If I were to, say, mop my floor with warm water (that is hotter than the air temperature in the room), then let the floor dry naturally, would the air temperature in the room go up or down as a result? Suppose the water is maybe 10degC warmer than the air and the relative humidity is fairly low (eg. 30% relative humidity).
Does it change if the water is really hot, eg. close to boiling? I am curious if more energy is taken out of the air in order to evaporate the water than is transferred from the hotter water to the air.
The enthalpy of vaporization is huge compared to the specific heat capacity. It takes around 6 times as much energy to evaporate water as it does to raise the temperature of liquid water all the way from 0 deg C to 100 deg C. So the net effect after the floor has dried is certainly cooling. I’m not sure whether it would significantly affect the air temperature, I would think most of the heat would be extracted from the floor, which is a huge heat sink.
The temperature of a body of liquid water reflects the average kinetic energy of all of its molecules, but that energy varies widely from one molecule to another. Some of the molecules are energetic enough to spontaneously transition to vapor, while others are sluggish enough that if considered in isolation, they would be colder than the bulk temperature of that body of liquid water.
So what happens is that the energetic liquid water molecules do indeed spontaneously transition to vapor, taking their excessive energy with them and leaving behind a pool of sluggish/cold liquid water molecules.
A pool of liquid water that is above ambient temp will at first warm the air, because the vapor it emits will be above ambient temp. But this evaporation cools the remaining pool of liquid water, and at some point it will be below ambient temp and will begin taking heat out of the ambient air (as well as whatever surface it’s in contact with).
The main point here is this: when evaporation happens, the thing that cools down is the liquid water left behind.
You’re right, but the air is heavier than most folks would suspect. A square meter of 3/4" plywood subfloor weighs about 9.5 kilograms, whereas the three cubic meters of air above it weigh about 3.6 kilograms. Air can also circulate throughout your home: you may have only mopped the kitchen, but free and forced convection will eventually distribute that heat and moisture throughout your home, probably more quickly than the subfloor will. In a 2000-square-foot (600-square-meter) home, the air weighs about 720 kilograms, versus maybe 100 kilograms for the kitchen subfloor.
I’d be interested to hear some one propose a way to determine this using direct measurements. It seems problems like this must arise in engineering contexts.
My initial thought is to simply measure the temperature directly above and below the surface of the floor as the water evaporates, while keeping the air as still as possible. Would something that simple allow an accurate determination as in the question?
As I noted earlier, the first thing to cool down will be the remaining liquid water. This water will then draw heat from both the air above and the floor below. If you eliminate convection to maximum extent possible, the air immediately above the water will equilibrate to the water temperature sooner than the floor does, since it’s got lower conductivity and mass than the floor - but the floor will give more thermal energy to the water, since it’s got more mass and conductivity than the the air.
To simplify, I will assume that the water is sprayed into the room air rather than evaporating from the floor. This will simplify if the floor is insulated or not etc etc
So, lets take a 4m length x 4m width x 3m high room (13ft x 13ft x 10ft) room (well insulated). Say the air temperature is 20C (68F) and it has 30% relative humidity :
Volume of air in room = 48 m3 (1700 ft3), Pressure = 1 atm (14.7 psia)
If the room had 100% relative humidity, then the amount of water contained in the air would be 0.82 kg (1.82 lb). Since you said, 30% relative humidity, the amount of water in the air is 0.25 kg (0.55 lb).
1> Spray additional 0.25kg (.55lb) of water, at 30C (86F), into the air of the room (Room air initially at 20C (68F)) : Final Room Air Temperature = 11.2 C (52F)
2> Spray additional 0.25kg (.55lb) of water, at 95C (203 F), into the air of the room (Room air initially at 20C (68F)) : Final Room Air Temperature = 11.7 C (53F)
As you can see from above, the difference is water temperature does not change the final temperature appreciably.
I had those calcs handy because last week my daughter and I were gardening and noticed something interesting. We live in Texas and it gets hot here. The water sitting in the garden hose would scald you, if you let it touch your skin.
However, if you set the hose nozzle on mist and stand a bit away from the stream, the mist is always cool : no matter if the scalding hot water is coming out or the cool tap water.
Evaporation quickly cools down the mist, and the water temperature doesnt play a big part.