Comparing temperature on varying scales

In summary, the resistance of a wire can be calculated using the equation ##R = R_0(1 + at + bt^2)##, where ##t## is the temperature in degrees celcius measured on the ideal gas scale and ##R_0## is the resistance at the ice point. To find the temperature on the resistance scale at a given temperature on the ideal gas scale, the equation ##R = \frac{R_0}{273.15K}{T}## can be used, where ##T## is the absolute temperature on the resistance scale.
  • #1
CAF123
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Homework Statement


The resistance of a wire is given by ##R = R_0 (1 + at + bt^2)##, where ##t## is the temperature in degrees celcius measured on the ideal gas scale and so ##R_0## is the resistance at the ice point. The constants ##a## and ##b## are 3.8 x 10-3 K-1 and -3.0 x 10-6 K-2 respectively. Calculate the temperature on the resistance scale at a temperature of 70oC on the ideal gas scale.

Homework Equations


Relationship between thermometric variable ##X## and temperature ##T##

The Attempt at a Solution



The given eqn for R is the relationship between the thermometric variable (R) and t. Consider a system where the thermometric variable ##X## varies linearly with the temperature ##T##. Then ##X = pT##, for some constant ##p## fixed upon determining the temperature at a particular point. It is given that at ##R_0##, T = 273.15K, (ice point) so then p = R0/273.15K, so ##R = \frac{R_0}{273.15K}{T}. ## When T = 70oC, on this scale, R= 1.26Ro.

So the temperature on the resistance scale at a temperature of 70oC on the ideal scale is found by $$1.26R_0 = R_0 (1 + at + bt^2)$$, cancelling and solving gives two values of t, neither of which agree with the answer. Physically, I don't see any sense in there being two solutions so I was wondering if there is an error in my method, particularly in considering the linear scale.

Many thanks,
 
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  • #2
CAF123 said:
... so ##R = \frac{R_0}{273.15K}{T}. ## When T = 70oC, on this scale, R= 1.26Ro.

The 70oC is for which scale?
 
  • #3
Hi TSny,
TSny said:
The 70oC is for which scale?

The 70oC is on the ideal gas scale: t = T(K) - 273.15. I converted 70o C using this and substituted into ##R_0/(273.15 K) T## Is it incorrect? I assumed that for such an ideal gas scale ##X\,\alpha\,T##
 
  • #4
By definition, ##\small T## in ##\small R = \frac{R_0}{273.15K}{T}## is the absolute temperature on the resistance scale. But the 70oC is stated to be the temperature on the ideal gas scale.
 
  • #5
TSny said:
By definition, ##\small T## in ##\small R = \frac{R_0}{273.15K}{T}## is the absolute temperature on the resistance scale. But the 70oC is stated to be the temperature on the ideal gas scale.

When you say '...is the absolute temperature on the resistance scale', do you mean (in this case) '...is the absolute temperature on a resistance scale'? (Since it is given for this particular resistance scale R grows quadratically with T)

On the ideal gas scale, the temperature is simply then T(K) = 343.15 K.

Could I have a hint on how to proceed?
 
  • #6
R is given to grow quadratically with the celsius temperature t associated with the ideal gas scale.

By definition, R grows linearly with the absolute temperature T defined on the resistance scale associated with this particular wire.

Is the 70oC (343.15 K) a value of t or a value of T?

Does the question ask for a value of t or a value of T?
 
  • #7
TSny said:
R is given to grow quadratically with the celsius temperature t associated with the ideal gas scale.

By definition, R grows linearly with the absolute temperature T defined on the resistance scale associated with this particular wire.

Okay, so I had them mixed up. Why is it 'by definition'? I do see in my book that it mentions resistance as an example that grows linearly with temperature, however.

Is the 70oC (343.15 K) a value of t or a value of T?

Does the question ask for a value of t or a value of T?

t for the former and T for the latter.
 
  • #8
CAF123 said:
Why is it 'by definition'?

I believe the equation ##R = \frac{R_0}{273.15K}{T} ## or, equivalently, ##T = \frac{273.15K}{R_0}R ## defines the "resistance temperature scale" for that particular piece of wire.

In general, ##T = \frac{273.15K}{X_0}X ## would define a temperature scale based on the thermometric property ##X## of some object. (The triple point of water is often used as the reference point rather than the ice point.)
 
  • #9
So ##R= R_0(1 + at + bt^2)##on the ideal gas scale. When t = 70 degrees Celsius, R ≈ 1.95R0. I want to find the equivalent temperature on the resistance scale where t = 70oC on the ideal scale. The resistance scale is ##R = R_0/(273.15 K) T##. Inputting R = 1.95Ro and solving for T, I obtain T = 532K, which is incorrect.

TSny said:
. (The triple point of water is often used as the reference point rather than the ice point.)

Yes, in my book they use the triple point of water as the fixed point for temperature scales defined using different thermometric variables. They only use the ice point in conjunction with the steam point when temperature was originally (pre 1954) expressed as ##X = pT + q##, and using these two fixed points allowed determination of p and q.
 
  • #10
CAF123 said:
So ##R= R_0(1 + at + bt^2)##on the ideal gas scale. When t = 70 degrees Celsius, R ≈ 1.95R0.

I don't get the factor of 1.95 when I use t = 70 and the values given for a and b.
 
  • #11
TSny said:
I don't get the factor of 1.95 when I use t = 70 and the values given for a and b.
I converted t = 70 degrees celcius to Kelvin so that on the RHS I would have the right units.
 
  • #12
In the equation ##R = R_0(1+at+bt^2)##, ##t## should be interpreted as a temperature measured relative to the ice point. So, it's really a temperature difference. Temperature differences have the same value for both the Celsius and Kelvin scales.

If you don't interpret it this way, note what you would get for ##R## at the ice point: ##\small t = 0 ^oC = 273.15 K##. What would you get if you substituted 273.15 K for ##t## in the ##R## equation? Would you get the value of ##R## at the ice point: ##R_0##?
 
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1. What are the most commonly used temperature scales?

The most commonly used temperature scales are Celsius, Fahrenheit, and Kelvin.

2. How do these scales differ from each other?

Celsius and Fahrenheit are both based on the freezing and boiling points of water, but Fahrenheit has a smaller degree increment. Kelvin is based on absolute zero, where there is no molecular movement.

3. How do you convert between these scales?

To convert from Celsius to Fahrenheit, use the formula (°C x 9/5) + 32 = °F. To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. To convert from Fahrenheit to Celsius, use the formula (°F - 32) x 5/9 = °C. To convert from Fahrenheit to Kelvin, first use the formula (°F - 32) x 5/9 = °C, then add 273.15 to the Celsius temperature.

4. Why do different countries use different temperature scales?

The choice of temperature scale is largely based on historical and cultural reasons. For example, the United States uses Fahrenheit because it was developed by a German physicist who immigrated there. Other countries, like most of Europe, use Celsius because it was developed by a Swedish astronomer. Kelvin is used in scientific applications because it is based on absolute zero and is used in thermodynamics.

5. Are there any other temperature scales besides Celsius, Fahrenheit, and Kelvin?

There are other lesser-known temperature scales, such as Rankine and Réaumur. Rankine is similar to Fahrenheit but uses absolute zero as its starting point. Réaumur is similar to Celsius but uses a different degree increment. However, these scales are not widely used or recognized.

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