Suppose there is a wet floor under normal conditions, how much can we alter the rate of evaporation of the wet floor by trying out different types of lights ex. Incandascnet, cfl, fluorescent etc ? I assume that these lighting solutions produce all types of EM waves but they must have distributed it unsymmetrically over the spectrum, now waterhas the frequency that equals that of microwave so my first guess was that turning on the light which emits maximum microwave would produce the fastest results. But then incandescent bulb produces lots oh heat which can heat up the water and then make it evaporate, how to compare and what would be the best solution ?
[Physics] How much does lighting of room effect evaporation
evaporation
Best Answer
I strongly suspect that the choice of bulb will make no noticeable difference in evaporation rate. The total energy density striking the floor due to a 100 W bulb, even assuming that all of the energy is converted to light, is on the order of 120 $\mu \mbox{W}/\mbox{cm}^2$, which is tiny, and real bulbs are only on the order of 1% to 5% efficient, with most of the rest of the energy carried away as heat via convective cooling of the lamp surface by air. So, it's probably more like at most 20 $\mu \mbox{W}/\mbox{cm}^2$, and in either case, this is a tiny power density.
In contrast, you need power densities on the order of 10,000 $\mu \mbox{W}/\mbox{cm}^2$ to be able to see radiation make a noticeable color change on thermal-sensitive liquid crystal sheets, and those are thin, unlike the floor, which is thick and thus can also dissipate heat via conductive cooling. So the radiation due to a single light bulb is unlikely to have any impact on the heating and subsequent evaporation of water on a wet floor.
Likewise, the amount of light emitted by each bulb which is on-resonance with water transitions (microwave and infrared absorption lines) is probably not relevant. This is because in the condensed phase, rotational and vibrational relaxation times (which redistribute energy from the excited states into thermal energy) are on the order of milliseconds or shorter, whereas the evaporation process takes on the order of days or longer, so the system will more or less be at thermal equilibrium. So while a few water molecules may every now and then get excited, they'll hop back down immediately and shed the energy as heat, so it probably won't be any different than the nonresonant heating.
A large room fan blowing air around, however, may speed things up noticeably.