Re: experimental latent heat of evaporation

[Ludwik Kowalski <kowalskil@MAIL.MONTCLAIR.EDU>]

Not knowing what the "d.c." means (in the context of my power supply),
and suspecting the worse, I used the oscilloscope. The amplitude of 60
Hz ripples, at my highest current (2.6 A and 115 V) was only 2 V. That
was good enough for me. I do not remember how the voltmeter was
connected. I did not pay attention because the heater resistance was
about 40 ohms. I assumed that the resistance of the ammeter is
negligible in comparison. But this has to be checked next time. My
highest electric power was 215 V * 2.6 A = 299 W (very intense
boiling); my lowest electric power was 61 W ( 1 A, 61 V and temperature
about 65 C). I naively expected that data collected at school will show
me how L depends on heating power. But I was surprised when
calculations were performed at home. Next time I will calculated L
after each experiment.
Ludwik Kowalski

On Tuesday, Sep 28, 2004, Michael Edmiston wrote:

> Here is one possibility to consider.  Add an oscilloscope across the
> resistor and see if you have a lot of ripple on the DC.  If the DC
> power
> supply is not regulated, there could be significant ripple, and the
> ripple will be worse at higher current.
>
> If you have a lot of ripple your voltmeter and ammeter probably are not
> reading the correct values.  You need the rms values of the voltage and
> current, but typical multimeters set to DC voltage and DC current will
> not measure the true rms values of a fluctuating voltage and current.
>
> Here is another possibility.  Make sure your voltmeter is measuring the
> voltage across just the resistor and not the voltage across the
> resistor
> and ammeter.  There is a voltage drop across the ammeter and thus some
> heat dissipated there.  If your voltmeter is reading the voltage across
> the power supply, as opposed to the voltage across the resistor, you
> will overestimate the power in the resistor, and it will be worse when
> the current is higher.
>
> You also have slight voltage drop in the lead wires, but that is minor
> compared to the voltage drop across the ammeter.  I think lead-wire
> resistance is not likely your problem.  However, to eliminate wire
> errors, run a separate set of wires from the resistor to the voltmeter
> so you are measuring just the voltage drop across the resistor, or
> better... the voltage drop across everything that is under water.
>
> Summary... If there is a lot of ripple, you probably have incorrect
> values for both current and voltage.  If there is not a lot of ripple,
> but you are measuring the voltage at the power supply, you have an
> incorrect voltage.  And, of course, you could have both problems.
>
> Michael D. Edmiston, Ph.D.
> Professor of Physics and Chemistry
> Bluffton University
> Bluffton, OH 45817
> (419)-358-3270
> edmiston@bluffton.edu