Say we have liquid water. We are given specific heat of water $C=4.2kJ(kg*K)$, a number of molecules in a mol $N_A=6*10^{23}$. The atomic weight of water is $18g/mol$, and the Boltzmann's constant is $\frac{7}{5}*10^{-23}J/K$. Calculate the thermodynamic degrees of freedom of liquid water.

I am quite confused by this exercise.

Water molecules consist of one oxygen atom and two hydrogen atoms. So one water molecule obviously has 3 degrees of translational, and 2 vibrational and rotational freedom. So 3+2+2=7.

I don't understand why I was given all these useless information, or did I miss out on something. I admit Boltzmann is something i don't understand and I don't know why we would ever need Boltzmann's constant.

I just realised this is from the equipartition theorem.

where we can use: $E_{kin}=\frac{f}{2}K_B T$

So $Q=mC\Delta T=\frac{f}{2}K_B T \rightarrow f=\frac{2mC}{K_B}$

Can I take $m=1kg$ here so the units will work out? But then the exercise wouldn't give me the information on atomic weight and Avagadro's constant?

Edit: Ah I can get the mass of one water molecule by dividing the atomic weight of water by Avagadro's constant

  • $\begingroup$ Yes, it's called the Entropy equation. $\endgroup$ Commented Jun 20, 2022 at 15:04

1 Answer 1


The heat capacity only includes degrees of freedom that are actually accessible to the system. In a quantized system, if the first excited state has energy much larger than $kT$, then thermal interactions can’t activate that degree of freedom and it doesn’t contribute.

You have already done this implicitly by treating your water molecule as three atoms, rather than three nuclei and ten electrons. When it gets hot enough to drive electronic excitations, you tend to dissociate the molecule. You could also have asked about the nuclear degrees of freedom, which are frozen out for temperatures below a mega-eV.

A famous example is hydrogen gas, whose rotational degrees of freedom (first excited state: 15 meV) are frozen out at its triple point (30 kelvin = 2.5 meV), but accessible at room temperature (25 meV). Cold hydrogen has a heat capacity like a monoatomic gas, while warm hydrogen has a heat capacity like a rigid rotor. I used to know whether the vibrational states are accessible before hot hydrogen dissociates, but I have forgotten.

You have predicted how many degrees of freedom a isolated water molecule could have. The question is how many degrees of freedom a water molecule effectively has, in liquid, which may include some neighbor-interaction stuff.

  • $\begingroup$ Ehmmmm what i wrote in the question is actually just part of the question, i actually didnt think that would matter...In the question, it also mentioned that interation with other molecules come into the equation. But we never had that stuff...If we include this then the question itself is not giving enough information is it? $\endgroup$ Commented Jun 20, 2022 at 18:20
  • $\begingroup$ @JerryHolmes You start with the specific heat per mole, then find the specific heat per particle, which is proportional to the number of (effective) degrees of freedom. This is how you discover whether liquid water has extra-molecular degrees of freedom. $\endgroup$
    – rob
    Commented Jun 20, 2022 at 19:10
  • $\begingroup$ The specific heat given here is per kg as usual. If i calculate the specific heat per mole and specific heat per particle. wouldnt they just differ by a factor of $N_A$? $\endgroup$ Commented Jun 20, 2022 at 20:11
  • $\begingroup$ That's a little homework-ish for me; try asking in Physics Chat. $\endgroup$
    – rob
    Commented Jun 20, 2022 at 21:02
  • $\begingroup$ Sorry I just want to ask one more question, so im not sure if a single water molecule has 7 or 9 degrees of freedom, some people say 7 some say 9, but i just calculated it again, i got 18 for liquid water, is this even plausible? $\endgroup$ Commented Jun 21, 2022 at 4:33

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