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Re: [Phys-L] Figuring Physics solution Jan 2018

This is pretty much how I explain it, except instead of talking about the
particles "slowing down", I describe the thermal (internal) energy as a
reservoir. In a liquid, intermolecular bonds (van der Waals attraction,
hydrogen bonding, etc.) are continuously forming and breaking. The energy
to break those bonds comes from the molecules with higher kinetic energies
within the reservoir, and the energy released when the is reabsorbed the
reservoir again.

This also explains why water at 0°C in a thermos doesn't freeze--the energy
released from forming intermolecular bonds is reabsorbed as thermal energy,
which is then available from the reservoir to break other intermolecular
bonds. And it explains colligative properties, such as why putting salt on
icy roads melts the ice even though it cools the liquid well below the
normal freezing point of pure water. When bonds between higher-energy
molecules are broken, thermal energy is absorbed (which cools the bulk
liquid). The salt inhibits formation of the intermolecular bonds,
preventing the bonds from re-forming and releasing the thermal energy back
into the reservoir.

This is a simplistic description, but more than adequate for my high school
classes. To demonstrate this phenomenon, I pour boiling water on a
washcloth, and invite students to hold their hands near it to feel the
warmth. I then spin the washcloth rapidly for about 20-30 seconds to
evaporate the water, and the washcloth becomes cool to the touch--noticeably
cooler than the ambient air in the room.

Jeff Bigler
Lynn English HS; Lynn, MA

-----Original Message-----
From: Phys-l [] On Behalf Of Robert
Sent: Tuesday, January 23, 2018 7:04 AM
Subject: Re: [Phys-L] Figuring Physics solution Jan 2018

Let me see if I can take a stab at this.

I'll start with the simple situation of a gas in a one-dimensional container
(so the particles only travel along that one dimension),
made up of particles of identical speeds under no attraction and totally
elastic collisions. That way the speeds remain identical.

I then add a small attraction so that if I remove the "top" of the
container, the fluid doesn't necessarily spit out the top. This can then be
considered a liquid. The speeds won't always be identical at any given time
but on average they are.

At any given time, only the particle at the top is exposed to the vacuum
above and thus only that particle can "leave" the liquid state and fly away
as a free particle. That particular particle need not be the "fastest" of
them all, does it? Does it even need to be going faster than the "average"
of them in order to leave?

I'd argue it doesn't. It just needs to be going fast enough to overcome the
attraction. And, while undergoing the escape process, it slows down AND the
particle it was attracted to slows down. This leads to a cooling of both,
independent of whether it was initially going faster than those it "left
behind" or not.

In other words, the real physics is on the cooling that results from the
"breaking" of the bonds (so to speak). Focusing on the speed of those left
behind (what JD calls "cancelling the sixes") allows students to ignore the
real physics.

From: Phys-l [] on behalf of John Denker via
Phys-l []
Sent: Monday, January 22, 2018 7:16 PM
Cc: John Denker
Subject: Re: [Phys-L] Figuring Physics solution Jan 2018

On 01/22/2018 03:03 PM, Jeffrey Schnick wrote:

Why are you writing about a uniform-velocity gas?

Velocity is a typo. I should have said speed. My bad.

As for the rest, I was trying to be as charitable as possible to Hewitt. In
computer science there is a class of problems called NP-complete that are
(probably!) rather hard. Hewitt's question belongs to a class I call
ESP-complete, because the only way to begin to answer it is to read the mind
of the person who asked it.

Hewitt didn't say anything about a uniform-velocity gas or liquid in
the question under discussion.

Not in the "question" strictly speaking ... but his proffered /answer/ makes
it clear that he imagined the escape probability to depend on speed, perhaps
normally, perhaps always.

It wouldn't be fair to students to expect them to start from Hewitt's answer
and work backwards, but it's more than fair to Hewitt.

Even /with/ the answer I find it impossible to make sense of the question,
even under the most charitable assumptions.

In particular, he seems to think his liquid can evaporate with no latent
heat, so calling it a liquid instead of a dense gas seems to be a
distinction without a difference.


I did not previously publicly discuss the interpretation that says the
hypothetical liquid has a wide range of speeds, but all speeds evaporate
equally. So here goes.
This doesn't make any sense either. You could go in there with a siphon and
remove a parcel of liquid, with the original liquid speed-distribution, and
this would leave the temperature unchanged ... but we call that siphoning,
not evaporation. At some point you have to convert that parcel from liquid
to vapor, which involves the latent heat somehow. So either something
cools, or you provide heat from outside, or ... I give up. It was meant to
be a simple question, and nobody has yet suggested anything resembling a
simple-yet-correct interpretation.


If anybody can come up with a set of assumptions that allow the question to
be mapped onto real physics, without contradictions or gross complexity,
please speak up.
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