While evaporative cooling is not strictly a heat transfer mechanism since it involves a change of state and a loss of mass (that of the evaporated gas), it does result in a rapid temperature change in cases like the cooling of a cup of coffee, and in practice we don't generally worry about actual volume of coffee loss while it's steaming, but after this discussion we might worry about the heat loss that it causes. [br][br]It turns out that the process of evaporation of a liquid is not at all trivial to handle from a theoretical perspective. Many texts incorrectly suggest that a Maxwell-Boltzmann velocity distribution explains it, but this is an over-simplification since liquids are not ideal gases. There are many aspects of a real liquid that are not correctly addressed by an ideal gas. [br][br]In any case, without going into those discussions, it turns out an empirical equation (one derived from experience and experiment) can be generated to predict the heat loss of a liquid with a surface from which a portion of the liquid is evaporating. The rate of heat loss from an evaporating water-based liquid (like coffee or a swimming pool) is given by: [br][br][center][math]\frac{dQ}{dt}=-L_v\frac{25+19v}{3600}A(x_s-x)[/math][/center] where [math]x_s=4.147\times10^{-3}e^{0.06281T_c}[/math], using the temperature of the liquid in degrees celsius. The term x is given by [math]x=x_sr[/math] evaluated at the environmental temperature, with [math]r[/math] given by the relative humidity of the environment. It should be noted that in order to get units of watts, we need to use [math]L_v=2.27\times10^6 \frac{J}{kg}[/math], which is not a conventional combination of units for this latent heat of vaporization. The [math]v[/math] in the numerator of the expression is the speed of air flow over the liquid surface in standard units of m/s. Experience tells us that blowing on food or hot drinks speeds the cooling. This term quantifies that increased rate of cooling. You also know that when you are wet, that if a wind is blowing, you feel much colder. That is the same process of evaporative cooling at work.[br][br]To arrive at a rate of change of temperature we need to relate heat and temperature as we have done before. As we did with other heat loss mechanisms, we can write [math]\frac{dT}{dt}=\frac{1}{mc}\frac{dQ}{dt}[/math]. [br]Doing this and also inserting the functions for [math]x_s[/math] and [math]x[/math], we get a final expression:[br][br][center][math]\frac{dT}{dt}=-L_v\frac{25+19v}{3600mc}A(4.147\times10^{-3}e^{0.06281T_c}-r4.147\times10^{-3}e^{0.06281T_{environment}}).[/math][/center]

[color=#1e84cc][i]At what rate will the temperature be changing (K/s) due to evaporation from a cup of 300 ml of coffee with an open coffee cup of radius 4.0 cm if the coffee is 90 C and the surrounding air is 20 C with a slight wind of 1.0 m/s blowing past since the person is walking with the coffee? The relative humidity is 30%.[br][br]The density of coffee (assumed to have the same density as water) is 1g/ml, so our coffee has a mass of 0.300 kg. The surface area from which coffee evaporates is just the area of a circle, which gives A=0.0050 m[sup]2[/sup]. Plugging into the equation above for rate of change of temperature gives dT/dt=-0.13K/s, which is the same as Celsius degrees per second. This rate should alarm you if you like your coffee hot! It'll lose one degree centigrade in just over seven seconds and ten in the first minute and a half![/i][/color]

[color=#1e84cc][i]Sunlight reaches earth's surface with an intensity, I (power per unit area) of 1050 W/m[sup]2[/sup] in absolutely ideal conditions where there are no clouds in the sky, minimal humidity and from a location where the sun is directly overhead. In conditions like in southern California during the day even, this number is closer to 700 W/m[sup]2[/sup]. If the person in the previous problem is walking on a sunny day and the coffee is in the sunlight, how does this affect the rate of temperature change in the previous example?[br][br]Here the coffee is receiving heat from the sunlight at a rate given by [math]P_{absorbed}=\epsilon IA.[/math] The sunlight is absorbed in proportion to the emissivity of coffee-colored water, which is around 0.96. So only 96% of the sun's energy is absorbed. If we solve for the rate of change of temperature and compare it to the previous number we can discern whether the sun compensates for the evaporative cooling. Using numbers, we find that the heating rate of the coffee due to sunlight is [math]\frac{dT}{dt}=\frac{P_{absorbed}}{mc}=0.0028K/s.[/math]The overall cooling due to the combination of these two effects is additive. [/i][i]This will always be true when we have combined effects. Please note that the sunlight will give us a positive rate of change of temperature and that the evaporation will be negative. [/i][/color][math]\frac{dQ}{dt}=\epsilon IA=\epsilon mc\frac{dT}{dt}[/math][math]\frac{dT}{dt_{total}}=\frac{dT}{dt_{evap}}+\frac{dT}{dt_{sun}}[/math][br][br]Given the present numbers, the sunlight will do almost nothing to offset the evaporative cooling. You have probably noticed that stepping out of a pool, there is little comfort due to the warm sunlight until the water is done evaporating off your skin.