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Q & A: units for temperature and energy

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Most recent answer: 11/30/2012
Q:
It is said that temperature of a body is the average of the kinetic energies of all the molecules in the body. But then, why do we consider temperature a different physical quantity altogether as [K] and not a derivative of the initially proposed 3 fundamental quantities, length [L], mass[M], and time [T] as with the same dimensional formula as energy? What is the reason behind such a consideration?
- Ujan Chakraborty (age 15)
Kolkata,West Bengal, India
A:

The general definition of temperature isn't quite that, but as you say in effect a temperature gives a characteristic microscopic energy scale.  For example, in an ideal classical monatomic gas, the average kinetic energy per atom is (3/2)kT, where T is the absolute temperature and k is Boltzmann's constant. "k" is used to convert conventional temperature units, say Kelvin, to conventional energy units, say Joules.

So what you're really asking is: why not just set k=1 and measure T in whatever units we're using to measure energy? The answer is purely a matter of historical accident. Temperature scales were devised before anybody understood the relation to microscopic energies.

When I do thermodynamic calculations, I usually set k=1 and either use "Kelvin" as a unit of energy or "Joule" or "erg" as a unit of temperature, depending on what's convenient for the particular problem. Many modern texts also define a temperature τ=kT to avoid having to use separate units for temperature and energy. Since the usual energy units are in turn defined in terms of length, mass, and time units, this follows your suggestion.

There's nothing special about length, mass, and time, however. Any three independent dimensional quantities can serve as the base for our units. In the most fundamental units we have developed, the Planck system, we define Planck's constant (h/2π), Newton's gravitational constant (G), and the speed of light (c) all to be 1. We can express length, mass, time, etc. as combinations of these three, obtaining Planck units for these quantities.

Mike W.


(published on 11/30/2012)

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