To increase the temperature of 1kg of water by 1C you need 4200J of energy. However, the KE gain is only $\frac{3}{2} k_B \Delta T \cdot 6.02\cdot 10^{23} \cdot \frac{1000}{18} = 692.3$J. Where does the other $5/6^{th}$ of the energy provided go?

  • $\begingroup$ The equation you used for K.E. is it valid for liquid? $\endgroup$ Apr 7 '17 at 14:40
  • $\begingroup$ Perhaps you are looking at the problem from the wrong perspective. We need molecular orbitals analysis; take a look at how IR are "stored", etc [Assignment of the IR vibrational absorption spectrum of liquid water] www1.lsbu.ac.uk/water/water_vibrational_spectrum.html $\endgroup$
    – Mihai B.
    Apr 7 '17 at 17:57
  • $\begingroup$ The specific heat of water is very close to the value of the Dulong and Petit law when one includes the hydrogens: three times the gas constant per mole av atoms. $\endgroup$
    – user137289
    Apr 15 '17 at 20:22

Water molecules in the liquid phase do not only have translational degrees of freedom. The "missing" energy goes into those other degrees of freedom, such that the molar heat capacity is actually about $9R$ and so water acts as if it had 18 degrees of freedom.

In the gas phase, a bent, triatomic molecule like water has 3 translational degrees of freedom (dof), 3 rotational dof and 6 vibrational dof (3 kinetic and 3 potential) - which would correspond to a molar heat capacity of $6R$.

In the liquid phase, water has some of these additional degrees of freedom, but more importantly is strongly affected by hydrogen bonding. Each water molecule can form up to four hydrogen bonds with other water molecules. Some of the additional energy required to heat water goes into reducing the potential energy of these bonds or breaking them altogether and this acts to increase the effective number of dof.

So to sum up - the additional energy goes into making water molecules rotate and vibrate and into loosening and breaking hydrogen bonds.

Some further reading:



  • $\begingroup$ So are you convinced that provided there is not much attraction between particles then such differences will not be much apparent? $\endgroup$ Apr 9 '17 at 16:29
  • $\begingroup$ @Utkarshfutous A molecular liquid has more degrees of freedom than a monatomic gas, even without hydrogen bonding between molecules. The OPs "calculation" would I expect still show a discrepancy in the absence of H-bonding, but it would be smaller. $\endgroup$
    – ProfRob
    Apr 9 '17 at 17:15
  • $\begingroup$ I've heard the hydrogen argument before but don't quite understand how it can be true. The strength of bonds depends primarily on how close the molecules are to each other. Water does not significantly expand when heated. $\endgroup$ Apr 11 '17 at 2:19
  • $\begingroup$ The "hydrogen argument"? @EdwardGaremo The fact that water molecules form hydrogen bonds is not arguable? Why does the breaking of bonds require expansion? The bonds are made and broken on short timescales and certainly will hinder the translation and rotation of the molecules. In any case, that is only part of the answer to this question - the majority of degrees of freedom are not in translational motion. $\endgroup$
    – ProfRob
    Apr 11 '17 at 5:51
  • $\begingroup$ Hydrogen bond argument. And I wasn't saying it wasn't correct, I was simply asking for an explanation to how it can be correct considering the water doesn't expand by much i.e. the water molecules do not move away from each other much i.e. they don't gain much potential electric energy. $\endgroup$ Apr 12 '17 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.