Explain the specific heat capacity of water based on the kinetic theory of matter.

(N/A) If we consider a water molecule as a solid,it consists of three atoms ($2$ hydrogen and $1$ oxygen) vibrating about their mean positions.
From the equipartition of energy,the energy associated with one molecule of water is:
$= 2 \times \frac{1}{2} k_{B} T = k_{B} T$ (for kinetic energy).
Considering the vibrational energy in three dimensions (each atom has $3$ degrees of freedom for vibration,contributing $k_{B} T$ per atom):
Total energy of one molecule of water $= 3 \times k_{B} T = 3 k_{B} T$ (for kinetic part) and $3 \times k_{B} T$ (for potential part).
Total energy of one molecule of water $= 3 \times (k_{B} T + k_{B} T) = 6 k_{B} T$ (simplified model for solid-like behavior).
Total energy of one mole of water:
$U = 6 k_{B} T \times N_{A} = 6 RT$.
Molar specific heat of water:
$C = \frac{dU}{dT} = \frac{d}{dT}(6 RT) = 6R$.
$C = 6 \times 8.31 = 49.86 \text{ J mol}^{-1} \text{ K}^{-1}$.
(Note: The actual specific heat of liquid water is much higher due to hydrogen bonding,which is not accounted for in simple kinetic theory models.)