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I Kelvin thermodynamics temperature scale

  1. Apr 28, 2016 #1
    The thermal efficiency of a reversible heat engine is solely a function of the temperature of the two reservoirs.
    η = f(θHC) = 1 - (QC/QH)
    (QC/QH) = 1 - f(θHC)
    (QC/QH) = Ψ(θHC)

    The simplest function that can be used is T1/T2
    (QC/QH) = T1/T2

    In order to define the Kelvin scale we assign a value to one of the temperatures and that is the triple point of water, 273.16K. Thus any other temperature is defined as
    T = 273.16(Q/Qtp)

    My question is, why was the triple point of water used in this equation and can we use any other temperature?

    How does this relate to the definition of Kelvin:"The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water"

    If we use another function, say (T1)2/(T2)3, can we stilldefine the Kelvin scale?

    In practice, is this equation useful? We could just measure the temperature using a sensor instead of finding the heat transfer at both reservoirs before calculating the temperature.
  2. jcsd
  3. Apr 28, 2016 #2

    Simon Bridge

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    The triple point is the reference for the Celcius scale, not the Kelvin scale.
    Prior to that, the Fahrenheit scale used the freezing point of a particular mix of brine.

    The Kelvin scale uses the intercept of the slope of volume vs temperature with the temperature axis for helium gas.

    The reason for choosing the reference is to do with ease of getting the value along with the reproducability of the value and politics.
    Water is picked because there is a lot of it easy to get hold of and handle ... brine over pure water because pure water would have been hard to get: but has the disadvantage of requiring care to determine the salinity... so different thermometers using the F scale will have the same number for slightly different temperatures.
    You could use the freezing point of pure water ... but that is different for different pressures. The triple point is easy to see and very precise so use that instead.
    Both these scales have the advantage of having sane numbers for everyday temperatures.

    The Kelvin scale comes from a desire to use the Universe itself as a reference. Find the lowest temperature possible in principle and call that zero.
    On top of which - finding absolute zero is now fairly straight forward and can be done accurately.

    There are other temperature scales in use. i.e.
    Note: you actually need two reference temperatures to set up a scale.

    Politics comes into it because standard units and measures are decided by a committee.
  4. Apr 28, 2016 #3

    Actually, the triple point of water is used as the one point needed to define the Kelvin as a thermodynamic temperature scale. Water is easily available and the triple point is not hard to achieve in laboratory conditions. This is why it was chosen. Of course other linear thermodynamic scales could be defined, and some have been (See the Rankine scale.) I suppose one could come up with a non-linear scale, but why one would want to do this, I have no idea.
  5. Apr 28, 2016 #4

    Simon Bridge

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    I was just musing that in practise we use the triple point and call it +273.15K, or whatever it is, instead of 0.

    You may want a non-linear scale if the physical thermometer has a non-linear response - in a cvgt we are exploiting the change in height of the column being directly proportional to the change in mean kinetic energy of the gas... chosen because of the resulting linearity I guess. iirc the connection beteween temperature and kinetic energy came after the invention of the cvgt so it comes down to what makes for nice maths.
  6. Apr 28, 2016 #5


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    The Kelvin unit was based on the Celsius scale, a change of 1 deg C = 1 K. Ultimately, it was an arbitrary decision.
  7. Apr 29, 2016 #6

    Simon Bridge

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    Strictly: ##1^\circ C = 274.15K## :P (though the distinction would be worth marks in an exam.)
    K is C with a different zero point.

    "'Lets make the centigrade scale start at absolute zero,' they said - so they ended up with Kelvins. Horrible isn't it."
    -- Marvin (the Paranoid) Android (guest lecturing thermodynamics section of physics 101).
  8. Apr 29, 2016 #7
    Okay, but I don't see how T = 273.16(Q/Qtp) brings about a linear scale. Suppose you fix the hot reservoir at 273.16K. The efficiency of the heat engine would vary depending on the temperature of the cold reservoir because η = (TH-TC)/TH so that both Q and Qtp will change
  9. Apr 29, 2016 #8


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    What @Khashishi was referring to is the unit of temperature difference, not the unit of absolute temperature. In the units of temperature difference, 1 K = 1 °C.
  10. Apr 29, 2016 #9
    I played around with the numbers a bit and can see what is going on.
    Let's assume x joules of energy is moving from the hot reservoir at 273.16K to the heat engine.
    η = 1 - TC/273.16
    Energy moving into the cold reservoir = x - x(1 - TC/273.16)
    = xTC/273.16
    T = 273.16(Q/Qtp)
    = TC
  11. May 1, 2016 #10

    Simon Bridge

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    @DrClaude : quite right - and that distinction is sometimes worth marks in an exam. ;)
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