So we had this massive ice storm about ten days ago and much of the city (and environs) was without power for days after including the Gauss household.
One of the lesser tragedies of the disaster was that the roast that we were going to cook that night. Or was it? I figured there was nothing to lose . . . I would place the beef in the oven and . . . wait. With a little luck, and a bit of patience, I thought it just might spontaneously cook itself.
I mean, is it not the case there is a non-zero likelihood that all the molecules of the roast beast (or enough of them) will spontaneously acquire the needed thermal energy? Worth a try, no? I knew it was a long shot but, hey, as I said, what was there to lose. I had all day.
Well, to tell the truth, I am beginning to harbour some doubts about the very notion of ‘spontaneous acquisition of energy’. It’s been ten days now and the damn thing isn’t even warm to the touch.
How much longer do you think I’ll need to wait? I have to be back at work soon.
You’ll probably need to know that it’s a 5 kg roast (we do things in a big way up here). And, I’d say the ambient temperature inside is around 20 degrees celsius (yup, power’s back but I want to see my experiment through to the [del]heat death of the Universe[/del] end)
This isn’t so much a quantum mechanical problem as it is a thermodynamic problem. It’s conceivable that energy can flow from colder objects like the air in the room into warmer objects like the roast, resulting in cooler air and a roast that’s cooked to medium. It’s just very very unlikely.
How unlikely? Well, let’s model the roast as 5 kg of liquid water. Let’s model the air in the room as a heat bath at 20 C. The difference in entropy between a roast at 20°C and one at 65°C is
∆S = C ln(293 K/338 K),
where C is the heat capacity of 5 kg of water. Calculating, this is about -3000 J/kg. Using the microscopic interpretation of entropy, this means that the “cooked” configuration is more likely than the “uncooked” configuration by a factor of
e^(-3000 J/kg / k)
where k is Boltzmann’s constant. This works out there being a one in 10[sup]10[sup]26[/sup][/sup] chance (more or less) of the roast being in the “cooked” state relative to the “uncooked” state.
This is a phenomenally unlikely event. Even if the system samples one possible state per Planck time (10[sup]-44[/sup] seconds, which is usually thought to be the shortest meaningful length of time possible), this would still take 10[sup]10[sup]26[/sup] - 44[/sup] ≈ 10[sup]10[sup]26[/sup][/sup] seconds to cook.
(This is a kludgey, back-of-the envelope calculation, and I welcome any corrections from the other boffins on the board. Mainly I was just worried that KarlGauss might be getting a bit peckish, and wanted to make sure he had a snack to tide himself over until the roast was done, since it was going to be a while.)
A better estimate for the relevant timescale would probably be to take the size of the beef or oven, and divide that by the thermal speed of a beef or air molecule. Planck times certainly aren’t relevant here.
Ah. When I said “quantum”, I was thinking of along the line of the Time-Energy Uncertainty Relation ((ΔT) (ΔE) ≥ ℏ/2) which may not even be possible according to Baez in any case.
By the way, I think the meat is turning color. Must be starting to cook!
True, true; my only defense is that it was late and I was just trying to get an ultra-conservative lower bound. The point stands, though, that 10[sup]10[sup]26[/sup][/sup] times any reasonable amount of time is a ridiculously long time.
Let’s be practical. Power out and a roast to cook? Wrap the roast in aluminum foil, put it on top of your car’s engine block, idle the car for 2 hours, then throw the roast in the garbage. It’s tidier and less effort if you leave the aluminum foil on it.
I was visiting my son in Seattle on US Thanksgiving. Turkey all ready to roast. Suddenly, no power. He has a gas range, but only electric oven. No problem. Entropy reversed with help from a propane bar-b-q. Bird spontaneously cooked. Rest of dinner cooked on the gas range. We weren’t even in the dark. He has a gas generator that can power everything except the oven and dryer.
Well, the problem is widely known: there’s no time-observable in quantum mechanics like there is a position-observable; time enters the Schrödinger equation as an independent parameter. Thus, the expression ‘Δt’ entering the uncertainty relation can’t have the usual meaning as a measurement uncertainty; but then, what else does it mean?
There’s various levels of sophistication on which one can attack the question, and mostly, in true physicist fashion, we just stay on the level of a good handwave. But it is possible to do things more carefully, which is just what Mandelshtam and Tamm did; they were able to derive an uncertainty relation between the energy of a system and, roughly, the lifetime of the state the system is in. That is, the quicker a system decays, the less well its energy is defined, and the other way around. Using this, one can ultimately justify most of the talk using the formal uncertainty relation between energy and ‘time’.
It’s a sufficiently fine detail that it’s seldom relevant to any discussion on this board. There is actually an energy-time uncertainty, but it’s not relevant all that often.
I have accidentally put cream in my coffee, and I would like to remove it. Foolishly, I stirred the coffee.
How long will it take for all the cream to huddle together so that I can carefully remove it with a spoon? I am thinking a cube about 1.5cm/side ought to be OK. It needs to be near the top, but I will leave the spoon in so as to minimize surface disturbance when the moment arrives. Maybe I should use a straw…
(This is actually based on a running argument I have with someone who refuses to accept that there is a nonzero chance that all the cream will huddle together, or “unmix” as it were)
You think you’ve got problems. I burnt the toast this morning. Thick clouds of soot set off the smoke alarm. I opened the window to relieve my aching eardrums and create a breathable atmosphere. But now i want my toast back. How long must I wait before my toast is reconstituted and back at the right temperature for buttering. I will leave the windows open just in case.
I don’t know the exact timescale for that one, but a day would probably do it. Cream doesn’t dissolve in coffee; it just becomes suspended. So it’ll naturally tend to float back up to the top.
As an aside, if you stir something into glycerin, and then unstir it by reversing the motion of the stirrer, you can get back to very close to the original configuration.
My teenage daughter’s room is a mess. Clothes on the floor, bed unmade, homework, magazines and nail polish bottles strewn everywhere. Entropy is reversible under certain conditions, so how long until the bed makes itself, the clothes hang themselves back up and the misplaced objects return to their proper positions?
I’ve tried inputting energy into the system in the form of telling her to clean her room. This method is so inefficient as to be non-functional. The only thing that really seems to act as a catalyst is the threatened loss of cell phone privileges, but it would be much less stressful for all if the room would just clean itself.
The speed that molecules at that temperature are typically moving. v ~ sqrt(kT/m), to within dimensionless factors, where k is Boltzmann’s constant, T is the temperature, and m is the mass of the molecules.
Though I suppose you could argue for using the speed of light, instead, since there will be some radiative heat transfer, too. Honestly, with numbers as big as we’re dealing with, it really doesn’t matter.