Depends on the sensitivity of the scale/ balance used. Most standard balances the final answer would be 10 grams. There are scientific balances that are sensitive enough to be affected by the turbulence caused by the hummingbird. For these balances, the weight would not be constant.
Imagine the weight on the balance if the box was removed.

You fail at physics! :D

It will be 20 grams + or - the force causing the vertical acceleration of the hummingbird.

Maybe easier to visualize.

Very large box of water on a scale. Next add a large heavy boat to the box. The boat is floating on the water. Do you add the weight of the boat?

Yes, assuming that the box is big enough that the water does not overflow. If the water level is right at the top of the box before the boat is put into it, then the weight will stay the same because the boat will displace a volume of water equal to its weight, and that displaced volume of water will overflow out of the box.

Come on, I passed both semesters of advanced physics with C's. :)

I originally visualized the problem with an open top box (open system), your answer makes sense for a closed box (closed system). Different views on the question.

It does not matter if the box has a top or not as long as the hummingbird is hovering over the scale.

The hummingbird would have to be close enough for the downforce to be felt by the balance if the box is open.

OK, you are right that the pressure waves will spread out and become less intense so depending on the AGL altitude of the hummingbird the scale would have to be greater than a certain area, roughly circular, to catch all of the downwash.

That is if the box has no sides, in which case it is not a box at all. If the box has sides and the hummingbird's altitude is less than the height of the sides then the box does not need to be any specific size and the weight will be the same with or without lid (assuming the weight of the "box" is the same whether it has a top or not).

Well said. This can go on and on. What if the sides of the box are mesh and allow air flow to pass through? It is still a box. :)

Maybe we are viewing this from two different angles and lost in translation.
Next we can discuss weight vs mass. :)
As long as we don't rehash 2 or 288. Hehe

if the box is closed the hummingbird will die (should I say you suck at biology?)

after it dies, the water wt. of its body will slowly permeate out of the box and the scale will then read...

Wow, now I can add Biology to my personal list of "dismal sciences" of Economics and Thermodynamics. :rolleyes:

I prefer to think Schoedinger's cat is alive, by the way. I like hummingbirds and cats. :)

Weight vs. Mass should be easier. Weight is the attractive force exerted between two massive objects (even if one is much greater than the other -Earth vs. hummingbird).

Now what mass is is harder to say. That would be the realm of the Higgs Boson I believe.

The weight of the box, plus the dead weight of the bird.

That was an attempt at humor. ;)

It's hard for me to sense humor sometimes with internet posts and not being in person. Plus I like talking science and puzzles about physics/science. :)

No problem, I couldn't agree me about that.

That mesh box thing really makes things interesting, though. :) Figuring out how much pressure escapes through them would be more work.

That is what makes these questions fun to discuss.

winner! winner!

tiny lil' tastes like chicken dinner!

Scott, are you gonna ask this kind of question every Friday? :)

I think the answer requires a stoichiometry-correct answer, which means you have to include the air in the wood. I think the percentage of air in the wood would be a percentage that is consistent... volume/volume, or mass/mass. So if you know the volume (or mass) of the wood, you can multiply by the percentage to get the volume (or mass) of air in the wood.

Which brings us back to... more beer.

_