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Thermal Expansion of a Solid

  1. Jun 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Usually when temperatures are written as "T", they are given in kelvins and when written in "t" they are given in celsius. The equation dL = aLdT used T for temperature yet, in the experiment temperature is measured as celsius. Is there any difference? Why?


    2. Relevant equations
    dL = aLdT

    where a is coefficient of Linear Expression


    3. The attempt at a solution

    This is just an experiment preparatory question. I was not sure about this but my answer would be that there is no large difference as we all know the equation:

    Kelvins = Celsius + 273

    Although since there is always a reading error with even temperature then the conversion would also accompany a calculated error. Although I am unsure. Could someone direct me to the right direction or just confirm that I am correct?
     
  2. jcsd
  3. Jun 5, 2008 #2

    mgb_phys

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    I wouldn't rely on t vs T - it's not necessarily widespread.

    If you are talking about changes in temperature it doesn't matter. Similairly any equations involving simply adding or subtracting a temperature works for celcius or kelvin. It's fairly easy to show this with some algebra.

    I'm not sure what question you are asking.
    The error in the change in length will be the same percentage as the error in the measurement of temperature. It's easy to show this by calculating the answer for a particlular dT and then for dT+(some small increment) and comparing the dL.
     
  4. Jun 5, 2008 #3
    Well I have to do an experiment, and in the experiment I will measure temperature using a thermometer in celsius. Now since the equation at the top though uses Kelvins for temperature and not celsius, does it make any difference in measuring with celsius and not kelvins?

    From your answer, I understood that since we are taking the difference there is no problem.
    I seem to have forgotten that little point somehow, I believe that would be a good enough answer seeing as the difference in kelvins or celsius would be the same.
     
  5. Jun 5, 2008 #4

    mgb_phys

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    Correct it's very easy to prove, imagine making a cup of coffee!

    t1 = 17C = (273 + 17) K
    t2 = 100C = (273 +100) K

    dT = t2-t1 = (273 + 100) - (273+17) = (273-273) + (100-17) = 83
    So it really doesn't matter about the 273!
     
  6. Jun 5, 2008 #5
    Thank you for the help mgb.
     
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