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Materials expand with increase in temperature. As far as I know this property for fluids was put in use to make initial thermometers. We know that expansion of fluid is given by the following formula (at a given pressure) (Wikipedia)

$$\Delta V = \alpha_V (T) V \Delta T$$

Now from this equation we get that the expansion of fluid is not linear. Therefore for a unit change in temperature (i.e., $\Delta T = 1\ \text {unit}$) at different volumes and temperature the expansion of fluid is different. This means the thermometer so made is not a linear one (in which equal spacing represent equal change in temperature).

So

  • How do we know now that our temperature scale now is a linear one?

If possible do tell how this problem was overcome by physicists?


Note that my problem is not that we cannot divide the scale on my own equal sections but rather if the fluid isn't expanding linearly then how would we be able to get the correct measurement.

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    $\begingroup$ The scientists guessed wrong. It's the example I finally turned to when I needed to prove to an obstinate programmer that a spec could be wrong after he said "define wrong" one too many times. $\endgroup$
    – Joshua
    Feb 3 '20 at 4:47
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    $\begingroup$ Note that Celsius, Reaumur, Fahrenheit were additionally "wrong" regarding the zero point ... $\endgroup$ Feb 3 '20 at 13:45
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    $\begingroup$ Related: "Timeline of temperature and pressure measurement technology", Wikipedia. $\endgroup$
    – Nat
    Feb 4 '20 at 5:02
  • $\begingroup$ Are you more concerned about the $V$ factor or the $\alpha_V$ factor contributing to the non-linearity? $\endgroup$
    – Rick
    Feb 10 '20 at 12:38
  • $\begingroup$ @rick non linearity as a whole. $\endgroup$
    – user249968
    Feb 10 '20 at 15:02
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There are physics answers to your question.

To answer your worry at the level of the thermometer scale we see in our house thermometers, where the temperature is given equal intervals from -30C to +50C look at the table here. It gives the volumetric coefficient of expansion per degree. For alcohol, which are the usual cheap house thermometers, it is less than .002. I cannot tell differences in reading the degrees even within 10% , so there is no problem in the linearity because the change is too small.

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    $\begingroup$ I don't get how you get from that table to "the whole scale would change at most 1%". $\endgroup$
    – JiK
    Feb 2 '20 at 13:48
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    $\begingroup$ So that tells that the volume of ethanol will change around 10 % when it goes from 0C to 50C. How is that related to possible nonlinearity between the extreme values of your thermometer? $\endgroup$
    – JiK
    Feb 2 '20 at 16:06
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    $\begingroup$ In other words, you seem to be arguing that the volume of ethanol at 25 C is so close to the average of the volumes at 0 C and 50 C that you can't see the difference in a thermometer. I don't understand how it follows from the number 0.002, and what you mean by "10 %" there, since the 10 % is the total chance, and you definitely can see the difference of 0 C and 50 C in a thermometer. $\endgroup$
    – JiK
    Feb 2 '20 at 16:14
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    $\begingroup$ Ah, are you perhaps assuming that the expansion coefficient is constant, and talking about the exponentiality that you get from $dV/dT \propto V$? The question includes an explicit temperature-dependence of the coefficient, which this answer apparently ignores. Did I understand that correctly? $\endgroup$
    – JiK
    Feb 2 '20 at 17:01
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    $\begingroup$ Alas, no, that can't be it. A thermometer calibrated at -30C and 50C that follows $V(T) \propto \exp(0.002 T / 1 \mathrm{C})$ would have a 17.35 % change from -30C to 50C and a 8.33 % change from -30C to 10C; that would mean the linear thermometer says +8.4 C when it's really 10C. $\endgroup$
    – JiK
    Feb 2 '20 at 17:10
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Simplistically, you can always define a linear temperature scale. For instance, you could call the freezing and boiling points of water (at some standard pressure) 0 and 100 and then construct a linear scale between them.

But it is reasonable to then ask whether the linear scale is then telling you anything fundamental about the observable physical properties of the object. i.e. is there some fundamental property that changes linearly with temperature?

For example, you might first determine your initial scale with the help of a gas thermometer, which determines temperature according to the volume of the gas. You would mark off the 0 and 100 points with reference to the boiling and freezing points and then determine intermediate temperature values as linear increments of the volume of gas.

Suppose you then determined the scale independently using the thermal noise voltage in a resistor, again marking the 0 and 100 points and determining the intermediate points as linear increments of the noise voltage.

Would the temperatures from the two thermometers match in between the 0 and 100 points? The answer is, in general, that they would not, but the disagreement would not be terrible. This is in spite of the Ideal Gas Law (for gas thermometers) and the Johnson-Nyquist equation (for the noisy resistor) both being linear. Real gas thermometers and resistors are not ideal enough.

Statistical mechanics and thermodynamics relate temperature to the internal energy of an object. In the simplified classical theories I learned at undergraduate level, that relationship is linear. This is not always true if you add quantum considerations and other complexities, but even if the "real" relationship were linear it is practically difficult to measure the internal energy.

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    $\begingroup$ No, at the fundamental theoretic level there is no linear relationship between internal energy and temperature. $\endgroup$
    – user137289
    Feb 1 '20 at 21:52
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    $\begingroup$ @Pieter actually yes, I mistook internal energy for kinetic energy. $\endgroup$
    – Ruslan
    Feb 3 '20 at 8:31
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    $\begingroup$ @Tony Under ordinary conditions, the greatest discrepancies occur at phase transitions: a lot of internal energy is added while the temperature remains constant. It is surprising how some university teachers (and the occasional textbook!) manage to ignore that. $\endgroup$
    – user137289
    Feb 3 '20 at 19:49
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There are already great answers, but I would like to address how one could operationally define temperature.

As Tony already pointed out, we define a temperature scale that is linear. The scale itself is the definition of temperature. And to make a straight line, you need two points. So take two points, say boiling and freezing of water under standardised conditions as the reference points.

But if we are defining in terms of a scale, we need a way to measure it. For this, we use the fact that temperature we are defining needs to be related to how hot an object is. And when we heat a gas, it expands. So we can fill up a gas in a balloon (say) that can expand and contract and can withstand boiling and freezing water. Since we can measure the volume of the balloon (gas), we can use this to measure temperature.

But we also need to assign meaning to points in between the two reference points. For this we notice one extraordinary thing! If we take an equal mixture (by volume) of hot and cold water and submerge the balloon in it, the resulting volume of the gas balloon is exactly the average of the volume at the two reference points! This way we can mix waters in various proportions to mark out the rest of the scale. This is how I imagine it was defined initially.

This wonderful result is due to the:

  1. Usage of gas/material whose expansion coefficient is constant in the interested temperature range

  2. Specific heat of water being constant (up to reasonably great approximation) between its freezing and boiling points. heat capacity versus temperature graph Once we have an operational definition and a way to measure the temperature, we can study the subject formally by making postulates based on empirical observations. That is the power of abstraction!

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  • $\begingroup$ Since last year, temperature is defined by the value of Boltzmann's constant. $\endgroup$
    – user137289
    Feb 1 '20 at 21:55
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    $\begingroup$ This. No other answer so clearly drives home the point that the OP conflates the two separate issues of (1) defining a temperature scale and (2) creating thermometers that accurately measure temperature according to that scale. And then it responds well to both. $\endgroup$ Feb 2 '20 at 19:52
  • $\begingroup$ Is your balloon example correct, though? I imagine the tension in the balloon material grows with the volume. Also, the balloon isn't flat so it will get compressed differently by water at different points. There's not much chance that your example is linear enough to be useful for measurements. $\endgroup$ Feb 3 '20 at 6:45
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    $\begingroup$ Balloon is just a placeholder for a container of gas that lets us measure the volume. A container with a movable piston would perhaps circumvent the issues posed by you? $\endgroup$ Feb 3 '20 at 7:05
  • $\begingroup$ @user3518839 it sounds better, yes. $\endgroup$ Feb 3 '20 at 8:19
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The temperature scale we use nowadays is not based on thermal expansion of liquids or any other property of a substance which varies monotonically with temperature. Different types thermometers such as mercury thermometers, platinum resistance thermometers, etc., do not agree with each other at all temperatures. This is because the reading depends on the thermometric substance used. Nowadays we use absolute temperature scale which does not depends on any substance. Ideal gas thermometric scale is close to the absolute temperature scale.

And if you are measuring the pressure variation with temperature of an ideal gas thermometer then it varies not only monotonically with temperature but also linearly in accordance with the ideal gas law.

As you pointed out, length of a fluid in a thermometer does not linearly increase with temperature (the temperature here is based on the absolute temperature scale). But the variation of length was assumed to be linear. Usually, thermometers were calibrated at two particular temperatures - the ice point and the steam point at 1 atmospheric pressure, which could be easily produced in a laboratory. When the temperature of some substance is in between these two values, the length of the fluid in the tube will also lie in between the two extremes. In order to give a numerical value, the temperature variation was assumed to be linear. But now, it's evident that any thermometric scale other than the ideal gas thermometric scale depends a lot on the substance being used.

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I think the quick answer is that, although $\alpha_V (T)$ is not constant, it only varies with $T$ slightly, especially when $T$ does not place the fluid close to a state change. Thus the expansion of the fluid can be approximated by using $\alpha_V (T) = \text {constant}$ which yields linear expansion.

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  • $\begingroup$ This seems to address the other half of the non-linearity not covered by Anna'a answer, but is missing why the V factor doesn't contribute significantly. $\endgroup$
    – Rick
    Feb 10 '20 at 12:44
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Since last year, the temperature scale is defined by the defined value of the Boltzmann constant.

This has not affected the practical measurements of temperature yet. That is still governed by ITS-90. In the region where liquid thermometers are used, there is the fixed point of the triple point of water. Platinum resistance thermometers are then used for scale.

For a longer answer, one would need to explain what temperature is.

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A theoretical temperature function $T$ must at least meet the following two criteria:

1) Two systems that are internally in thermal equilibrium and are suddenly brought into thermal contact with each-other (though not in mechanical contact) show no heat-transfer iff they share the same temperature. If a energy or heat-transfer does take place, it is from the system with the higher temperature to the lower temperature. Suppose then that system $A$ and $B$ share the same temperature and $B$ and $C$ share the same temperature (in other words: bringing $A$ and $B$ into thermal contact doesn't yield a heat-current and neither does bringing $B$ and $C$ into contact), then $T_A=T_B=T_C$: $A$ and $C$ also have the same temperature. So "having the same temperature" is a transitive and reflexive relation and therefore an equivalence relation (0'th law of thermodynamics) among systems that are internally in thermal equilibrium.

2) Temperature should be an integrating factor for the heat 'vectorfield': recall that the first law of thermodynamics reads $dU = \delta Q + \delta W$ where $U$ is a system's internal energy. This first law is shorthand for a decomposition of the gradient of the function $U$, whose independent variables are a finite number of pairs of thermodynamic potentials, with the first member of each pair having an extensive nature (volume, charge, magnetisation,...) while the second member of each pair has an intensive nature (external pressure, external electric potential, external magnetic field,...): $\{V_j,p_j\}_{1\leq j\leq n}$. Let me stress however that temperature and entropy are not among these variables: the variables $\{V_j,p_j\}_{1\leq j\leq n}$ have an unambiguous pre-thermodynamic 'mechanistic' interpretation. In any case, the work-differential $\delta w$ equals $-\sum_{j=1}^n p_jdV_j$. Rewritting this in perhaps more familiar calculus notation, this means $\vec{w}(V_1,p_1,...,V_n,p_n)=-\sum_{j=1}^np_j\vec{e}_{V_j}$. In any case, $T$ is desired to fulfill the role of integrating factor for that other term, $\delta Q \leftrightarrow \vec{Q}$, so that the 1st law acquires the form $$\nabla U = \vec{Q}+\vec{w}=T\nabla S -\sum_{j=1}^n p_j \vec{e}_{V_j}$$ where the function $S(V_1,p_1,...,V_n,p_n)$ is called the system's entropy.

If $\vec{w}$ is of the stated form, then its curl is non-zero for every $(V_1,p_1,...,V_n,p_n)$. Taking the curl of the first law (i.e. $\nabla U = \vec{Q}+\vec{w}$), then yields the conclusion that the curl of the heat-vectorfield $\vec{Q}$ (i.e. the heat conceived as a vectorfield over the variables $\{V_j,p_j\}_j$) also vanishes nowhere.

Now comes a positive surprise: if $T$ is an integrating factor for such a "everywhere-rotational" vectorfield $\vec{Q}$ and suppose that $T'$ is another integrating factor (for the same vectorfield $\vec{Q}$), i.e. $T\nabla S = \vec{Q}=T'\nabla S'$ and if we demand that the functions $T$ and $T'$ share the same contours (in order to respect the zeroth law, which we discussed earlier), then one can show that there is a constant $\lambda>0$ s.t. $$T'=\lambda T$$ Our two requirements have fixed a temperature scale that is unique up to linear rescaling/linear recalibration.

I believe this argument and reasoning can ultimately be traced back to Carathéodory's 1909 seminal work but I'm not 100% sure whether that is correct and/or a complete historical account.

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Physicists were using gas-based thermometers, where linear range is looking like -150 to 2000 Celsius. They were quite lucky that gases are not very easy to liquify in XVII century.

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