Heat capacity of a single water molecule

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.43.

The specific heat capacity of water is ${1\mbox{ cal g}^{-1}\;^{\circ}\mbox{C}^{-1}=4.186\mbox{ J g}^{-1}\;^{\circ}\mbox{C}^{-1}}$, so how does this translate into the heat capacity per water molecule? The molar weight of water is ${18.01528\mbox{ g mol}^{-1}}$ so in 1gram of water there are

$\displaystyle N=\frac{6.02\times10^{23}}{18.01528}=3.3416\times10^{22}\mbox{ molecules} \ \ \ \ \ (1)$

The heat capacity of a single molecule is therefore

 $\displaystyle C$ $\displaystyle =$ $\displaystyle \frac{4.186}{3.3416\times10^{22}}=1.2527\times10^{-22}\mbox{J }^{\circ}\mbox{C}^{-1}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 9.08k \ \ \ \ \ (3)$

where ${k}$ is Boltzmann’s constant.

Thus to raise the temperature of a sample of water by ${1^{\circ}\mbox{ C}}$, we need to raise the energy of each molecule by around ${9k}$. If we assumed that all the thermal energy of water is due to quadratic degrees of freedom, then because each degree of freedom contributes ${\frac{1}{2}kT}$ to the energy, we would need 18 degrees of freedom in a water molecule. As we’ve seen, water has around 10 such degrees of freedom, so the thermal energy must also be stored elsewhere.