Can entropy in a closed system remain constant when you heat a liquid? This is in regards to an exam question that I did not understand and got wrong. 

The question was: When you heat acetone from 200 K to 400 K, what happens to the entropy?
Answers: Increases, decreases, more info needed

I put more info needed, but this was incorrect. My logic was that if you have a sealed container completely filled with acetone (there is no gas or empty space in the container) and you heat it, there are no new states that the liquid particles can occupy so there is no change in entropy. Is this reasoning incorrect?
The quantitative differential classical thermodynamic definition of entropy is $$\mathrm{d}S=\mathrm{d}Q/T$$

Therefore providing heat $Q$ to the acetone increases its entropy as
$$\Delta S=\int_{T_\mathrm{1}}^{T_\mathrm{2}} \frac{C(T)}{T}\cdot \mathrm{d}T$$

where $C$ is the heat capacity of the system.

The statistical thermodynamics definition has an alternative definition to $$S=k_\mathrm{b} \cdot \ln \Omega$$ as $$S=-k_\mathrm{b} \cdot \sum_\mathrm{i} p_\mathrm{i} \cdot \ln p_\mathrm{i}$$ as the sum over all states.

The higher energy states of electrons, of bond vibrations and molecule rotations are not occupied at low temperatures, or are occupied unevenly.
In such a case, even for the same number of states, more evenly distributed probabilities at higher temperatures lead to higher entropy.