Why liquids don't stay hot/cold indefinitely in a thermos

  • Level: Undergrad 
  • Thread starter Thread starter Mentallic
  • Start date Start date
  • Tags Tags
    Liquids
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
6 replies · 4K views
Mentallic
Homework Helper
Messages
3,802
Reaction score
95
I'm curious as to why liquids don't stay hot/cold indefinitely in a thermos (the double cylinder which are separated by a vacuum). Thermal energy is being lost, but how? Is it being converted to another form and possibly radiated out like the sun does? But then I'd wonder why the container itself doesn't contain+reflect the radiation back.
 
Physics news on Phys.org

Discussion



Well, for one, you have the top part which is not separated from the exterior by a vacuum. Secondly, the vacuum isn't perfect, energy still gets through, albeit slowly.
 


So provided we can make a thermos without a top (all of the inner cylinder is separated by a vacuum) and this vacuum approaches perfection, then the heat will dissipate ever more slowly? A theoretically perfect vacuum would in effect have no heat loss?
 


No, because there would still be radiation losses.
There are three mechanisms: Convection ("particle transport"), conduction (things touching) and radiation. Something suspended in a perfect vacuum would not cool down due the first two, but it would still cool down due to radiation losses.
Btw, the reason why most thermos bottles are shiny on the inside is because the surface is covered by a reflective metal layer in order to minimize radiation losses.
 


You also have radiative energy from the substance that also results in energy loss. Both energy transfers work both ways as well, it can absorb radiative energy as well.
 


f95toli thanks for that explanation. Then let's assume the only possible energy loss is through radiative means by allowing for a near perfect vacuum separation. Is it possible to completely halt this process? I've heard that the higher frequency radiation types can be stopped by thick layers of lead. Could this be used? Or would the heat just conduct through the metal and escape the surface as radiation?

Intuitively, I would believe that the difference between temperatures of the inner cylinder and outside environment would depend vastly on the rate of energy loss. e.g. A coffee at 50oC would take much longer to drop 10oC (to the room temperature 25oC outside) than a coffee at 200oC to drop 10oC.
So how effective would a thermos be to contain/repel the heat of a substance at extreme temperatures?

Basically I'm asking because I'm trying to understand how ineffective it is to replace telephone lines, cables etc. with a superconductive material (<100K) which is insulated by the basic thermos type principle.
 


Mentallic said:
f95toli thanks for that explanation. Then let's assume the only possible energy loss is through radiative means by allowing for a near perfect vacuum separation. Is it possible to completely halt this process? I've heard that the higher frequency radiation types can be stopped by thick layers of lead. Could this be used? Or would the heat just conduct through the metal and escape the surface as radiation?


There is no such thing as perfect insulation. Eventually everything in a closed system will be at the same temperature, the only thing we can do is to slow down this process.
Lead is good at stopping alpha and beta radiation (which are ordinary particles, helium nuclei and electrons) but it is useless for infrared radiation (photons) which is what we are talking about here/


Intuitively, I would believe that the difference between temperatures of the inner cylinder and outside environment would depend vastly on the rate of energy loss. e.g. A coffee at 50oC would take much longer to drop 10oC (to the room temperature 25oC outside) than a coffee at 200oC to drop 10oC.
So how effective would a thermos be to contain/repel the heat of a substance at extreme temperatures?

There is no simple answer to that question since it depends on how the heat is transferred. The amount of thermal radiation emitted from an object is proportional to T^4; i.e. it is a very strong function of temperature. If the heat is transferred via conduction the process is exponential and depends on the difference in temperature.


Basically I'm asking because I'm trying to understand how ineffective it is to replace telephone lines, cables etc. with a superconductive material (<100K) which is insulated by the basic thermos type principle.

Commercial superconducting cables use a combinaton of a "thermos" (good thermal insulation) and active cooling; essentially cold gas that is circulated in the cable. A cryocooler at one or both ends is used to keep the gas cold. Modern cryocoolers are quite efficient so this process doesn't really use much energy and the cryocooler only needs to "compensate" for the inperfect insulation; the superconductor itself is not generating any heat. Also, the fact that there are no resisitve losses (which can be siginifant in high current copper cables) means that the NET energy lost during transfer can be smaller for a superconducting cables then for a normal metal one.

The main problem with superconducting cables have been mechanical: the cables are quite complicated and there have also been practical problems with e.g. the minumum bending radius. long-term stability etc. Many of these problems have now been solved and there have been a number of succesfull demonstrations which is why we are now starting to see superconducting cables being used in "real life".

Also, there is absolutely no point in using superconducting cables for telecom and other low power applications where resistive losses are negligable anyway.