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Should I average the following when calculating heat capacity?

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Make that specific heat capacity xD

    Well anyways, I did 5 trials trying to calculate the specific heat capacity of copper. I measured the temperature of metal, initial temperature of water, final temperature of water, calorimeter mass, mass of water etc.

    When I'm calculating the specific heat capacity with mc(triangle)t = mc(triangle)t

    should I average mass of the water?

    Should I average the temperature difference?


    2. Relevant equations

    Also I am required to make uncertainty calculations as well as percent difference.

    How do I carry uncertainty into percent difference? (don't know if that makes sense or not)

    3. The attempt at a solution

    Well I did some rough calculation with the above averaged and came pretty close :)
     
  2. jcsd
  3. Sep 5, 2009 #2

    kuruman

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    You should not average the mass of the water and should not average the temperature difference. I assume that your five measurements were independent of each other, that is you refilled your calorimeter with water and reheated the piece of copper. Averaging these is meaningless. The only average that makes sense is the five values of the specific heat that you got from your measurements.

    [tex] Pct. Diff=\frac{Final - Initial }{Initial }\times 100[/tex]
     
    Last edited: Sep 5, 2009
  4. Sep 5, 2009 #3
    ok thanks, but i wasn't asking for the percent difference, I was asking how to carry the +/- 's (uncertainty) into the percent difference equation.

    Oh yea, if I would to do them individually how would I average the uncertainty?
     
    Last edited: Sep 5, 2009
  5. Sep 5, 2009 #4

    kuruman

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    With five measurements, you could take your percent uncertainty to be about two standard deviations from the mean.
     
  6. Sep 6, 2009 #5
    I have no idea what that is lol, never applied statistics to physics and don't know how =/

    But is it necessary to carry the +/- into the percent difference?

    I've never seen a a% +/- x% difference before
     
  7. Sep 6, 2009 #6

    kuruman

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    Find the mean or average of your specific heat measurements.
    Subtract the average from each of the five measurements.
    Square each difference.
    Add the squares.
    Take the square root of the sum.
    Divide by the number of measurements.
    Multiply by two.
    Find what percent of the average the last number is. That is what I would call the "plus or minus" percent difference.
     
  8. Sep 7, 2009 #7
    alright! thanks for the very clear instructions sorry if I'm noob :D
     
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