Answer
See the detailed answer below.
Work Step by Step
a) We know that the heat energy will not transfer from a cooler system to a warmer system, so the beaker of water must be cooler than the room temperature.
b) In such systems, ice plus 100$^\circ$C steam, some of the steam will convert to water and some of the ice will convert to water too.
So the scenario of finding a large ice cube and some 100$^\circ$C steam in a sealed container with no net change in energy is not realistic because it violates the second law of thermodynamics.
The second law of thermodynamics states that in any process, the total entropy of a closed system and its surroundings must always increase or remain constant. Entropy is a measure of the degree of disorder or randomness in a system. Therefore, for the scenario to be possible, there would need to be an increase in the entropy of the system and its surroundings that offsets the decrease in thermal energy of the water molecules that turn into ice. This could occur, for example, if the container were placed in contact with a much larger system that absorbs the excess entropy and heat generated by the process.
