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.

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