I have never seen this so I went and tried it and it is quite surprising to me that it works! My card was bending a little and there was actually a visible bead of water about 1/4 inch along two of the sides between the small square plastic container and the paper index card. Surface tension of the water seems to be what is keeping the water from flowing out sideways.
Just to put some math to it.... Say your mug has a diameter of 75 mm, which gives a cross sectional area of A = 4400 mm^2 = 0.0044 m^2, and a height h = 100 mm. That's a volume of V = A*h = 440 mL. Let's say it's f = 2/3 full of water, for a water height of hw = 66.7 mm, and a water volume Vw = fV = 293 mL. That means there's an air height of (1-f)h = 33.3 mm, and an (initial) air volume of Va1 = (1-f)V = 147 mL of air inside. The water has a mass of m=Vw*ρ, where ρ=1 g/mL = 1000 kg/m^3, so m 293 g. We'll assume the paper and air are both negligible mass. So, to hold that in the cup, you need a force of F = m * g, where g = 10 m/s^2, F = 2.93 N.
As the water and card drop a little, the gas in the mug will expand, lowering the pressure from atmospheric to some internal pressure Pi. But we still have atmospheric pressure acting on the bottom of the card, Po = 100 kN/m^2. That creates a pressure difference, Po - Pi, between the inside and outside the mug. The pressure difference, acting on the area of the mug cross section, creates the force. (Note, the pressure acting up on the card outside of the mug cross section is balanced by the pressure acting down on the other side of the card outside the mug cross section.) We have F = (Po - Pi)A = 2.93 N. So, Pi = Po - F/A, F/A = 665 N/m^2, and Pi = 99.33 kN/m^2.
If we assume there's no energy transfer to/from the gas as it expands (the system is adiabatic), which seems reasonable, then PV^γ is constant, where P is pressure, V is volume, and γ is the heat capacity ratio, which is 1.4 for air. We can use that to find how much the volume of air inside the cup needs to expand to create that pressure. With an expanded air volume Va2, we have
So, the volume expansion of the air to create the force necessary to hold the water up is pretty small, about 0.5%. Since the coffee mug is a cylinder, that translates directly to the change in height of the air. So we're talking about going from 33.3 mm of air to 33.5 mm. And that 0.2 mm change in height means that's how much of the water is exposed to air at atmospheric pressure, trying to push into the mug. I won't (read: am not smart enough to) compute how small that gap has to be for the surface tension of the water to prevent "swallowing" an air bubble, but, intuitively, that's small enough.
Algebraically, after substitutions,
Va2/Va1 = ((Po - F/A)/Po)^(-1/γ)
ha2/ha1 = (1 - f*h*ρ*g/Po)^(-1/γ)
The change in air height is
Δh = (1-f*h) * ((1 - f*h*ρ*g/Po)^(-1/γ) - 1)
Note, that's entirely independent of the cross section of the mug (assuming it's constant vertically, or at least the same at the mouth and at the air gap). For a 100 mm tall mug, that's going to range between 0 to 0.2 mm back to 0 of height change in the air as you fill the mug from empty* to full, which, again, certainly seems small enough. But as the height of the mug increases, the change in air height grows - if you're drinking one of those yard-long margaritas in New Orleans to celebrate the Super Bowl, that change in height will range between 0 to 19 mm back to 0, and surely that's big enough to leak.
*Also note, in my earlier comment, I had asked, what's the dividing line for the amount of water necessary to hold the card up? This analysis doesn't find one, but that's because we assumed the card mass was negligible. As the amount of water drops, this is no longer a good assumption.
It's basic physics but basic physics isn't always quite how you expect it to be.. .
1 air pressure is a pretty big force: it's holding up the water in the glass .
2 how big? You can work it out (roughly). The air is around 8km deep (at the top of Everest you're above most of the air) and weighs about 1 gramme per litre or 1kg per cubic metre. So the air above 1 sq m weighs around 8000kg. Water weighs 1kg per litre. So 8km of air can potentially hold up 8m of water. (Roughly)
3 so why doesn't this enormous force squash us? (8000kg or 8 tons per sq metre). Because it operates in all directions. Including upwards, as we saw with the water glass. The force actually comes from the atoms in the air zooming around in all directions and banging against the sides of the container. (And they carry on doing this even if you screw a lid on)
4 So you left out an important force. The pressure in the air trapped above the water, pressing down on the top of the water. (It also presses outwards and upwards on the glass container but that's not important).
5 when some body of air EXPANDS the hydrostatic pressure is reduced. (Fewer zoomy atoms in a given volume). If you let the volume of that air expand by 10% as the water drops, the pressure drops by 10% (approx, the expanding air cools, slightly altering this). The 10% difference between the air pressure on the bottom of the water and on the top is enough to hold up the water.
> so why doesn't this enormous force squash us? (8000kg or 8 tons per sq metre). Because it operates in all directions.
I would quibble with that a bit, to say that it operates inside us as well as outside us. If you take a Styrofoam cup, made of closed cell foam, where the foam is little bubbles of gas at basically atmospheric pressure, and take it down to the bottom of the ocean, it will get squashed, because the pressure can't equalize between the water and the foam cells.
That's not a huge change in pressure. -600 mbar or so. Certainly enough to explode a bag of potato chips, but you get a +600 mbar change about 6 m underwater (although expansion probably isn't the same as compression). I've tried to find video of a styrofoam cup going underwater with a depth/pressure scale and couldn't do it in a quick googling. But there's this (cup starts at 1:00): https://youtu.be/Jh6-0aqft1k?si=TntuZ2eWQd1NSzzp Goes to much higher pressure, 270 bar, and he says he does it quickly.
I thought it had something to do with the vacuum being left at the top creating a pressure imbalance? When you turn the mug upside down and the air is forced out from the bottom (since it's not a perfect seal on the side, so the opposite of your hypothesis), it leaves a kind of low pressure vacuum above the water such that the air pressure from below wants to rush in to fill that void.
For what it's worth, you can easily get the experience with negligible water mass with a semi-rigid rubber sheet; I'm using one of those rubber jar lid grippers. I got the gripper wet, and now I can press it up against the cup and it will stick... But if I press it slightly INTO the cup, it sticks MUCH better. If I set it down on a paper towel, and press the cup down onto the gripper, it does NOT stick... But if I place something small in between the gripper and the paper towel, so that when I lower the cup onto the gripper, the gripper has been pushed slightly into the cup, I can then use the cup to lift the gripper.
I would suggest thinking about, if this doesn't work with ZERO water, and does work with the cup FULL of water, what is the dividing line between working and not working, and why?
It doesn't need to be completely full-- that's actually one of the things we were talking about at dinner. You can have a moderately large air bubble at the top (you just need to be more careful when inverting the cup).
Another thing at play here is surface tension of the water. Over a small distance (between the edge of the cup and the card), the air isn't putting enough force on the water surface to bend it in enough to close on itself, and "swallow" a bubble of air, which would then go to equalize the pressure and allow the card and water to fall. Similar effect to why you can hold a medicine dropper vertically without water falling out, but you can't hold a cup upside down without the card covering the bottom.
I have never seen this so I went and tried it and it is quite surprising to me that it works! My card was bending a little and there was actually a visible bead of water about 1/4 inch along two of the sides between the small square plastic container and the paper index card. Surface tension of the water seems to be what is keeping the water from flowing out sideways.
Just to put some math to it.... Say your mug has a diameter of 75 mm, which gives a cross sectional area of A = 4400 mm^2 = 0.0044 m^2, and a height h = 100 mm. That's a volume of V = A*h = 440 mL. Let's say it's f = 2/3 full of water, for a water height of hw = 66.7 mm, and a water volume Vw = fV = 293 mL. That means there's an air height of (1-f)h = 33.3 mm, and an (initial) air volume of Va1 = (1-f)V = 147 mL of air inside. The water has a mass of m=Vw*ρ, where ρ=1 g/mL = 1000 kg/m^3, so m 293 g. We'll assume the paper and air are both negligible mass. So, to hold that in the cup, you need a force of F = m * g, where g = 10 m/s^2, F = 2.93 N.
As the water and card drop a little, the gas in the mug will expand, lowering the pressure from atmospheric to some internal pressure Pi. But we still have atmospheric pressure acting on the bottom of the card, Po = 100 kN/m^2. That creates a pressure difference, Po - Pi, between the inside and outside the mug. The pressure difference, acting on the area of the mug cross section, creates the force. (Note, the pressure acting up on the card outside of the mug cross section is balanced by the pressure acting down on the other side of the card outside the mug cross section.) We have F = (Po - Pi)A = 2.93 N. So, Pi = Po - F/A, F/A = 665 N/m^2, and Pi = 99.33 kN/m^2.
If we assume there's no energy transfer to/from the gas as it expands (the system is adiabatic), which seems reasonable, then PV^γ is constant, where P is pressure, V is volume, and γ is the heat capacity ratio, which is 1.4 for air. We can use that to find how much the volume of air inside the cup needs to expand to create that pressure. With an expanded air volume Va2, we have
Po*Va1^γ = Pi*Va2^γ
Va2/Va1 = (Pi/Po)^(-1/γ) = (99.33/100)^(-1/1.4) = 1.0048
So, the volume expansion of the air to create the force necessary to hold the water up is pretty small, about 0.5%. Since the coffee mug is a cylinder, that translates directly to the change in height of the air. So we're talking about going from 33.3 mm of air to 33.5 mm. And that 0.2 mm change in height means that's how much of the water is exposed to air at atmospheric pressure, trying to push into the mug. I won't (read: am not smart enough to) compute how small that gap has to be for the surface tension of the water to prevent "swallowing" an air bubble, but, intuitively, that's small enough.
Algebraically, after substitutions,
Va2/Va1 = ((Po - F/A)/Po)^(-1/γ)
ha2/ha1 = (1 - f*h*ρ*g/Po)^(-1/γ)
The change in air height is
Δh = (1-f*h) * ((1 - f*h*ρ*g/Po)^(-1/γ) - 1)
Note, that's entirely independent of the cross section of the mug (assuming it's constant vertically, or at least the same at the mouth and at the air gap). For a 100 mm tall mug, that's going to range between 0 to 0.2 mm back to 0 of height change in the air as you fill the mug from empty* to full, which, again, certainly seems small enough. But as the height of the mug increases, the change in air height grows - if you're drinking one of those yard-long margaritas in New Orleans to celebrate the Super Bowl, that change in height will range between 0 to 19 mm back to 0, and surely that's big enough to leak.
*Also note, in my earlier comment, I had asked, what's the dividing line for the amount of water necessary to hold the card up? This analysis doesn't find one, but that's because we assumed the card mass was negligible. As the amount of water drops, this is no longer a good assumption.
It's basic physics but basic physics isn't always quite how you expect it to be.. .
1 air pressure is a pretty big force: it's holding up the water in the glass .
2 how big? You can work it out (roughly). The air is around 8km deep (at the top of Everest you're above most of the air) and weighs about 1 gramme per litre or 1kg per cubic metre. So the air above 1 sq m weighs around 8000kg. Water weighs 1kg per litre. So 8km of air can potentially hold up 8m of water. (Roughly)
3 so why doesn't this enormous force squash us? (8000kg or 8 tons per sq metre). Because it operates in all directions. Including upwards, as we saw with the water glass. The force actually comes from the atoms in the air zooming around in all directions and banging against the sides of the container. (And they carry on doing this even if you screw a lid on)
4 So you left out an important force. The pressure in the air trapped above the water, pressing down on the top of the water. (It also presses outwards and upwards on the glass container but that's not important).
5 when some body of air EXPANDS the hydrostatic pressure is reduced. (Fewer zoomy atoms in a given volume). If you let the volume of that air expand by 10% as the water drops, the pressure drops by 10% (approx, the expanding air cools, slightly altering this). The 10% difference between the air pressure on the bottom of the water and on the top is enough to hold up the water.
> so why doesn't this enormous force squash us? (8000kg or 8 tons per sq metre). Because it operates in all directions.
I would quibble with that a bit, to say that it operates inside us as well as outside us. If you take a Styrofoam cup, made of closed cell foam, where the foam is little bubbles of gas at basically atmospheric pressure, and take it down to the bottom of the ocean, it will get squashed, because the pressure can't equalize between the water and the foam cells.
Yes. Interesting to take a styrofoam cup up Everest.
That's not a huge change in pressure. -600 mbar or so. Certainly enough to explode a bag of potato chips, but you get a +600 mbar change about 6 m underwater (although expansion probably isn't the same as compression). I've tried to find video of a styrofoam cup going underwater with a depth/pressure scale and couldn't do it in a quick googling. But there's this (cup starts at 1:00): https://youtu.be/Jh6-0aqft1k?si=TntuZ2eWQd1NSzzp Goes to much higher pressure, 270 bar, and he says he does it quickly.
I thought it had something to do with the vacuum being left at the top creating a pressure imbalance? When you turn the mug upside down and the air is forced out from the bottom (since it's not a perfect seal on the side, so the opposite of your hypothesis), it leaves a kind of low pressure vacuum above the water such that the air pressure from below wants to rush in to fill that void.
P. S. Basically the same physics as why sucking on a straw makes water go up the straw.
For what it's worth, you can easily get the experience with negligible water mass with a semi-rigid rubber sheet; I'm using one of those rubber jar lid grippers. I got the gripper wet, and now I can press it up against the cup and it will stick... But if I press it slightly INTO the cup, it sticks MUCH better. If I set it down on a paper towel, and press the cup down onto the gripper, it does NOT stick... But if I place something small in between the gripper and the paper towel, so that when I lower the cup onto the gripper, the gripper has been pushed slightly into the cup, I can then use the cup to lift the gripper.
I predict this will only work with cold water, not hot (and don't burn yourself trying).
Worked fine with the hottest water I could get out of the kitchen tap.
So it's not the water cooling the headspace...
I would suggest thinking about, if this doesn't work with ZERO water, and does work with the cup FULL of water, what is the dividing line between working and not working, and why?
(You've missed a force....)
It doesn't need to be completely full-- that's actually one of the things we were talking about at dinner. You can have a moderately large air bubble at the top (you just need to be more careful when inverting the cup).
So, what's happening in that air bubble as gravity tugs the card and water down?
See the addendum; I didn't see this comment until after I typed it out, because my notification emails have gone screwy.
Another thing at play here is surface tension of the water. Over a small distance (between the edge of the cup and the card), the air isn't putting enough force on the water surface to bend it in enough to close on itself, and "swallow" a bubble of air, which would then go to equalize the pressure and allow the card and water to fall. Similar effect to why you can hold a medicine dropper vertically without water falling out, but you can't hold a cup upside down without the card covering the bottom.