I had a very long car ride yesterday, and I passed a small beer store with a sign that said “coldest beer in town”. I got to thinking…if you made sure the amount of liquid and can volume were such that it wouldn’t explode when frozen, how long would it take a can of beer frozen to absolute zero to return to room temperature (let’s say 68 degrees) when placed in a room of that temperature?
0º Kelvin?
Finally, I get an easy one.
4 hours, 38 minutes, 16 seconds. (Plus or minus 4 seconds.)
Yes.
If you want to go the simple route, you’d just use Newton’s Law of Cooling. But then it’s important to understand that the beer will never reach exactly room temperature, but it’ll get as close as you want. It’ll be something like, after 1 hour, it will be 67°F, then after two hours it will be 67½°F, then after three hours it’ll be 67¾°F. You can’t really ask how long it takes to get to room temperature, but you can ask how long it takes, say, to get within a degree of room temperature.
Thanks!
It’d have to take longer than that. I’ve frozen bottles of soda to normal freezing temperatures and it took longer than that to thaw out.
Simplifying a lot:
It would take the same time as a can at room temperature placed in an atmosphere at absolute zero would take to reach that temperature.
Note that there is a discontinuity when it freezes/thaws.
I am not going to bother doing the calculations but, if you are willing to accept certain simplifications, they are relatively simple. I have calculated heat sinks for electronics and they are extremely simple. You determine the specific heat transfer for that sink (can) which is directly proportional to the temperature difference. The temperature curve would be a simple time constant exponential like the charging of a capacitor through a resistor.
Yeah, real simple.
Some of us got C’s in High-school algebra. Just give us a number.
I’m not so sure it would be the same. That absolute zero atmosphere is not going to do a lot of heat transfer quickly.
No, it’s 0 K. The units in the kelvin scale are Kelvins, not degrees.
Yup, it’s the same. A capacitor takes the same time to charge as it takes to discharge. A can of beer takes the same time to go from 40 to 50 as it takes to go from 50 to 40. Heat transfer is directly proportional to temperature difference and inversely proportional to the resistance to heat transfer specific to that object.
Why would the resistance of air at room temperature be the same as air at abs. zero?
It wouldn’t be, but nevertheless the situation is completely reversible. The variables may change, but they change at the same rate, regardless of the direction of the temperature change.
I told you I am simplifying but that’s the way it works in general terms. If you have a heater in the room, the transfer of heat is proportional to the temperature differential, twice the difference, twice the heat transfer. That’s the way it works in simplified terms.
Correct me if I’m wrong, but I thought that we couldn’t get to absolute zero (at least yet). I know we can get really really close though. Is this still true?
It is true, and it will always be true. Because of quantum mechanics and thermal noise, we’ll never be able to reach absolute zero. We can currently get to something like a billionth of a Kelvin.
IANAPhysicist, but wouldn’t the gases in air at (or very near) absolute zero have solidified, leaving vacuum and some nitrogen and oxygen powder on the table? And wouldn’t that mean that RT-0 is not exactly the same as 0-RT?
In other words, taking the supercooled beer out of your 0.000001K fridge and setting it out on the table is going to have the advantage of convection heat transfer with the air the whole time.
But when you put the RT beer into the fridge, convection will stop when the air solidifies, leaving the vacuum, and then the only heat transfer will be by radiation.
But I’m just guessing, I could be wrong.
Reversible? You don’t have the same endpoints.
Although the can has the same temperature at the endpoints, in one case the atmosphere remains at room temperature throughout the transition. In the other case, the atmosphere starts at the low temperature. I shudder to think how that would work out–but it’s still not the reversed case.
Well, it’s not reversible for absolute zero, of course, because that’s a special temperature. But say you stayed within the temperature range where air was gaseous and beer was liquid, so you don’t have any pesky changes of state. Then any situation is reversible. Say you put a 0°C can of beer in a 100°C room. They might both end up (after an arbitrarily long time) at 99°C. If you tried the same experiment except with the temperatures reversed - 100°C beer and 0°C room - they’d both end up at 1°C. Furthermore, it would take the same amount of time. If it took the beer in the first example 4 hours to reach 98°C, then it will take the beer in the second example 4 hours to reach 2°C.