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Re: experimental latent heat of evaporation



L!

"See my reply to Herbert."


I think I'm missing a post(s). i.e. what reply?


"How would two values of L (at 100 C and at 80 C) be calculated in this
method?"


keep the outer container at 79 C.


Note, you can make glass insulated using insulation in a can.

"I do not see why 10 minutes is too long. My container is constantly
heated to maintain the desired temperature. The time can be as long as
I want, in principle. To measure the mass of evaporated water
accurately I need at least several minutes."



I thought a systematic error in a variable depending on time, e.g. using an estimation of "heat" loss, would increase with time, but it's a percentage, and, therefore, longer times will increase the accuracy of the steam's measurement while not increasing the "heat" loss error.


"Fortunately, retired people often have a lot of time. I would very much
appreciate if someone else could perform similar experiments. This can
lead to publication in The Physics Teacher."

Not this one, I'm just tired (after several hours of house work each day).

bc

Ludwik Kowalski wrote:

On Tuesday, Sep 28, 2004, Bernard Cleyet wrote:



I suggest using a double Dewar (if you don't want to do the reverse
xpt.: "work" done on water by condensing steam -- much easier xpt).



See my reply to Herbert.



The
inner container has the water and the resistor. The next one, water w/
another resistor whose temp. is maintained just below boiling, thereby
neither container's water does "work" on either. Boil off some water
(in the inner); remove the inner container, and cool quickly * in
cold
water. Wipe off the outside water and determine the mass loss. Or do
continuously w/ a scale, as you did.



How would two values of L (at 100 C and at 80 C) be calculated in this
method?



* better? put some cold water in the outer container. The thermal
cap. of the resister is negligible?



I assumed so in a "quick and dirty" attempt to see if two Ls are
different. I would not make this assumption if the goal was to
determine L as accurately as possible. Minimizing Qc, by using a Dewar
(instead of my open beaker) is a very good idea. I will try this,
possibly very soon.



The std xptl. method is to use a boiler and steam trap (actually water
trap); run the hose from the trap into a Dewar containing water ~ 10
deg. below RT. When the temp is 10 above RT remove tube (shake off
condensate into the Dewar) and determine mass gain.

bc

p.s. ten min. is a long time for a non insulated container xpt.
For your R value you used the tangent at the beginning of cooling?
Glass, window or pyrex, at what temp.?



Yes, I am using the tangent. It is the tangent at the beginning of
cooling or a tangent at a lower constant temperature, for example, 80
C. I do not see why 10 minutes is too long. My container is constantly
heated to maintain the desired temperature. The time can be as long as
I want, in principle. To measure the mass of evaporated water
accurately I need at least several minutes.


clue:

http://www.engineeringtoolbox.com/24_154.html

note: much more variation than I expelled.

p.p.s. "Frankly", it's not obvious why you have such a great variation.
Was the mass loss constant? The Joule "heating" was, no?



Yes, the mass was being evaporated at a constant rate; I checked this
in each experiment. The voltage and the amperage were very steady in
each experiment. Large fluctuations are totally unexplained. Next time
I will try to measure L several times at the same heating power. If
fluctuations disappear then I will try another power, also several
times. I do not believe that the true L changes when the heating power
is changed. Unfortunately, this becomes a real research project.
Fortunately, retired people often have a lot of time. I would very much
appreciate if someone else could perform similar experiments. This can
lead to publication in The Physics Teacher. Please contact me in
private, if you (or your students) are interested.

Ludwik Kowalski <kowalskil@mail.montclair.edu>



On Monday, Sep 27, 2004, at 23:20 America/New_York, Ludwik Kowalski
wrote:


Prompted by good comments I went to the school laboratory this
afternoon and attempted to measure L. Unfortunately, my results
fluctuated widely and I have no idea why. The average result from 8
attempts was 1344 J/gram (instead of 2260); the smallest value was
800
and the largest was 3030. The method I "invented" is straight
forward;
it has probably been used by others. Expecting critical comments I
will
post a short description of the method tomorrow. Perhaps those who
are
familiar with thermal measurements will help me.
Ludwik Kowalski


Well, I did not wait till tomorrow. Here is my piece.

A 600 cc beaker (mass = 182.5 grams) is filled with 400 grams of
water.
Then it was placed on a wooden tile supported by a scale. That scale
was used to determine the amount of water, m (typically 10 grams),
evaporated during a specified time t (typically 10 minutes). A
resistor
of 40 ohms, connected to a d.c. power supply, was immersed into water.
The heating power was determined from the measuring current and
voltage. The experiment started when water in the beaker (always close
to 400 grams) reached a constant temperature. It consisted of
measuring
m and t. I assume that the constant temperature implies

electric energy supplied = thermal energy lost.

In other words,

V*I*t = Qe + Qc = m*L + Qc (1)

where Qe is the energy needed to evaporate m grams of water and Qc is
the energy lost in time t due to conduction, convection and radiation.
The value of Qc depends on the constant temperature, for example 100 C
or 70 C, on the geometry of the setup and on the known heat capacity
(m1 grams of water and m2 grams of glass). More specifically,

Qc=(c1*m1 + c2*m2)*R*(t/60) (2)

where c1 and c2 are specific heats of water and glass and R is the
experimentally measured cooling rate at the chosen constant
temperature. Division by 60 was necessary because values of R were in
degrees per minute (t is in seconds). At the boiling temperature, for
example, R=3.5 degr/min, while at 80 C it is 2 degr/ min. These values
were obtained from the cooling curve (temperature versus time). One
cooling curve was taken before the the experiment and another was
taken
after the last experiment. The two curves were practically identical.
The heat of evaporation was calculated as

L =( V*I*t - Qc)/m (3)

That is it. I am certain that V*I*t values do not fluctuate by more
than 5%. Likewise, the error in m is most likely smaller than 5%.
Thus,
if there were no errors in Qc then the values of L should be
reproducible within 10% (most likely to within 5%). But my values of L
fluctuate more widely. I suspect that this has something to do with R.
I see nothing wrong with my method of determining R; two cooling
curves
were practically identical. Is my method of finding L acceptable? Do
you see something wrong in its use?

At the beginning I said that eight attempts to find L were made. What
I
did not say was that each attempt was made at a different heating rate
(some with bubbling and some without bubbling). I did not expect
troubles and I wanted to see if L depends on the rate of bubbling. I
see no consistent pattern in the results. I suspect that the values of
L would fluctuate even if the temperature was the same in all eight
experiments. Please help.
Ludwik Kowalski