Effect of doubling kinetic energy on phase transitions

Question

The kinetic energy of the molecules in a sample of $\ce{H2O}$ in its stable state at $ \pu{–10 ˚C}$ and $\pu{1 atm}$ is doubled. What are the initial and final phases? [US National Chemistry Olympiad, 2010]

Answer: $\ce{solid -> gas}$

My attempt: Since water freezes at $\pu{0^\circ C}$ at $\pu{1 atm}$, then the water clearly begins in a solid state. In the solid state, most of the atoms are locked within a crystal of hydrogen bonds and therefore have extremely small kinetic energies. Thus, doubling the kinetic energy should not have any impact on the phase, since gases have significantly more kinetic energy than the solid/liquid phase.

Why does the solid have enough kinetic energy such that doubling it results in sublimation?

Answer 1

The atoms are not locked in place. They vibrate back and forth. They do so with plenty of velocity and thus kinetic energy, even though the vibrations switch back and forth so rapidly that the atoms do not travel a large distance in any one direction.

Answer 2

I don't think there is a way to solve this without some knowledge of the equipartition theorem. The reasoning goes something like this.

In ice, the only significant method of storing energy is in bulk vibrations of the lattice. Thus, the heat capacity of ice must be at least $3RT$ because each of the vibrational modes of the lattice contributes $RT$ to the heat capacity by the equipartition theorem. In reality, the heat capacity will be larger because we ignored all intermolecular vibrational and librational modes and the intramolecular modes. If you look it up, the heat capacity of ice is closer to $4.5RT$. This does not affect our reasoning though.

Now, if we think about the gas, the heat capacity should be $3/2RT+3/2RT=3RT$ for the translational and rotational motions, then there should be a small vibrational contribution less than $3RT$. In the ideal gas situation, the heat capacity will be exactly $3RT$ so let's proceed with this assumption.

Now, note that rotational and translational energies are purely kinetic, so we have $3RT$ units of kinetic energy. In ice, half of this vibrational energy will be kinetic and half will be potential if we assume harmonic oscillators. Anharmonic effects should basically cancel out so no big worries there. Thus, doubling the kinetic energy leaves us with at least $3RT$ in kinetic energy. Therefore, the solid should turn into a gas.

I don't know how we can conclude that it won't end up as a liquid besides by arguing that water is quite structured as a liquid in the sense that any instantaneous snapshot will look like quite similar to other ones, while this is not true for gases. This basically means that translational motion and rotational motions are not important enough to scramble everything.

There may be a more elegant solution. This really doesn't take long to write down the energies for though, so this might be what they had in mind.

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Note that this problem can be confusing if we think about adding energy to the system and populating modes. Instead, the problem just says that initially there is some kinetic energy and finally there is twice as much. What state can handle having this much kinetic energy? That tricked me for a bit but if it didn't confuse you then just ignore this addendum.