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

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Discussion Overview

The discussion revolves around the energy required to evaporate 1 liter of water in 5 minutes, considering various factors such as initial temperature, impurities, and the efficiency of energy transfer. Participants explore theoretical and practical aspects of the problem, including the implications of different heating methods.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks a ballpark figure for the energy needed to evaporate 1 liter of water with impurities in a short time frame.
  • Another suggests using the heat of vaporization of water multiplied by mass to find the energy required, stating that time is irrelevant to energy calculations.
  • Some participants raise the importance of the initial temperature of the water and the sensible heat required to reach boiling temperature.
  • There is a discussion about the efficiency of energy use in different heating scenarios, such as simmering versus rapid boiling, with one participant arguing that energy loss to the environment is a significant factor.
  • One participant proposes that spraying water through a nozzle with hot gas could be the fastest method for evaporation.
  • Another participant calculates that approximately 2.3 Megajoules are needed to evaporate 1 kilogram of water, leading to a discussion about the power requirements over 5 minutes.
  • There are corrections regarding the energy calculations, with some participants providing different values for the heat of vaporization and the resultant power needed.
  • One participant notes that practical limitations, such as heat loss and the need for a broad pan for efficient vaporization, complicate the theoretical calculations.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of time in energy calculations, the efficiency of various heating methods, and the practical feasibility of evaporating water in the specified timeframe. No consensus is reached on the best approach or the exact energy requirements.

Contextual Notes

Participants mention the need for additional information regarding energy delivery methods and practical constraints, such as heat loss and surface area considerations, which remain unresolved.

orangeglow
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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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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"...)
 
Consideration of the initial temperature of the water?
 
DarioC said:
Consideration of the initial temperature of the water?

room temp.
 
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?
 
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.
 
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.
 
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.
 
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.
 

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