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


by orangeglow
Tags: energy, evaporate, liter, minutes, water
orangeglow
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#1
Apr30-10, 11:26 PM
P: 9
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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russ_watters
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#2
May1-10, 01:09 AM
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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"...)
DarioC
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#3
May1-10, 01:26 AM
P: 187
Consideration of the initial temperature of the water?

orangeglow
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#4
May1-10, 01:38 AM
P: 9

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


Quote Quote by DarioC View Post
Consideration of the initial temperature of the water?
room temp.
orangeglow
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#5
May1-10, 02:05 AM
P: 9
Quote Quote by russ_watters View Post
(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?
0xDEADBEEF
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#6
May1-10, 04:31 AM
P: 824
Quote Quote by orangeglow View Post
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 500C it's temperature will stop at 100C 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.
russ_watters
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#7
May1-10, 09:40 AM
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Quote Quote by DarioC View Post
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.
russ_watters
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#8
May1-10, 09:42 AM
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Quote Quote by orangeglow View Post
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.
DarioC
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#9
May1-10, 06:50 PM
P: 187
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.
orangeglow
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#10
May1-10, 09:52 PM
P: 9
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!
uart
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#11
May1-10, 11:43 PM
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Quote Quote by orangeglow View Post
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).
Phrak
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#12
May1-10, 11:56 PM
P: 4,513
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.
orangeglow
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#13
May2-10, 01:19 AM
P: 9
Quote Quote by uart View Post
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.
DarioC
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#14
May2-10, 03:28 AM
P: 187
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.
russ_watters
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#15
May2-10, 08:26 AM
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Quote Quote by uart View Post
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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