Instead of temperature drop, we have to to consider amount of heat transferred to the building from the wildfire. The temperature of the structures will rise towards the ignition point depending on the temperature and closeness of the heat source. Cooling can then slow down the heating or in best case stop it completely.
The heat transfer is is a complicated thing to calculate for real, especially in this kind of environment where winds are probably turbulent and heat is transferred in many forms. Luckily there has been research on the subject and we can use those results for estimating the needed cooling.
From the practical point of fire safety, one of the most important thing seems to be distance of the closest fire front from the house. There is an article on this subject "Reducing the Wildland Fire Threat to Homes: Where and How Much?" by Jack Cohen, nicely summarized in [http://www.saveamericasforests.org/congress/Fire/Cohen.htm]. The article contains a graph of radiant heat flux as function of distance from the wild fire front as well as wood ignition times as function of the distance.
Knowing the radiant heat flux and energy needed for vaporization of water, it is possible to derive an equation for cooling effect of the water:
$q = \frac{m H_{vap}}{A}$
where
- q is radiant heat flux [kW/m^2]
- m is amount of water used per second [kg/s]
- H_vap is heat (or enthalpy) of vaporization of water [kJ/kg]
- A is area of the walls and roof of the house [m^2]
As an example, let's consider a house with outer surface area of 500 m^2. For water, the heat of vaporization is 2257 kJ/kg. The amount of water we can spend is 12 gallons per minute, that is 0.76 liters per second. From this we can work that the maximum cooling effect produced by the cooling system (all water vaporized instantly) would be:
$q = \frac{m H_{vap}}{A} = \frac{(0.76\: \mathrm{kg/s})(2257\: \mathrm{kJ/kg})}{500\: \mathrm{m^2}} = 3.43\: \mathrm{kW/m^2}$
When comparing this to the model of the article where heat flux from for example 20 meters away is is around 45 kW/m^2, and heat flux from 22 meters away is around 40 kW/m^2, we can say that the cooling would have approximately the same effect as moving the tree line by two meters.
The model is known to overestimate the heat flux so the actual distances may be smaller, but anyway the cooling effect has approximately the same effect.
Things to consider:
- I assumed that we can't know what side of the building will be closest to the fire or that house will be surrounded so all sides need to be cooled.
- Distances given in the graph in the article are for wood. For other materials, the distances will be larger or smaller. Thickness and density of the material also matters.
- According to the article, in a full-blown forest fire the burning happens very fast. If the house can stand the fire for two minutes, it won't probably ignite as the fire has moved on.
- Clearing the surrounds of the house and having nonflammable materials would be much more effective way of shielding the house. From the link I gave: "Given nonflammable roofs, Stanford Research Institute (Howard and others 1973) found a 95 percent survival with a clearance of 10 to 18 meters and Foote and Gilless (1996) at Berkeley, found 86 percent home survival with a clearance of 10 meters or more." This might of course have effect on how nice and cozy the yard is.
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
No.
Water is $H_2O$ which has a molecular weight of 18.
Nitrogen is $N_2$ which has a molecular weight of 28.
Oxygen is $O_2$ which has a molecular weight of 32.
Argon is $Ar$ which has an atom weight of 40.
So a water molecule has a mass that is less than that of all the significant components of air.
There is a dynamic equillibrium. When there is a body of water, with air containing water vapor above it, gas phase water molecules DO continuously condense into the liquid water. One way to verify this is to leave a $D_2O$ bottle exposed to the air for a period of time, and look at its proton NMR spectrum before and after.
In other words, if the liquid water is in equillibrium with moist air, the rate at which liquid water molecules enter the gas phase is equal to the rate at which gas water molecules enter the liquid phase. If the system is out of equillibrium, net evaporation or net condensation occurs, until equillibrium is reached.