# Comparing temperature on varying scales

1. Jul 6, 2013

### CAF123

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Relationship between thermometric variable $X$ and temperature $T$

3. 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,

2. Jul 6, 2013

### TSny

The 70oC is for which scale?

3. Jul 6, 2013

### CAF123

Hi TSny,
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. Jul 6, 2013

### TSny

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. Jul 6, 2013

### CAF123

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. Jul 6, 2013

### TSny

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. Jul 6, 2013

### CAF123

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.

t for the former and T for the latter.

8. Jul 6, 2013

### TSny

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. Jul 6, 2013

### CAF123

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.

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. Jul 6, 2013

### TSny

I don't get the factor of 1.95 when I use t = 70 and the values given for a and b.

11. Jul 6, 2013

### CAF123

I converted t = 70 degrees celcius to Kelvin so that on the RHS I would have the right units.

12. Jul 6, 2013

### TSny

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: $R_0$?