Fastest way to evaporate water?
Greg Kuperberg
fastest way to boil a quart of water on a stove?" E.g., do you cover
it until it boils, or leave it uncovered the whole time?
People have responded with a lot of advice, which is fine as far is goes.
Unfortunately people are contradicting each other and there is too little
theory in the discussion for me to tell fact from falsehood.
Here are two extreme cases which may be of interest: If you want
to boil lead, you should certainly not fan it as it heats.
You should probably insulate it and vent it to let the vapor escape.
Stirring may or may not matter.
If you want to boil liquid nitrogen, you should certainly leave it
uncovered and stir it and fan it as you heat it.
dca...@home.com
to keep it covered till the water gets to 100C then remove the lid.
I'll test it the next time I make spaghetti :>
On 12 Nov 1998 05:06:29 GMT, Greg Kuperberg <gr...@math.ucdavis.edu>
wrote:
Roy McCammon
>
> For those who have lost track, the original question was, "What is the
> fastest way to boil a quart of water on a stove?" E.g., do you cover
> it until it boils, or leave it uncovered the whole time?
The question was more or less "what is the fastest way to evaporate
half the water away from broth".
Ralph Frost
>
> For those who have lost track, the original question was, "What is the
> fastest way to boil a quart of water on a stove?" E.g., do you cover
> it until it boils, or leave it uncovered the whole time?
>
What do you think, Greg?
Dan Christensen
> For those who have lost track, the original question was, "What is the
> fastest way to boil a quart of water on a stove?" E.g., do you cover
> it until it boils, or leave it uncovered the whole time?
>
> Unfortunately people are contradicting each other and there is too little
> theory in the discussion for me to tell fact from falsehood.
My impression from the discussion is that one of the posters was
implicitly ignoring heat lost to the environment. Under that
assumption, after a specific amount of heat has been transferred from
the burner to the pot, all the water will be gone. Therefore, it is
best to maximize the transfer of heat, and this is accomplished by
leaving the lid off. Because of the assumption about heat loss, this
isn't a great model of reality, but it is a useful simplification that
I think we understand completely.
Another poster didn't explicitly bring up the issue of heat loss
to the environment, but it seemed that he was using this intuitively
to conclude that the lid should be left on until a certain point and
then removed.
My intuition is that the lid should be left on to reduce heat loss
to the environment, but that it should be periodically removed very
briefly to reduce the vapor pressure in the air space in the pot.
This is an approximation to the best parts of the above two arguments.
(I am assuming that the ambient temperature is lower than the boiling
point of the substance to be boiled to avoid the extreme case of
liquid nitrogen.)
Another extreme situation is when the heat source is very weak. With
no heat source we want the lid off all the time. And I suspect that
in general the optimal percentage of time that the lid is on will
depend on the strength of the heat source and the temperature of
the water; in particular, it will vary as the experiment progresses.
Maybe it's time to place bets and have someone do an experiment? :-)
Dan
--
Dan Christensen
j...@math.jhu.edu
Patrick Van Esch
> My impression from the discussion is that one of the posters was
> implicitly ignoring heat lost to the environment. Under that
> assumption, after a specific amount of heat has been transferred from
> the burner to the pot, all the water will be gone. Therefore, it is
> best to maximize the transfer of heat, and this is accomplished by
> leaving the lid off.
>
> My intuition is that the lid should be left on to reduce heat loss
> to the environment, but that it should be periodically removed very
> briefly to reduce the vapor pressure in the air space in the pot.
This cannot be the case. In fact, the same reasoning (keep the
temperature of the water as LOW as possible - hence lid off) that
makes the heat transfer from heat source TO pot maximal, also MINIMIZES
the heat loss from pot to environment.
cheers,
Patrick.
Stein Vidar Hagfors Haugan
<Joke mode - exaggeration follows>
Oh - so by maximizing heat loss, the temperature goes down, and
we're minimizing the heat loss...? But then the temperature
would go up, so we're loosing more heat... and then....
</Joke mode>
Seriously, I think Dan Christensen has a very good point. Heat
loss hasn't been considered carefully enough in this discussion,
in several ways. Patricks comment *is* in some way valid, too.
The thing is, we have several "heat loss" mechanisms here, and
we haven't yet discussed the *heating* mechanism at all!
It seems that some people implicitly assume a heat source that's
automagically kept at a fixed temperature. Is this a realistic
assumption? I would think not - at least for an electric stove,
where assuming a constant power input is more like it. Hence, if
the heat sink is inefficient, the temperature of the warming
plate goes up. (I should say that when talking about kitchen
appliances, my knowledge of English is a bit ... vague, to say
the least, so bear with me...).
The way I see it, we have a system that can be described this
way:
Q H E
Heat Input -> Warming Plate -> Cooking vessel -> Evaporation
| |
| P | V
v v
Heat loss Heat loss
(to stove & surroundings) (radiation & heating air)
Of course, I could have left out something, but this is a good
start, anyway.
Now, the physical variables of this system are:
Q : Heat input
Tp : Temperature of warming plate
P(Tp) : Heat loss (to surroundings) from warming plate
Tv : Temperature of cooking vessel
V(Tv) : Heat loss from cooking vessel
E(Tv) : Energy "lost" (Good Thing (TM)) by evaporation
H(Tp,Tv): Heat transfer from plate to cooking vessel
Clearly, V(Tv) and E(Tv) are dependant on the "geometry",
i.e. lid off or lid on. Note, however, that if you get P(Tp) and
V(Tv) both to be zero, you'll automatically maximize the value
of E - unless you have a pressure cooker and you want the water
to be removed in an explosion at the end :-)
Also, given:
d/dt(Tp) \propto Q - P - H
d/dt(Tv) \propto H - V - E
and given plausible postulates/approximations on the functional
shape (and relative magnitudes!) of P(Tp), V(Tv), E(Tv), and
H(Tp,Tv) given various options (lid on/off), we should get The
Answer (tm) (pun intended :-), by equating
\int_0^tm E dt = m
where m is the amount of water to be boiled away..or rather the
energy required to boil away that amount. The geometry with
the smallest tm wins.
I leave it as an exercise to the readers to estimate the
functions (or to perform experiments to determine the answer).
Regards,
Stein Vidar H. Haugan
Charles B. Schroebel
> Greg Kuperberg <gr...@math.ucdavis.edu> writes:
> > For those who have lost track, the original question was, "What is the
> > fastest way to boil a quart of water on a stove?" E.g., do you cover
> > it until it boils, or leave it uncovered the whole time?
[snip]
[snip]
> the water; in particular, it will vary as the experiment
progresses.
>
> Maybe it's time to place bets and have someone do an experiment? :-)
>
> Dan
Use a pressure cooker type device, but with much higher pressure
ratings(well insulated, if possible), leave the lid on, apply all the heat
you have (calculate it based on initial T and the heat of vaporization)
and then (being far away from the now 300+C pot) vent the system
to atmospheric.
If you leave the lid on in a standard, non insulated pot, the water
vapor will condense on the upper part of the pot and the lid (that's
why we need to insulate) and the heat of vaporization will be transferred
to the pot and the water will reflux; by cycling (removing the lid) you're
just decreasing the reflux but you're not gaining anything else,
leaving the lid open allows for radiative loss from the water surface
and the lower part of the exposed pot. Covering the pot before the
water boils has little effect for a standard-sized pot--would make
a bigger difference if it were Lake Michigan.
[do not cook spaghetti with the cover on because you want to boil
vigorously and if you do that with the lid on you'll make a mess,
and remember to add a tablespoon or two (i use more, it's insoluable)
to keep the strands from sticking together. I'd be willing to watch
your experiment and even test the product :-)]
covering the pot lets you simmer at minimum energy input.
try it.
charles b. schroebel
box 1205 baltimore, md 21203-1205
cschroeb@umaryland.
geht noch ein?
noch ein geht immer noch!
Patrick Van Esch
> > temperature of the water as LOW as possible - hence lid off) that
> > makes the heat transfer from heat source TO pot maximal, also MINIMIZES
> > the heat loss from pot to environment.
>
> No, I don't think so. The lid provides some insulation, and also
> reduces heat loss by convection. Hot water in an enclosed pot
> can lose heat more slowly than somewhat cooler water in an open pot.
Yes, but with the lid on, you cannot evaporate water.
The best way to evaporate water is to have a HUGE contact area between
the water and the air, and to have the water essentially at the same
temperature as the environment. In that case, there is NO heat loss.
So if you succeed in adding only the latent heat required for evaporation,
all the water will be evaporated. This will never be the case when you
work at higher temperatures.
cheers,
Patrick.
David Combs
>and remember to add a tablespoon or two (i use more, it's insoluable)
>to keep the strands from sticking together. I'd be willing to watch
>your experiment and even test the product :-)]
>
>covering the pot lets you simmer at minimum energy input.
>
>try it.
>
Add a tablespoon or two OF WHAT?
Thanks
Alexander Y. Vlasov
>> the water and the air, and to have the water essentially at the same
>
>boil down my broth to half its volume in the least amount of time?".
Seems there is some difference between boil and evaporate: if to consider the
question in Subject formally, I could suppose a "standard" way for fast
(few seconds) evaporation of water in physical research style:
====
||
+---==========++==========---+
| +-======================-+ |
| | thermoinsulting tube | |
+---+ +--+ (with valve!) +---+ +---+
| | | |#########| |
|-o--o-o-| |~~|absorbent|~~|
|- water-| |~~|# carbon#|~~|
+--------+ |~~+---------+~~|
|~~ liquid N2 ~~|
+---------------+
So, it is possible to evaporate water with using liquid nitrogen --- the right
part of diagram is simplest fore-vacuum pump, absorbent carbon after contact
with low temperature (77K) of liquid nitrogen absorbs a lot of gas and if
volume of water is small, it can disapers during few seconds (the problem is
that evaporation can be too effective and water not only boil, it can freez
at the same time, even if you use usual pump without any additional cooling
like in the picture --- excellent phenomena, I saw it).
Warning: It is better do not try to prepare food such a way before reading a
manual about the carbon pumps, it is a bit dangerous, because after heating
to room temperature it returns part of absorbed gas with possible destruction
of part of the experimental equipment if valve would not open when necessary.