Final answer:
Water expands into steam at a 100°C boiling point at 1.00 atm by absorbing 539 cal/g of heat. A mole of water requires 40.7 kJ of heat for vaporization. An open pot of water boils away as water vapor continuously evaporates into the surrounding air.
Step-by-step explanation:
Phase Change from Water to Steam
When water is heated to its boiling point, 100°C at 1.00 atm, it undergoes a phase change into steam. During this process, water absorbs 539 cal/g of heat for this phase change, known as the heat of vaporization. This amount of heat is required for 1 gram of water to convert from liquid to vapor without a temperature increase.
When 1 mol of water vaporizes at 100°C and 1 atm pressure, it absorbs 40.7 kJ of heat from the surroundings, allowing the transition to occur. Conversely, condensation of 1 mol of water vapor releases the same amount of heat. It's important to note that an open pot of water at 100°C will eventually boil away because the air above it is not pure water vapor, allowing continuous evaporation.
A good example of the principles of evaporation and boiling can be found in water containing air bubbles, where the vapor pressure increases with temperature until it exceeds the pressure in the air bubble, causing it to expand and rise to the surface as boiling occurs.