# Why does a warm weighing cup weigh less?

• dye45714
In summary, the job of a summarizer is to summarize the content of a conversation. The summarizer of this conversation is trying to explain why a metal cup might not be accurate when it comes to weighing oil and grease. The summarizer explains that, if you are looking for 0.0005g, that doesn't sound too crazy, after all.f

#### dye45714

One of my jobs at work is to determine the amount of oil and grease present in wastewater samples. In the beginning of the procedure I usually take a tray of oil and grease cups out of a 125 degree oven and place it in a desiccator for roughly an hour before weighing on a balance. After extracting the oil and grease out of the water and into the cup it is again dried in an oven and placed in a desiccator to cool before weighing back.

The first cup in the tray I use as a constant which has to be within + or - 0.0005g of it's initial weight. If I leave the cups cooling too long, my first cup will be up to 2-3mg heavier than where it started. If I pull them too early then my constant is light.

I realize this is a very basic question but why does this happen? Is it because the heat transferring to the gas surrounding the cup makes the gas rise and therefore exerts less pressure down on the cup when it is hot? When the cup is cold it can't excite the gas away as well?

I did not quite understood the process...is there a pool of wastewater and then you bring an empty cup and dunk into waste water to fill it up? or what?

the process for extraction doesn't really matter.

I initially weigh an empty metal cup that has been cooled in a desiccator after being removed from an oven. Later, I take the same cup which has had nothing added, dry it in an oven, cool it in a desiccator and weigh it again on a balance.

Very difficult to give an answer here.

My hypothesis is that, as is cools and sits it begins absorbing a tiny amount of water vapour.

Wait. The cup is metal? OK, that's a salient piece of information. I assumed it was paper.

yes. I'm pretty sure its stainless steel.

So if the cup has expanded and therefore less dense it affects the weight?

OK, I'm starting to understand. Cup 1 is the control. It does not get filled with anything.

Everything else aside, if you left out all other cups and even the entire oil measuring process, would this one cup still change weight?

yes. I'm pretty sure its stainless steel.

So if the cup has expanded and therefore less dense it affects the weight?

No.

It may be less dense because it has expanded but that has not changed its weight.

yes the control does change and you are correct to think it doesn't get put through any extraction process. it just sits there in the tray. It is only heated in the oven and cooled in a desiccator.
so basically, in order for me to get a weight within 0.0005g of its initial weight I have to weight it back at the same temperature.

yes the control does change and you are correct to think it doesn't get put through any extraction process. it just sits there in the tray. It is only heated in the oven and cooled in a desiccator.
What would happen if you did your entire process with only the control cup and no oil at all?

I'm trying to eliminate variables.

the control cup doesn't ever get anything added to it.

I weigh it then heat it up then cool it and weigh it again.

how long does it take you to weigh the cup? could the temperature of the cut be affecting the operation of the scale?...just reaching here...

the balance has sliding doors on it so the cup sits inside of an enclosed box. It only takes 10-20 seconds for me to get the cup out of the desiccator and onto the balance.

my point is whether the balance is a mechanical balance with some kind of delicate spring that might change its coefficient should it get warmer or something line that.

tell me if this is the wrong way to think about it?

If you could weigh a constant mass of gas at temp 35 degrees then that same mass of gass would weigh less at 40 degrees because it has expanded and is trying to rise.

Why would this not work for a metal?

I don't know I am just confused now..

its an analytical balance. the tray is not anywhere near its inner workings.

tell me if this is the wrong way to think about it?

If you could weigh a constant mass of gas at temp 35 degrees then that same mass of gass would weigh less at 40 degrees because it has expanded and is trying to rise.

Why would this not work for a metal?

I don't know I am just confused now..

hhhmmm...when you are looking for 0.0005g, that does not sound too crazy, after all.

It should actually be possible to rule that in or out...all you need to do is find the density of air and find out how much volume of air weighs 0.0005g...is it possible that the cup increased in volume that much? Be careful, though, the volume of the empty cup does not include its hollow inside, in this case, it is totally submerge in air and the volume is only the thickness of the cup time its total surface area...

the control cup doesn't ever get anything added to it.

I weigh it then heat it up then cool it and weigh it again.

Yes. But it is still in the vicinity of the other cups with oil and such. I'm trying to deduce what would happen if that were completely eliminated as a variable.

just to give some examples:

typical initial weight is 20.3020g
a cold cup can weigh as high as 20.3055g
a warm cup as low as 20.2985g

it is true that the control is heated in an oven along with the other cups sitting in a tray. This is to drive off all the moisture. Then they are all placed together in a tray in a desiccator. I am pretty sure I would get the same results by using the control all by its lonesome.

it is true that the control is heated in an oven along with the other cups sitting in a tray. This is to drive off all the moisture. Then they are all placed together in a tray in a desiccator. I am pretty sure I would get the same results by using the control all by its lonesome.

I think the issue is that you may have actually stumbled onto a bona fide error. I don't believe it should be lighter when warmer, i.e. there may be an error in the procedure that's contaminating your results.

even the warmth of my hands warming the cup up to body temp from an initial chilly AC cooled room temp shows a difference on the balance.

maybe it is all just a difference in moisture.

even the warmth of my hands warming the cup up to body temp from an initial chilly AC cooled room temp shows a difference on the balance.

Yes, you're introducing several sources of condensing moisture here.

Could it possibly be buoyant force from the cup heating the air in the measuring device?

just to give some examples:

typical initial weight is 20.3020g
a cold cup can weigh as high as 20.3055g
a warm cup as low as 20.2985g

Uh really ? Well to what temperature have you heated your metal cup ?

maybe it is all just a difference in moisture.

Your point is but agreeable. I think I concur with it.

Could it possibly be buoyant force from the cup heating the air in the measuring device?

Your point is not relevant because if air is heated it becomes less dense and hence escapes. When air is dense and cool it exerts more force (well compare it with the volume of cup) but the force which it exerts is equal in all direction inside cup which gives you zero net force. Hmm let me think uhh. What about loss in energy of cup huh ? Mass changing in energy ? No I don't know to what temperature has OP heated his metal cup. So this goes irrelevant.

All I could find relevant is Dave's point in page 1 I guess. Mass of air hmm. When air is cold it sinks in cup so total mass of cup = Mass of air + Mass + cup. But when air gets heated it escapes so mass of cup decreases ( air containing moisture). After all Weight = Mass x Gravity.

Hope this helps.

:)

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Could it possibly be buoyant force from the cup heating the air in the measuring device?

Close and I have seen this myself. If you have a balance like this

[PLAIN]http://www.kern-sohn.com/images/shop/img-lr-abt.jpg [Broken]

then you need to weight items at room temperature. The hot cup sets up a convection current and that pushes on the bottom of the platform. My first week in chemistry lab demonstrated this "source of error".

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Close and I have seen this myself. If you have a balance like this

then you need to weight items at room temperature. The hot cup sets up a convection current and that pushes on the bottom of the platform. My first week in chemistry lab demonstrated this "source of error".

This sounds plausible and gives the results observed.

The cup is immersed in a 'fluid' called air.

The cup could be imagined to be suspended in a true liquid fluid where the cup has neither positive or negative bouyancy. If the cup expands in volume for the same mass, it is therefore less dense and it will rise in the fluid and have a negative weight.

Therefore because the warmed cup is less dense it is more bouyant in air.

http://msl.irl.cri.nz/sites/all/files/training-manuals/TG07-July-2009.pdf

Calibrating Standard Weights

The standard weights considered here are usually
calibrated in terms of a so-called conventional mass
value – the mass of a weight of density 8000 kg/m3 that,
in air of density 1.2 kg/m3 at 20 °C, would balance the
weight being calibrated. For each accuracy class, R 111-
1 specifies how close the density of the standard weight
must be to 8000 kg/m3 to avoid significant errors due to
air buoyancy.