Thermodynamics and thermometers (Thermistor)

  • #1
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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)

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?
 
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Answers and Replies

  • #2
DrClaude
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What should be the answer to B? I am not sure what is meant by the thermodynamic scale of temperature
https://en.wikipedia.org/wiki/Thermodynamic_temperature

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?
The answer for A comes out of thin air. It will be difficult to help you without more information.
 
  • #3
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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
 
  • #4
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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
I am confused in part B. I have done part A
 
  • #5
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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.
 
  • #6
gneill
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I am confused in part B. I have done part A
Consider whether the revelation of the new data point given in part (b) calls into question any assumptions made in part (a).
 
  • #7
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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
 
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  • #8
gneill
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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
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.
a. Calculate the temperature of the water according to calibrated thermistor assuming the thermistor varies linearly with temperature?
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.
 
  • #9
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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.
Thank you for explaining that. I wasn't even sure what the question was asking.
 
  • #10
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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.
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."
 
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  • #11
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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."
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.
 
  • #12
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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.

"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????
 
  • #13
gneill
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"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????
The problem statement declares the value, so it's another implied assumption:

"The temperature of water on the thermodynamic scale of temperature is 286K".
 
  • #14
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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".
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
 
  • #15
gneill
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it is not implied by some accurate method....it is given....there are no inaccuracies in these numbers
Fine. Either way it is to be taken as an accurate value. It doesn't change the problem.
 
  • #16
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Fine. Either way it is to be taken as an accurate value. It doesn't change the problem.
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.
 
  • #17
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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.
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?
 
  • #18
gneill
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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?
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).
 
  • #19
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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.
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.
 
  • #20
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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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