1 K is defined as (1/273.15)th of the temperature of the triple point of water. At least, that's how it's defined in my book. But which scale is the triple point of water being measured in?

Celsius? Fahrenheit?

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    $\begingroup$ It doesn't matter. That's the point of that definition. $\endgroup$
    – Buzz
    Dec 17, 2020 at 21:53
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    $\begingroup$ You can use any scale, it will just be a dummy reference. The scale the authors probably intended as the answer to this question is the Kelvin scale. That the above definition reads as tautologous is because the authors phrased the definition badly. What they probably meant to say was that the temperature at which water is at its triple point is (was) defined to be 273.16 K in the Kelvin scale. (And it should be 273.16 K, not 273.15 K). $\endgroup$
    – Zorawar
    Dec 18, 2020 at 0:11
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    $\begingroup$ It can't be Celsius or Fahrenheit, because neither of those starts at absolute zero. $\endgroup$
    – Simon B
    Dec 18, 2020 at 21:40
  • $\begingroup$ @SimonB : Well, ..., not quite (delta temperatures are not temperatures). But if you're going to take that point of view, then you must promote Rankine. $\endgroup$ Dec 20, 2020 at 3:09
  • $\begingroup$ @Zorawar Actually, this is an excellent question: The definition does not say how you would set up an experiment to compare two temperatures $T_1$ and $T_2$ to find a temperature $T_{mean} = \frac{T_1 + T_2}{2}$. Or the other way round, given two systems at $T_1$ and $T_2$, what is the temperature of the combined system? $\frac{T_1 + T_2}{2}$? Or $\root{T_1\cdotT_2}$? Even if the two systems are the same amount of the same substance, neither of these two temperatures need to result. I remember having that same question when I sat through my thermodynamics lectures. $\endgroup$ Dec 20, 2020 at 14:06

5 Answers 5


To answer this question it may help to take an example from a more familiar area of physics, and then discuss temperature.

For a long time the kilogram (the SI unit of mass) was defined as the mass of a certain object kept in a vault in Paris. Then the gram can be defined as one thousandth of the mass of that object, and so on. If you now ask, what units are being used to state the mass of the chosen object? then it does not matter as long as they are proportional to the scale of units you want to adopt. So if someone were to tell you the mass of the special object in pounds (e.g. 2.2 pounds) then you would still know that one gram is a thousandth of that.

With temperature it goes similarly. There is a certain state of water, water vapour and ice all in mutual equilibrium. That state has a temperature independent of other details such as volume, as long as the substances are pure and they are not crushed up too small. So that state has a certain temperature. It has one unit of temperature in "triple point units" (a temperature scale that I just invented). When we say the Kelvin is a certain fraction of that temperature, we are saying that a thermometer whose indications are proportional to absolute temperature must be calibrated so as to register 273.16 when it is put into equilibrium with water at the triple point, if we wish the thermometer to read in kelvin. For example, if the thermometer is based on a constant-volume ideal gas then one should make the conversion factor from pressure in the gas to indicated temperature be a number which ensures the indicated temperature is 273.16 at the triple point. You then know that your gas thermometer is giving readings in kelvin, and you never needed to know any other units. (Note, such a thermometer is very accurate over a wide range of temperature, but it cannot be used below temperatures of a few kelvin. To get to the low temperature region you would need other types of thermometer. In principle they can all be calibrated to agree where their ranges overlap.)

(Thanks to Pieter for a detail which is signaled in the comments and now corrected in the text, but I hope the comment will remain.)

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    $\begingroup$ That should be 273.16, as the triple point is at 0.01 C. $\endgroup$
    – user137289
    Dec 18, 2020 at 9:59
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    $\begingroup$ @Pieter Thanks! I was not sure about details such as chemical composition either. It is useful to have this precision. $\endgroup$ Dec 18, 2020 at 20:03
  • $\begingroup$ But if at the triple point of water we have exactly $273.16$, then the standard pressure ice temperature cannot be exactly $273.15$, but approximately. It would be physically meaningless. If one of the two is exact, the other must be approximated. Am I wrong? In the books they say $T_K = T_C + 273.15$ as if it were exact. @AndrewSteane $\endgroup$ Jun 25, 2022 at 13:34
  • $\begingroup$ (obviously I was talking about the melting temperature of the ice at standard pressure, I missed a word) $\endgroup$ Jun 25, 2022 at 15:02

That was the old definition.

Since May, the kelvin is defined by fixing the value of the Boltzmann constant: https://physics.nist.gov/cgi-bin/cuu/Value?k

This is consistent with the triple point of a specific kind of water (VSMOW) at 273.16 K.

It is also historically consistent with the still older definition of the size of the centigrade as 1/100 of the difference in temperature between freezing and boiling of water.

A different scale for absolute temperature is based on the size of a degree on the Fahrenheit scale. This is the Rankine scale where $1$ kelvin = $1.8\ ^\circ$R.

Edit: so your book was wrong. The triple point is at $273.16$ K which is $0.01\ ^\circ {\rm C}$ (as the triple point is slightly higher than the melting point of ice at atmospheric pressure).

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    $\begingroup$ OK, that's the old definition, but the old definition is what OP is interested in. This answer doesn't help clarify OP's misconception that the units used to measure the temperature of the triple point matter. $\endgroup$
    – kaylimekay
    Dec 17, 2020 at 12:48
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    $\begingroup$ OP might be interested to know that his book is slightly out of date. And if I had written this as a comment, someone would have complained that I should have written it as an answer. Always those tiring complaints on this site. $\endgroup$
    – user137289
    Dec 17, 2020 at 13:11
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    $\begingroup$ Check my answer here: hsm.stackexchange.com/questions/6794/… $\endgroup$
    – FGSUZ
    Dec 17, 2020 at 15:04
  • $\begingroup$ A nice answer but it would be even better to mention why you have $273.16$ rather than the OP's $273.15$. $\endgroup$
    – badjohn
    Dec 18, 2020 at 10:15
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    $\begingroup$ @badjohn Good suggestion. Done. $\endgroup$
    – user137289
    Dec 18, 2020 at 10:21

It may not be obvious from everyday experience with temperature, but it has a natural zero point, independent of any choice of scale.

Temperature is related to the internal motion of particles making up a substance--when all the internal motion ceases, the temperature is zero.

You can think of it like the concentration of dye in a tank of water. There's no ambiguity about what zero means: no dye means zero concentration. Accordingly, what you mean when you say "The concentration of dye in this tank is half the concentration in that one" does not depend on the units you use to specify the concentrations.

The confusion may arise from the fact that, unlike with most quantities that have a natural zero point (mass, kinetic energy, etc.) familiar temperature scales have an offset so that commonly encountered temperatures come out as smallish numbers.

So the answer to your question about which scale is used in the definition is: any one that doesn't impose such a shift.


The triple point of water exists at exactly one pressure-temperature point.

The measurement of temperature by the gas-scale, is done by finding NRT=PV at two different pressures, and reducing this to 0, on the assumption NRT = PV + kV² + ...

So 1/273.16 of the triple-point then says that 1 kelvin is 1/273.16 of the implied PV value when N=0.

So it's a naturally occurring event.

In the older days, the degree was defined as 0 = some cold point, 1 = some hot point, and the scale divided into a number of degrees.

Rømer set 0 = salt-ice water freezing, 1 = boiling water, divided into 60 degrees,

Fahrenheit built a thermometer that made Rømer's degrees too big, so he quartered them, and used a colder cold (basically the refigeration is at 0 °F). Rømer's multipoint scale was corrected, so pure water freezes at 32 and boils at 212.

Celcius scale is pure water freezes at 0 and boils at 1, divided to 100 degrees.

Réaumur's degree is an expansion of 1000 units of alcohol at freezing, which climbs 40 until it evaporates, but 80 is the boiling of water.

  • $\begingroup$ For historic accuracy it should be mentioned that Mr Celsius never tried that scale, he had freezing at 100 and boiling at 0. $\endgroup$
    – Džuris
    Dec 18, 2020 at 9:29
  • $\begingroup$ The anton scale was the first absolute scale. It involved two tubes of mercury, one of which was ended closed, and its pressure set to read 73 inches of mercury at the boiling point of water. Given that atmospheric pressure was 29 inches, Paris measure, the difference is due to PV=NRT, where the LHS is allowed to fall from 73 inches to whatever the room pressure less 29 inches might be. $\endgroup$ Dec 18, 2020 at 12:16
  • $\begingroup$ Of all of the scales, none satisfy both the use of fahrenheit / celcius (where normal cold to hot run from some 0 to 100), and an absolute scale. Instead, the best solution seems to be to use 1.5 times kelvin (or gorem), which lets water freeze at 410 and boil at 560 (so the range 400-500 runs from -6.67 C to + 46.67 C), and the 970 is the temperature of the hottest water can be. Gas-scale for cooking runs at 600 + 20 GM. $\endgroup$ Dec 18, 2020 at 12:20
  • $\begingroup$ @wendy.krieger Thanks..I learnt the most from your answer! $\endgroup$ Dec 18, 2020 at 15:45
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    $\begingroup$ refrigaration is meant: the cold-point here is because by adding a variety of salts to the liquid reduces the freezing point, so the liquid can still go around in the pipes while the inside of the chest would make ice of pure water. $\endgroup$ Dec 19, 2020 at 10:13

I read this question as asking about how we actually determine a complete scale from having only two single points on it. Which is far from trivial.

Historically, people first built thermometers, for instance by putting a liquid into a fixed volume with a thin pipe attached. Then they calibrated those thermometers by getting a reading for boiling water (100°C) and its freezing point (0°C). In between, they simply attached a linear scale. And whatever that scale said was 50°C, that was called 50°C.

This worked surprisingly well. If you have two glasses of water at different temperatures, and you mix them together, you get a temperature that is very close to the mean of the temperature as measured above. This in itself is coincidence, and depends on the near-constant heat capacity of the substance. However, if you take a substance that changes its heat capacity significantly, your experimental mean temperature will not be the mathematical mean.

A more precise definition of the temperature is only taught in statistical mechanics: Here temperature is defined as $$T=\frac{dE}{dS(E)}$$, where $E$ is the energy of the system, and $S(E)$ is the entropy of the system. The Boltzmann constant $k_B$, which is part of the definition of the entropy, connects the Joule to the Kelvin in this equation. As such, this definition serves to define the scale of the Kelvin, and by consequence all the other temperature scales that we have in use.


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