Boiling off Liquid Argon Using A High Powered Resistor

[russ_watters]

This passed without enough notice:

If this will be in a vacuum later then heat transfer from the environment to the dewar will be greatly reduced. You should take this into account.

At room temperature and pressure, the majority of that 1KW that V50 calculated is via convection. Removing the air removes the convection and the boiling rate will drop significantly unless that power is replaced. @Phys12, this takes you backwards!

 

[russ_watters]

The reason why I used the equation of heat capacity is because this is what my actual experiment was going to look like: I would first put my dewar in high vacuum and then fill it with liquid argon. After that, I made a crude approximation to find out the power I may need and assumed that initially the argon wouldn't be boiling at all (since it's not at room temperature as the dewar was closed and evacuated) and I would have to raise the temperature of it in order to make it boil. I didn't go through the process of first calculating the heat it takes to raise the liquid to its boiling point and then using specific heat to figure out how much of the liquid I'm converting to gas in what amount of time given the amount of heat I'm adding...

So, I think that calculation is wrong, but for a different reason. Reason being that I calculated only the heat needed to raise the temperature of the liquid by a few Kelvin, but not the heat required to convert that volume to gas (note that the calculation which yielded 3600 kJ was to raise the temperature of about 100 L of liquid argon to about 10K). I used 10K since the difference in the temperature of liquid argon and its boiling point is about a few Kelvin...

I only considered the part of raising the liquid's temperature to its boiling point...

To be precise, it would have giving me lower heat than I actually need since I first need to heat the liquid to its boiling point (which I calculated) and then convert it to gas (which I missed). I don't understand where I insisted that the wrong equation will give me the right answer...I may have insisted that I do not have a misunderstanding when it comes to knowing the difference between heat and temperature and between latent heat and heat capacity, which I think I'm right about.

I'm at a bit of a loss here, and I promise I'm not trying to be mean, but these statements reflect a near total misunderstanding of the relevant basic chemistry of heating and boiling a liquid. Things that should have been taught in high school.

What you've said you want to do is transfer liquid argon from a (presumably) room temperature and pressure storage tank to a room temperature evacuated dewar. Here's what happens:

First, the environment around the storage tank is room temperature, but the argon is at its atmospheric pressure boiling point of 87K. Principle: The boiling point of a chemical liquid is a function of pressure (higher pressure = higher boiling point). The boiling point has nothing to do with the temperature of the environment around it.

When you transfer the argon to the evacuated dewar, presumably by hose and controlled manually by a fill valve, the argon will have to cool the hose and dewar from room temperature, which heats and boils-off the argon (so your vacuum pump will need to deal with that). Definition: "heat" (verb) means to transfer or add thermal energy to something. Heat (noun: thermal energy in transit) flows from warmer things to colder things, but the objects do not need to be "hot" to transfer heat. You expressed surprise that your resistor was cold when you pulled it out of the liquid nitrogen: of course it was cold - it was just immersed in liquid nitrogen! But that doesn't mean it wasn't doing heating.

Simultaneous to the heating but more specific, the argon pressure drops from atmospheric pressure to near vacuum as it passes through the valve. Per the principle above, the boiling point of the argon *instantly* drops. Due to the mismatch in temperatures, argon now has an excess of thermal energy and a portion of it *instantly* vaporizes ("flashes").

Eventually, this process will reach a steady state where you have:
-Liquid argon flowing from the storage tank at atmospheric pressure.
-Liquid argon flows through the valve, its pressure and temperature drops, and a portion flashes to vapor.
-A liquid/vapor argon mix flows into the dewar where the liquid drops to the bottom and absorbs heat from whatever sources are available (radiation through the vacuum chamber, conduction, an electric resistor in the dewar, whatever) and boils.
-The vacuum pump draws-off very cold, low pressure argon vapor.

As you can see, *nowhere* in this process is the temperature of the liquid argon rising to get to its boiling point. Being *exactly at* its boiling point is its default state.

 

[cjl]

The goal is to recycle argon, while also increasing its purity. Over time, we expect the purity of argon to decrease and we're hoping this to help with that.

The only mechanism I can think of that would cause the purity to decrease though would be that water and oxygen would condense into your liquid argon, due to their higher boiling points and presence in the atmosphere. Unfortunately, though, if this happens, boiling off the argon and trying to purify the vapor that comes off is going to increase the concentration of impurities in the remaining liquid, not decrease it. This will happen because the argon, being the component of the mixture with the lowest boiling point, will preferentially boil off first, leaving the remaining liquid enriched in oxygen and water relative to the initial concentration.

If you really need to keep the argon pure, your best bet is to ensure that oxygen, water, and other impurities don't come in contact with the free surface of the argon in the first place by purging your container with argon after filling. Luckily for you, argon is denser than air, so even if you don't take any special precautions, the gas directly above your liquid should be nearly 100% argon just from normal boiloff as long as you don't have any air currents to disturb it.