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Thermodynamics and thermometers (Thermistor)

  1. Mar 22, 2017 #1
    1. The problem statement, all variables, and given/known data
    Q1
    Resistance at 0 degrees Celsius = 3840 ohms
    Resistance at 100 degrees Celsius = 190 ohms
    Resistance at ? degrees Celsius = 2300 ohms
    a. Calculate the temperature of the water according to calibrated thermistor assuming the thermistor varies linearly with temperature?
    Answer = 42 Celsius

    B. The temperature of water on the thermodynamic scale of temperature is 286K. BY reference to what is meant by thermodynamic scale of temperature, comment on your answer of a(i)

    3. The attempt at a solution
    What should be the answer to B? I am not sure what is meant by the thermodynamic scale of temperature and secondly, I am aware there is a discrepancy between the temperature recorded by both but what reason should I provide for the wrong answer given by the thermistor?
     
    Last edited: Mar 22, 2017
  2. jcsd
  3. Mar 22, 2017 #2

    DrClaude

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    Staff: Mentor

    https://en.wikipedia.org/wiki/Thermodynamic_temperature

    The answer for A comes out of thin air. It will be difficult to help you without more information.
     
  4. Mar 22, 2017 #3
    You are told that temperature (on the thermistor scale) varies linearly wit resistance. You are given the resistance at 0C and the resistance at 100C. Suso any value of resistance can be used to give a temperature in C by extrapolation
     
  5. Mar 22, 2017 #4
    I am confused in part B. I have done part A
     
  6. Mar 22, 2017 #5
    The temp of the water on the thermodynamic scale is given as 286K which is 13C.
    The thermistor scale is not "wrong" but to compare different scales one needs to be "calibrated" with another. So the temp on the thermistor scale should be marked 13C. The logic of this procedure is that the temp scale using the the thermistor would produce a calibration curve so that it can give temperatures on the thermodynamic temp scale.
     
  7. Mar 22, 2017 #6

    gneill

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    Consider whether the revelation of the new data point given in part (b) calls into question any assumptions made in part (a).
     
  8. Mar 22, 2017 #7
    I would query that this is a "new" data point! It is the same point (temperature) on a different scale
    There are no assumptions in part (a), it specifies the thermistor temp scale
     
    Last edited: Mar 22, 2017
  9. Mar 22, 2017 #8

    gneill

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    Part (a) gives two data points with temperatures on the Celsius scale. Using those two data points and an explicit assumption of linearity the temperature for a given measured resistance was determined, expressed on the Celsius scale.
    Now, part (b) gives the supposed actual temperature for that data point but expressed on the "thermodynamic scale", also known as the Kelvin scale. Presumably we are to believe that this measurement was made by some accurate method. Converted to Celsius (about 13 C) it differs markedly from the value interpolated on the curve proposed in part (a). They are asking why this discrepancy might exist.
     
  10. Mar 22, 2017 #9
    Thank you for explaining that. I wasn't even sure what the question was asking.
     
  11. Mar 23, 2017 #10
    You're 100% correct with your interpretation. The answer for b indeed is
    "Thermodynamic scale of temperature is a temperature scale which is independent of the thermometric properties of a substance. The reading acquired by the calibrated thermistor is incorrect because of non-linearity of resistance with the temperature and the proposed linearity is erroneous."
     
  12. Mar 23, 2017 #11
    pleased to see you have the idea.
    I made more posts but these were deleted because I apparently was giving too much help !
    Unfortunately in deleting my posts a book reference was also deleted.
    I recommend Nelkon & Parker Advanced level physics if you want to look into temperature scales and thermometers in more detail.
     
  13. Mar 23, 2017 #12

    "Presumably we are to believe that this measurement was made by some accurate method." I am puzzled by this statement...what accurate method are you referring to and how are we being lead to believe????
     
  14. Mar 23, 2017 #13

    gneill

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    The problem statement declares the value, so it's another implied assumption:

    "The temperature of water on the thermodynamic scale of temperature is 286K".
     
  15. Mar 23, 2017 #14
    it is not implied by some accurate method....it is given....there are no inaccuracies in these numbers
    The "inaccuracies" are due to the misunderstanding of how differences occur with different temperature scales
     
  16. Mar 23, 2017 #15

    gneill

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    Fine. Either way it is to be taken as an accurate value. It doesn't change the problem.
     
  17. Mar 23, 2017 #16
    Great....there is no problem, 13C on the thermodynamic scale is 42C on the thermistor scale....nothing to do with inaccuracy, nothing to do with electronics.
    Now I can sign off.
     
  18. Mar 23, 2017 #17
    Why no inaccuracy?? If 13C is given by one and second gives you 42C for the same substance there has to be a wrong value. Now, of course, we don't have to take into account the measurer's inefficiency in measuring the temperature. What we have to consider is the thermometer inaccuracy in providing the value. Hence the inaccuracy. Is my interpretation of the question wrong?
     
  19. Mar 23, 2017 #18

    gneill

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    Your interpretation is fine for answering the given problem in the manner the problem's author intended (as amply evidenced by your quoted answer in post #10).
     
  20. Mar 23, 2017 #19
    Exactly. If I saw the data as stated in the problem, I would conclude, "Hmm, that thermistor just isn't very linear at all." In other words, I would trust the thermometer much more than I would the thermistor - although every thermometer has to have some measurement uncertainty.
     
  21. Mar 24, 2017 #20
    The problem is not recognising how a temperature scale is DEFINED.
    Chose a physical property such as resistance of platinum or the pressure of a gas
    Measure the physical property at the two fixed points...0C and 100C (ice and steam point)
    The two properties are X0 and X100
    Measure the value of the physical property at the unknown temp...Xt
    The temp is tC = (Xt - X0)/(X100 -X0) x 100C
    This is why the two values for the same temp will not be equal.
    I recommend Nelkon &Parker for clear examples of this feature of a physical procedure for defining a temperature scale
     
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