How much energy is needed to evaporate 1 liter of water in 5 minutes?

In summary, the conversation revolves around finding the fastest and most energy-efficient way to evaporate water. The initial scenario includes having 1 cubic feet of space and 1 liter of water with impurities dissolved in it. The physicians suggest looking at the heat of vaporization of water and mention that time is irrelevant to the amount of energy needed. The conversation also delves into the effects of initial water temperature and the importance of surface area for efficient evaporation. After researching, it is concluded that it would take approximately 2.6 megajoules of energy to evaporate 1 kilogram of water at room temperature, which translates to around 8.7 kilowatts over 5 minutes.
  • #1
orangeglow
9
0
Hello physicians, I need you help with a little experiment. Basically I'm trying to find the fastest way to evaporate water with as little energy as possible.

Let's say I have 1 cubic feet to work with and I want to evaporate 1 liter of water as quickly as possible. I know the impurities matter so we have 1 cup of 'soil' completely dissolved into the water.

I'm looking for a ballpark figure. How much energy would I need to completely evaporate all the water in 5 minutes?
 
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  • #2
Welcome to PF.

Look up the heat of vaporization of water. Multiply by the mass. That'll be your answer.

(hint: time is irrelvant to "energy"...)
 
  • #3
Consideration of the initial temperature of the water?
 
  • #4
DarioC said:
Consideration of the initial temperature of the water?

room temp.
 
  • #5
russ_watters said:
(hint: time is irrelvant to "energy"...)

hmm, are you sure? why then is it that if you drive at a certain speed, you can get higher MPG. IE: traveling 1 mile @ 60mph = 30mpg, 1 mile @ 75mph = 25mpg.
OR evaporating 1 liter of water...
A) simmering it over a long period @ 30C
VS
B) rapidly boil @ 500C.

Are you saying that in A and B the amount of energy used is always going to be the same?
 
  • #6
orangeglow said:
hmm, are you sure? why then is it that if you drive at a certain speed, you can get higher MPG. IE: traveling 1 mile @ 60mph = 30mpg, 1 mile @ 75mph = 25mpg.
OR evaporating 1 liter of water...
A) simmering it over a long period @ 30C
VS
B) rapidly boil @ 500C.

Are you saying that in A and B the amount of energy used is always going to be the same?

This isn't even freshmen physics... The energy is the same. The question in both cases is how much of the energy goes into your target. When driving, friction is the main concern. Almost none of the energy goes into the kinetic energy of the car.

You cannot boil water at 500°C it's temperature will stop at 100°C until it is completely evaporated.

The energy needed to evaporate the water is fixed, the question is -- where does your energy come from and where does it go. When you boil water, a lot of the heat flows into the pot, the air, the stove itself. The amount depends on the temperature of the surroundings and also on the water temperature. The key for speed is surface area. If your pan is extremely hot you suffer from the Leidenfrost effect and the heat cannot reach the water.

I would think the fastest way to evaporate water would be to spray it through a nozzle with very hot and dry gas.
 
  • #7
DarioC said:
Consideration of the initial temperature of the water?
Yes, sorry - plus the sensible heat required to get it up to boiling temp: specific heat times delta-T.
 
  • #8
orangeglow said:
hmm, are you sure? why then is it that if you drive at a certain speed, you can get higher MPG. IE: traveling 1 mile @ 60mph = 30mpg, 1 mile @ 75mph = 25mpg.
OR evaporating 1 liter of water...
A) simmering it over a long period @ 30C
VS
B) rapidly boil @ 500C.

Are you saying that in A and B the amount of energy used is always going to be the same?
If you have an insulated system, yes, A and B are going to be the same (assuming you mean the energy source is at 500C). That example is not analagous to a car's efficiency.

The difficulty, though, is the "insulated" part. Using a stove, your system won't be insulated so you'll have a lot of trouble keeping it warm if your heating element is only at 30C and there will be a lot of energy exchange with the ambient air. But for general principle: boiling and evaporating use the same energy.
 
  • #9
Question cannot be answered as posed. The energy to bring up to boiling temperature and boil the water away has an answer. The 5 minute part makes it an engineering problem about how the energy is delivered and we have no information to work with for that part.

I suggest you might want to go to an old books shop and purchase an elementary book on physics and one for chemistry if you are going to play with such problems.
 
  • #10
After some research, I found that it takes approx. 2.3 Megajoules to evaporate 1 kilogram of water which is @ room temp. Since the mass of room temp water = approx. 1Kg, this means it would take approx. 7,500 watts-hours to completely evaporate 1 liter of water in 5 minutes. Please correct me if I'm wrong. That's some crazy wattage LOL!
 
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  • #11
orangeglow said:
After some research, I found that it takes approx. 2.3 Megajoules to evaporate 1 kilogram of water which is @ room temp. Since the mass of room temp water = approx. 1Kg, this means it would take approx. 7,500 Kilowatt-hours to completely evaporate 1 liter of water in 5 minutes. Please correct me if I'm wrong. That's some crazy wattage LOL!

No that's wrong. It is about 2.6 MJ/kg (raising from 20C), but that is only 0.722 KW-hrs (2600 kJ divide 3600 seconds).

Over 5 minutes this corresponds to 722W * 60/5 = 8.7 KW.

In practice you might need quite a bit more due to heat loss. Also you'll need it spread over a fairly broad pan to get efficient vaporization without any superheating (which would cost you more energy).
 
  • #12
I suppose the 8.7 KW answer is the one to evaporate water into 100% humidity. The other answer is zero Watts with the water spread over a thin surface.
 
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  • #13
uart said:
No that's wrong. It is about 2.6 MJ/kg (raising from 20C), but that is only 0.722 KW-hrs (2600 kJ divide 3600 seconds).

Over 5 minutes this corresponds to 722W * 60/5 = 8.7 KW.

In practice you might need quite a bit more due to heat loss. Also you'll need it spread over a fairly broad pan to get efficient vaporization without any superheating (which would cost you more energy).

oops, I meant 7,500 watts, not kilowatts.
 
  • #14
Confirm the 2.6 MJ. I got 2.574 from 25*C. Kilowatts look correct also. You can't actually do it in any practical way, but those ARE the numbers if you could.
 
  • #15
uart said:
Over 5 minutes this corresponds to 722W * 60/5 = 8.7 KW.
Note that the entire power output of a typical electric range/oven is about 8 kW, so in practice it isn't really possible to boil away 1L of water without unusual equipment.
In practice you might need quite a bit more due to heat loss. Also you'll need it spread over a fairly broad pan to get efficient vaporization without any superheating (which would cost you more energy).
I don't know that either of those would really be a big factor. If you insulate the pot, it won't lose much heat and on an electric range, the pot is touching the heating element, so there really isn't much heat loss. And superheating wouldn't be an issue because the pot boils from the bottom-up and bubbles of steam interact with/exchange energy with water on the way up.
 

1. What is the energy required to evaporate 1 liter of water in 5 minutes?

The energy required to evaporate 1 liter of water in 5 minutes depends on several factors, such as the initial temperature of the water, the humidity of the surrounding air, and the pressure. On average, it takes about 2260 joules of energy to evaporate 1 gram of water, which means that 1 liter of water would require approximately 2.26 million joules of energy to evaporate in 5 minutes.

2. How does the initial temperature of the water affect the energy needed for evaporation?

The initial temperature of the water has a significant impact on the energy required for evaporation. The higher the initial temperature of the water, the more energy is needed to break the bonds between the water molecules and turn them into vapor. Therefore, water at a higher temperature will require more energy to evaporate in 5 minutes compared to water at a lower temperature.

3. Does the humidity of the surrounding air affect the energy needed for evaporation?

Yes, the humidity of the surrounding air does affect the energy needed for evaporation. When the humidity is high, the air already contains a significant amount of water vapor, making it more difficult for the water molecules to escape into the air. This means that more energy is required to evaporate 1 liter of water in 5 minutes in humid conditions compared to dry conditions.

4. How does pressure affect the energy needed for evaporation?

Pressure also plays a role in the energy required for evaporation. Generally, as pressure decreases, the boiling point of water also decreases, making it easier for water to evaporate. This means that at lower pressures, less energy is needed to evaporate 1 liter of water in 5 minutes compared to higher pressures.

5. Is it possible to calculate the exact amount of energy needed to evaporate 1 liter of water in 5 minutes?

While it is possible to estimate the amount of energy needed to evaporate 1 liter of water in 5 minutes, it is challenging to calculate the exact amount. This is because there are many variables involved, such as the initial temperature, humidity, and pressure, which can all vary and affect the energy required. Additionally, the process of evaporation is complex and affected by other factors, such as air flow and surface area, making it difficult to calculate an exact value.

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