in an experiment 4 liters of water was observed to evaporate in 20 hours what is the rate of evaporation was observed in this experiment
Answers
(4/20)× 100
(1/5) × 100
=20
Mark as brainliest please
The answer is yes, the rate that water evaporates can indeed be calculated, but it depends on a few more things than you mention. The evaporation rate is influenced by
1) The temperature of the water at the air-water surface
2) The humidity of the air
3) The area of the air-water surface
4) The temperature of the air (more on this below)
In a real-world situation of evaporating water, none of these four quantities above remains constant because the process of evaporation itself changes them. Water evaporating takes quite a lot of heat away -- 540 calories per gram -- when it evaporates. That’s enough to cool down 540 grams of water by a degree, or 50 grams of water a little more than ten degrees. If you are not very careful to replace the lost heat energy during the evaporation, the temperature will go down. And even then the temperature right at the surface will be lower than elsewhere in the water and it will depend on
5) water currents convecting heat and the ability to keep the temperature constant at 100 degrees F.
For a similar reason, the air near the surface of the water will become more saturated with water as the water evaporates. The evaporation rate will depend on:
6) airflow past the water/air surface.
These factors explain why:
People sweat to stay cool.
People say "It’s not the heat, it’s the humidity."
People use fans to keep themselves cool on hot days.
Spraying a fine mist of water in the air will make it evaporate faster (increased total surface area).
, I see that the vapor pressure of water at 100 degrees F is 0.96 pounds/square inch. If the vapor pressure is less than 0.8, the water would not evaporate at all, and in fact in that atmosphere, water would condense.
Irving Langmuir developed a way to measure vapor pressure
by measuring the evaporation rate, which we will now flip around backwards. His reasoning is that the rate at which molecues are lost due to evaporation to a gas with no partial pressure of the evaporating substance is the same as the rate at which molecules of the substance would hit the surface if it were in equilibrium with the vapor (because in equilibrium the evaporation rate and re-condensation rate cancel each other out). His expression is:
(mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) )
The ambient partial pressure of water is the 0.8 PSI you gave me, and the vapor pressure is the 0.96 PSI (we’re going to have to convert this to different units to make it work out). T is the temperature in degrees Kelvin -- 310 degrees above absolute zero. R is the gas constant which is 8.314 Joules/(mol degree K). Naturally the units of measurement in formulas like these are a disaster. I would suggest using standard Metric (SI) units for everything. Multiply the pressures in PSI by 6895 to get pressurs in Pascals (Newtons/sq. M), and the molecular weight should be in kg/mol (to cancel out the moles in the gas constant R). For water this is 0.018 kg/mole.
A real situation involves the fact that the humidity near the interface is much higher than even a short distance away, and that the water vapor must diffuse away. This effect will slow the evaporation down quite a a lot because the evaporation rate is proportional to the difference between the vapor pressure and the partial pressure of the substance, and diffusion can only take water away so fast. As the water evaporates, the partial pressure of water in the gas right over the water will be nearly equal to the vapor pressure, and then it will drop as you go away from the surface, and how steeply this drops (which depends on the airflow rate and how long the water has been there evaporating) determines the rate at which water will diffuse away. Even with a fan blowing air past the surface, the process is limited by diffusion very close to the surface because a thin layer of air (called the "boundary layer") right next to the surface does not move relative to the surface. I won’t do the work on the diffusion because it depends too much on the details of the setup. The diffusion constant for Nitrogen is 0.185 cm**2/sec at room temperature and 1 atm.
Lowering the partial pressure of water will raise the evaporation rate as mentioned above. Lowering the air pressure will increase the diffusion rate. The partial pressure of water at 100 degrees F is so high in fact, that bubbles will form spontaneously in the fluid and cause it to boil rapidly if the water is placed in a vacuum (upsetting the surface area because of all of the bubbles). Increasing the temperature will increase the evaporation rate. It appears in the denominator (in the square root) above, but a much more important dependence comes in with the vapor pressure.