Quiz thread (don't be afraid of to be wrong)

ApeXaviour

EDIT: This thread was split from another where it began somewhat tangentially related the ideal gas equation... but I'll expand it to any intuitive-type question related to physics or chemistry.

Too often there is a culture in this country of students not answering for fear of being wrong. This is IMO anti-creative and a hindrance to learning. There's a great TED talk "On Being Wrong" that challenges this. So, I would say if you think you can make a reasonable attempt at an answer then go for it! This is a non-judgmental space.

(and in light of that there will be no tolerance for condescending replies, and similarly on the opposite end, none for trolling either)


Question: you have two sealed containers, identical in every way except one contains dry air while the other contains "damp" air.

Which is heavier?

FISMA

ApeXaviour wrote: »

Question: you have two sealed containers, identical in every way except one contains dry air while the other contains "damp" air.

Which is heavier?

Is this a pop quiz ApeXaviour?

If, we are talking about an ideal gas, I'll go with the dry air being heavier. All things considered equal, P, T, V...

Air is made up of mostly N₂ and O₂. If we have an ideal gas in a closed volume, then we should have the same number of moles and/or molecules.

What you are saying is to take out some N₂'s and O₂'s and replace with H₂O. N₂'s and O₂'s are more massive than H₂O's.

Thus, I vote dry air weighs more.

ApeXaviour

Hahah good man. I agree with your assessment.

I used to love that question back in the day because it would catch out the theoreticians, and thus momentarily quell my sense of inferiority.

FISMA

I think that one stood out because I remember the first time I answered it. Yep, I didn't think for a second and went with the humid air being heavier!

Nowadays when I ask this question I change it around a bit. These days, students are now Lance Armstrong's coach. Lance is about to do a time trial and as his coach students can pick the start time: in the morning when there is dry air or the afternoon when it is humid.

All things considered equal and don't worry about Lance's ability to stay cool. All they have to do is determine which situation is more aerodynamically advantageous.

It's a nice way to get people to understand: ideal gases, avo's number, moles, and masses.

FISMA

Here's one that I always like to share with students. It gets them thinking.

Have you ever noticed that sometimes you blow on your hands to warm them up whereas other times you can cool things down? Actually, in one exhale you could do both!

Why the difference? There's a hint in white below.

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Hint: think about: the shape of your mouth and bust out the Thermo book!

ApeXaviour

Keeping it going. Great idea! I'll split this thread.

I could hazard an answer for that but sure I'll leave it to someone else to give it a whack and I'll get back to my thesis!

FISMA

ApeXaviour wrote: »

Keeping it going. Great idea! I'll split this thread.

I could hazard an answer for that but sure I'll leave it to someone else to give it a whack and I'll get back to my thesis!

I was thinking the same thing - a thread called pop quiz. Someone asks a brainteaser and the person who answers correctly gets to ask another.

Might stir things up on the board for the summer!

Thanks

thecornflake

The dry air / "damp" air can be easily answered if you understand weather forecasts.

High pressures and low pressures and then some obvious conclusions using the ideal gas law.

or can it ?

When we have high pressure is it both temp and n increasing or is it mainly just temp ?

ianobrien

ApeXaviour wrote: »

Question: you have two sealed containers, identical in every way except one contains dry air while the other contains "damp" air.

Which is heavier?

I'm voting for the "dry" air, but.....

The ideal gas law holds for dry air, but not for damp air. For dry air, ideal gas law (with rearrangement) has air density as

Density = (Pressure)/(RT)

But, for wet air, the vapour pressure of water must be taken into account by the following

Density = Pressure(dry air)/R(dry air)T + Pressure(water vapour)/R(water vapour)T

The rub is that for 0psig, and 297K, the vapour pressure of water is -0.982.....

krd

FISMA wrote: »

I think that one stood out because I remember the first time I answered it. Yep, I didn't think for a second and went with the humid air being heavier!

Nowadays when I ask this question I change it around a bit. These days, students are now Lance Armstrong's coach. Lance is about to do a time trial and as his coach students can pick the start time: in the morning when there is dry air or the afternoon when it is humid.

All things considered equal and don't worry about Lance's ability to stay cool. All they have to do is determine which situation is more aerodynamically advantageous.

It's a nice way to get people to understand: ideal gases, avo's number, moles, and masses.

Morning air is denser because it's colder. That's why hot air balloonists prefer morning air.

I don't know if heavier air is more aerodynamically advantageous.

Does a cyclist use their body like the wing of an aircraft or is denser air less advantageous because its' heaviness provides more resistance to the cyclist?

FISMA

Alright everyone,

We are onto:
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Quiz #2
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Have you ever noticed that sometimes you blow on your hands to warm them up whereas other times you can cool things down? Actually, in one exhale you could do both! Explain - winner gets to make the next quiz.

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nk about: t

krd

Ok, I'm going to make some guesstimates

If your breath is on average 20 C

If your hands are cold - near 0 C - blowing on them will make them warmer.

If your hands are hot - near 50 C - blowing on them will make them cold.


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But in reality you can modulate the temperature of your breath. Inhaling and then exhaling quickly will give you a cold stream of air - perfect for blowing heat away from the surface of your hands. Or putting it another way, making the air around your hands colder, so the heat transfers away quicker.

If you want to warm your hands, you cup them and slowly exhale warm moist air. The heat will transfer from your breath to your hands.

thecornflake

As gasses expand they cool, when you blow cold air onto your hands, you usually make your mouth/lips smaller and a smalller higher pressure burst of air comes out relative to the hot air when you mouth is open wider.
When the gas expands it cools.

I think . . . .

ianobrien

FISMA wrote: »

******************************************************
Quiz #2
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Have you ever noticed that sometimes you blow on your hands to warm them up whereas other times you can cool things down? Actually, in one exhale you could do both! Explain - winner gets to make the next quiz.

******************************************************


nk about: t

I've two possibilities, the first of which is air velocity & "wind chill factor". ie blow onto your hands and you will get wind chill. Blowing quickly results in good ol' fashioned wind chill producing a drop in temp through convection. blowing slowly on your hands results in the moisture in your breath forming a humid zone around your hands, producing an heating effect.

Also, another possibility (a small one I reckon) is that if you hold your breath long enough, the air in your lungs will achieve steady state temp with the rest of your body, as opposed to the air temp of "normal" breaths every few seconds.

Morbert

ianobrien wrote: »

I've two possibilities, the first of which is air velocity & "wind chill factor". ie blow onto your hands and you will get wind chill. Blowing quickly results in good ol' fashioned wind chill producing a drop in temp through convection. blowing slowly on your hands results in the moisture in your breath forming a humid zone around your hands, producing an heating effect.

Also, another possibility (a small one I reckon) is that if you hold your breath long enough, the air in your lungs will achieve steady state temp with the rest of your body, as opposed to the air temp of "normal" breaths every few seconds.

Your first answer is closer to the real reason(s). But instead of moisture in the air, think about the moisture on your skin...

FISMA

There have been some really good answers that made me think - thanks for the responses.

I was trying to keep the "feelings" out of it. I was looking for a true change in temperature and not a "feeling" of cooling due to an evaporative effect.

I was going for the shape of your lips. However, there does appear to be a wind speed difference between open mouth and kissy lips, , sorry but I have no idea what to call it - the way your lips look when you're going to blow out birthday candles.

Anyhow...

When you exhale air from your lungs, the gas will be at a temperature not far off of body temperature.

The air is coming right out and it feels warm. When you want to warm your hands you exhale on them with an opened mouth, not constricted, like kissy lips.

When you exhale air through kissy lips, the air stream is pressurized.

The exiting air enters into an area of lower pressure and expands.

The expansion happens quickly. Heat travels slowly. So we can neglect heat transfer. Thus, we have adiabatic expansion - no heat flow.

You know how us Physics people love the LCOE - Law of Conservation of Energy! We love it so much that we often camouflage it: Bernouilli, Kirchoff, and the FLOT - First Law of Thermo to name a few.

Aside - isn't Schrodinger's Wave Equation a Statement of Energy Conservation?

So in your Physics book the FLoT says

ΔU = Q - W

Where U = Internal Energy, Q heat added to the gas, and W work done by the gas.


But your Chem book says

ΔU = Q + W

What's the difference? The Chemists are wrong!:D N'ah, just kidding! Seriously, Physics is interested in the work done BY the gas and Chem books look at the work done ON the gas, in general.

Sorry to lecture lads, but young Physicists often struggle with this sign convention stuff.

So in order to use the Physicist's (AKA correct:D) version we use the following sign convention: (1) heat added is positive, (2) heat lost is negative, (3) work done ON the system is negative, and (4) work done BY the system is positive. (3) and (4) are a bit weird but it accounts for the negative sign.

So, let's use the FLoT!

ΔU = Q - W

Since our process is adiabatic Q = 0 and we have

ΔU = - W

In most introductory texts, they are going to use Thermal Energy synonymously with Internal Energy - bad habit. It's sloppy Physics in my opinion.

There are many internal energies, thermal being one of them. Additionally, you can have chemical and so on. Just keep this in mind so you don't have to break a habit when you take Thermo at University.

In most beginning Thermo classes they are going to relate a change in internal energy to Temperature and use a formula like U = 3/2kT.

In our scenario, work is done by the gas, which is positive. That makes the RHS of our equation negative.

You can just read the FLoT from here

ΔU = - W

Δ(3/2kT) = - W

Pulling out constants

3/2k*ΔT = - W

Get rid of the constants and equal sign and replace with a proportionality sign "α"

ΔT α - W

So we see that in our system, the work done by the gas is solely at the expense of the temperature of that gas.

The expanding gas does work on the environment. The gas does work at the expense of its temperature - the temp decreases.

If I goofed on any signs or concepts, please advise. I should also mention that cavaet that our breath needs to be an ideal gas, correct?:rolleyes:

I think thecornflake was onto it. I just wanted a thermo statement!

What say ye mods? ApeXaviour, Professor Fink, and Morbert? Who's the next quizmaker?

Some other ground rules.

Let's try and make the latest quiz standardized using some format like the above with stars and all.

We should try and keep it fun, interesting, but always true to Physics! Everyone likes a brainteaser or a paradox!

Finally, how should we choose the winner?

thecornflake

It's easily seen if you blow cold air on your hand by making your lips smaller. Then move your hand right up to your mouth and do the exact same thing. As the gas does not have time to expand it will be warm.

krd

FISMA wrote: »

I was going for the shape of your lips. However, there does appear to be a wind speed difference between open mouth and kissy lips, , sorry but I have no idea what to call it - the way your lips look when you're going to blow out birthday candles.

It's called pursing your lips.

Anyhow...

Are you saying. That when you blow through pursed lips, heat energy is being transferred to the work needed for the gas to flow and expand - so the breath will be cooler than body temperature?

The heat is lost through work? So that the breath that reaches the hand is cooler?

Do you have to assume the breath leaving pursed lips is at body temperature for this idea to work?

I'm looking at the wikipedia equations for Adiabatic process.

This bit confuses me

Adiabatic Free Expansion of a Gas

For an adiabatic free expansion process, the gas is contained in an insulated container and a vacuum. The gas is then allowed to expand in the vacuum. The work done by or on the system is zero, because the volume of the container does not change. The First Law of Thermodynamics then implies that the net internal energy change of the system is zero. For an ideal gas, the temperature remains constant because the internal energy only depends on temperature in that case. Since at constant temperature, the entropy is proportional to the volume, the entropy increases in this case, therefore this process is irreversible.

Is this stating there's no temperature change. Am I not thinking this through enough.

ianobrien

I seem to recall (probably incorrectly) from Fluid Dynamics that when you have a flow through a restriction (your lips) into an area much larger than the other side of the restriction (your lungs against the atmosphere), that the change in temp/pressure in the downstream side of the restriction can be regarded as zero.

I'm still going for gas velocity over your hands and "wind chill". When there is a high gas flowrate, there will be convection heat loss. Also, the volume of air passing over your hands will absorb moisture, meaning heat loss from your hand (latent heat of evaporation).

For slow air speeds, the moisture in your breath will condense on your hands (the air speed is not high enough to carry the moisture away), meaning the hands gain heat from latent heat of condensation.

Morbert

ianobrien wrote: »

I seem to recall (probably incorrectly) from Fluid Dynamics that when you have a flow through a restriction (your lips) into an area much larger than the other side of the restriction (your lungs against the atmosphere), that the change in temp/pressure in the downstream side of the restriction can be regarded as zero.

I'm still going for gas velocity over your hands and "wind chill". When there is a high gas flowrate, there will be convection heat loss. Also, the volume of air passing over your hands will absorb moisture, meaning heat loss from your hand (latent heat of evaporation).

For slow air speeds, the moisture in your breath will condense on your hands (the air speed is not high enough to carry the moisture away), meaning the hands gain heat from latent heat of condensation.

You are right that "cooling by expansion" isn't responsible, thought it is often reported as responsible in older school textbooks. If you blow on a dry thermometer, it will not register any cooling effect.

What is responsible is entrainment and evaporation. When you produce a thin, high-velocity stream of air, it mixes with the ambient air. This air does not contain as much moisture as your breath, so when it comes in contact with your hand, it causes evaporation, which cools your hand.

thecornflake

right i think it's time for a new question, I got this question from the user "citrusburst"


We are onto:
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Quiz #3
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If you have a piece of metal, such as a metal washer. You then heat the metal washer. Does the centre hole become smaller, larger or remain the same due to the thermal effects ?



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ray giraffe

thecornflake wrote: »

Quiz #3
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If you have a piece of metal, such as a metal washer. You then heat the metal washer. Does the centre hole become smaller, larger or remain the same due to the thermal effects ?
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I think the hole gets larger.

Put a disc of metal (at same temperature) that fits exactly into the hole, then when it gets hotter both will expand just as a large metal disc would.

Therefore there should be no gap appearing between washer and disc and the disc should still slide out (no compression of the disc by the washer).

So just as the disc will expand, the space it occupies will expand.

I got this intuition from linear maps x -> kx in maths. (k>1 is a scalar, x is the vector associated with a point in space)
The map increases the distance between every two points in space by the same factor k: [latex]\parallel kx_1-kx_2\parallel=k\parallel x_1-x_2 \parallel[/latex]

Agent_99

thecornflake wrote: »

right i think it's time for a new question, I got this question from the user "citrusburst"


We are onto:
******************************************************
Quiz #3
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If you have a piece of metal, such as a metal washer. You then heat the metal washer. Does the centre hole become smaller, larger or remain the same due to the thermal effects ?



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I don't have any fancy quotes or calculus but I thing that the dimensions stay the same for the whole in the middle.

shizz

I have an inkling that the Hole will get smaller, as the increase in heat will make affect all of the metal disk. Although I'd imagine that that its rate of increase would be smaller than that of the outer diameter as it would be restricted by all other parts of the inner diameter increasing equally. I guess with this train of thought it might also be possible that the inner diameter of the metal disk doesn't change at all due to this restriction?

krd

The centre hole would expand in proportion to the rest of the disc. It would get larger.

FISMA

Morbert wrote: »

You are right that "cooling by expansion" isn't responsible, thought it is often reported as responsible in older school textbooks.
...
What is responsible is entrainment and evaporation. When you produce a thin, high-velocity stream of air, it mixes with the ambient air. This air does not contain as much moisture as your breath, so when it comes in contact with your hand, it causes evaporation, which cools your hand.

I'm not convinced Morbert. I am going to run some numbers to see what kind of temperature change would be expected from adiabatic expansion and see if it is a measurable or metaphysical.

You're going with the volume flow rate of air and its moisture content as being the greater cause, correct?

Have the other mods any ideas about this one?

I just did a simple test. If others could replicate and post the results, I would be grateful.

First, I exhaled as fast as possible with an open mouth. The breath was warm.

Next, I exhaled through pursed lips slower than above, but not that slow and the breath was cool.

Finally, I exhaled through pursed lips slowly and held my hand closer than above (which was about 6-8") - at about 1".

Would you agree that these results contradict entrainment and evaporation as being the main cause?

If anyone has some calculations, please advise, I think it would be a phun exercise.

Otherwise, if anyone has some data: moisture content of breathed air, air pressure inside lungs, and so on, I would appreciate them.

I would particularly like to see the % change of moisture content as a function of humidity. One would imagine that as humidity increases, the percent change of the breathed in air decreases.

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+1, yes, the hole gets larger. - next quiz!:D

krd

FISMA.......it would be interesting to see some thermal imaging. But since Argus aren't doing golf balls thermal imaging cameras - and it's whatever number of hundreds of quid to rent one.


Quick question - what modification would need to be done to the array in a typical golf ball cam to make it a thermal imagining device. And I'm not talking about the cam you get from Argos - I mean the semiconductor in the array element.

ianobrien

FISMA wrote: »

I am going to run some numbers to see what kind of temperature change would be expected from adiabatic expansion and see if it is a measurable or metaphysical.

I reckon that's your problem. You're looking for a temp change strictly, where I am looking at it in a thermal manner. It's the very same thing as sticking your hand out of a moving car window. Due to the evaporation of water from your hand, it gets cold, yet there is no change in air temp.........

(then again, I am looking at it from a hand approx 6 inches from my mouth. If the hand was against the lips, the hot air would heat the hand)

As for quiz#3, the hole gets larger. Ask any mechanic who has taken a blow-torch to a stuck nut or bolt.

Morbert

FISMA wrote: »

I'm not convinced Morbert. I am going to run some numbers to see what kind of temperature change would be expected from adiabatic expansion and see if it is a measurable or metaphysical.

You're going with the volume flow rate of air and its moisture content as being the greater cause, correct?

Have the other mods any ideas about this one?

I just did a simple test. If others could replicate and post the results, I would be grateful.

First, I exhaled as fast as possible with an open mouth. The breath was warm.

Next, I exhaled through pursed lips slower than above, but not that slow and the breath was cool.

Finally, I exhaled through pursed lips slowly and held my hand closer than above (which was about 6-8") - at about 1".

Would you agree that these results contradict entrainment and evaporation as being the main cause?

If anyone has some calculations, please advise, I think it would be a phun exercise.

Otherwise, if anyone has some data: moisture content of breathed air, air pressure inside lungs, and so on, I would appreciate them.

I would particularly like to see the % change of moisture content as a function of humidity. One would imagine that as humidity increases, the percent change of the breathed in air decreases.

Those results don't contradict entrainment/evaporation as the correct explanation. They are to be expected.

1. Exhale on a dry thermometer.
2. Blow on a dry thermometer.
3. Wave a dry thermometer around.

4. Exhale on a wet thermometer.
5. Blow on a wet thermometer.
6. Wave a wet thermometer around.

The effect of evaporation can be seen when 5. and 6. will result in cooling, greater than 2. and 3.

The effect of entrainment can be seen when 1. results in a temperature increase, but 2. does not.

The fact that expansion is not responsible can be seen when 2. does not cool the thermometer any more than 3.

FISMA

I was reading another thread which made me think of this old favorite. I will try and be more precise. Sorry about the last one. I was trying to avoid the feeling sensation of cooling (with respect to the person) and was going more for an actual change in temperature (with respect to the gas).

Anyhow, since this thread appears to have waned...

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Quiz #4
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It is a bit below freezing outside and you have two buckets: (A) contains very hot water and (B) contains cooler water, let's say lukewarm - room temperature.

You take the two buckets and place them outside. Later on you come back to check which one is freezing and you are surprised to find bucket A, which contained very hot water is beginning to freeze, but not B?

How is that possible? There's no law of conservation of momentun for cooling? Surely, bucket A has to get to the temperature bucket B was previously at. But by then, bucket B has cooled some more.

So what gives?

FYI - both buckets have the same volume and are filled with the same amount of water. The buckets are placed far enough from each other so that they do not interact.

No tricks like holes in the bucket - just a bit of science.

The question: how is this possible? (1) Give us the short answer followed by the Physics!

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Delphi91

FISMA wrote: »

I was reading another thread which made me think of this old favorite. I will try and be more precise. Sorry about the last one. I was trying to avoid the feeling sensation of cooling (with respect to the person) and was going more for an actual change in temperature (with respect to the gas).

Anyhow, since this thread appears to have waned...

******************************************************
Quiz #4
******************************************************

It is a bit below freezing outside and you have two buckets: (A) contains very hot water and (B) contains cooler water, let's say lukewarm - room temperature.

You take the two buckets and place them outside. Later on you come back to check which one is freezing and you are surprised to find bucket A, which contained very hot water is beginning to freeze, but not B?

How is that possible? There's no law of conservation of momentun for cooling? Surely, bucket A has to get to the temperature bucket B was previously at. But by then, bucket B has cooled some more.

So what gives?

FYI - both buckets have the same volume and are filled with the same amount of water. The buckets are placed far enough from each other so that they do not interact.

No tricks like holes in the bucket - just a bit of science.

The question: how is this possible? (1) Give us the short answer followed by the Physics!

******************************************************

It has got to do with Newtons Law of Cooling which states that the rate of change of temperature of an object is proportional to the temperature difference between the object and its surroundings.

Assume the warm water(A) is at a temp of, say, 75 degrees, the lukewarm water(B) is at a temp of 15 degrees and the air temp is -5 degrees. The difference is temp between the hot water and its surroundings is 80 degrees while for the lukewarm water it is only 20 degrees. Therefore, the warm water will cool, initially, 4 times faster than the cooler water.