But say you had a a jar of water sealed in argon at 1atm, which is larger than the vapour pressure of water, wouldn't some water evaporate?
Yes, but there will not be boiling, only evaporation from the surface.
there are no water molecules in the argon initially, so wouldn't some randomly leave the liquid, making the partial pressure on average non-zero?
Yes.
So why is it that at 100C there is a sudden dramatic change?
At this temperature the vapor pressure inside the bubble begins to be high enough to overcome the pressure in the liquid, so the bubble survives and rises.
Isn't the vapour pressure at e.g. 95C more than the partial pressure of water in the atmosphere as well?
Yes, and the water will evaporate to the atmosphere, but only on the surface. There will be no boiling, since the pressure in the liquid is higher than the pressure in the bubble could be, so any small bubble is crushed and the vapor turns back into liquid.
Firstly, it would be better to use actual, accurately measured numbers than the human 'experience': the human body is a poor thermometer and the mind plays tricks on us.
But that hot water droplets lose heat and thus cool down in cooler air is an established fact and a consequence of the laws of thermodynamics.
Regarding your three first bullet points, despite some limitations you point out, those do not mean a hot droplet of water doesn't cool in air: it does and partly in accordance Newton's cooling law.
As regards work done by the droplet (overcoming the viscous drag), if anything that would lead to heat generation, not cooling (but the effect is truly minuscule).
Kinetic or potential energy of the droplet have no effect on the droplet's temperature. Temperature is simply a measure of the average speed of the molecules of the water and that is not affected by these energies. Spinning water in an ultra-centrifuge does not make its temperature rise, for instance.
You have however overlooked one major cause of heat loss: evaporation. Your shower 'steams up' because hot water evaporates and that costs energy, known as the Enthalpy of vaporisation.
Millions of tons of water are cooled this way everyday in power plants world wide: the cooling towers drop hottish water from the top of the towers and evaporative heat cools down the water (the evaporated water escapes as steam clouds).
If your shower has been in operation for a long time and the bathroom's temperature is equal to the shower head's water temperature and the air is saturated with water vapour, then no cooling of the shower water would take place.
Best Answer
Water and other liquids have a partial vapour pressure, even at temperatures well below their boiling point. At boiling point that vapour pressure becomes equal to the atmospheric pressure and the evaporation manifests itself as bubbles of vapour leaving the bulk of the liquid, a phenomenon known as boiling.
Even at room temperature a saucer of water will evaporate, given some time, due to the relatively low vapour pressure at that temperature. Shower water, say roughly at $50\:\mathrm{C}$, already has a much higher vapour pressure, so the shower generates quite a bit of water vapour, which then condenses in the colder surrounding air, forming a white mist.
Even below $0\:\mathrm{C}$ water has some, limited vapour pressure: see the industrially exploited process of freeze drying.