# Specific heat capacity vs KE gain of particles

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?

• The equation you used for K.E. is it valid for liquid? – Utkarsh futous Apr 7 '17 at 14:40
• 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 – Mihai B. Apr 7 '17 at 17:57
• 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. – 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.