 right?
     
  7. Apr 13, 2005 #6

    Gokul43201

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    A positive number can never be equal to a negative number. So, what Dexter is doing is simply making sure they are both positive, and equating them.

    If you want to strictly follow convention, you would swap Tf and Ti inside the brackets and then put a minus sign in front of one of the Q's. This will give the same result (the two minus signs will cancel out).
     
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