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What metal or material will stay cold best after being in a freezer?
I am asking what metal or material will retain its coldness the longest after being kept in a standard home freezer. After being removed from the freezer, the metal will be subjected to a moderate heat source for 5-20 minutes. Is it a metal or something else, possibly ceramic? I am using this for a project.
3 Answers
Metal would not be suitable for your project since it is a conductor.
Ceramic is a good candidate since it is an insulator. Depending on how hot the 'moderate heat source' is, there are different materials to choose from. If it's not very hot (below say 100degC), the you can try polystyrene (the stuff that makes cups that holds hot beverages) as it is an excellent thermal insulator - the thicker the better.
If the heat source is quite hot, then go for ceramic since ceramic can withstand high temperatures, just make sure that the change in temperature around the ceramic isn't too sudden otherwise it may crack.
Ceramic is a good candidate since it is an insulator. Depending on how hot the 'moderate heat source' is, there are different materials to choose from. If it's not very hot (below say 100degC), the you can try polystyrene (the stuff that makes cups that holds hot beverages) as it is an excellent thermal insulator - the thicker the better.
If the heat source is quite hot, then go for ceramic since ceramic can withstand high temperatures, just make sure that the change in temperature around the ceramic isn't too sudden otherwise it may crack.
Your best bet is some sort of phase-change material, like water. Actually water is a really good material overall, because it has such a high specific heat.
Apologies for the English units here.
Let's say you had a heat sink that has 1 pound of water contained in a 1 pound steel vessel, and you froze it to 28 F and needed to keep a final temperature of less than 40 F.
The Aluminum would absorb:
0.2 BTU / lb °F * 1 lb * (40 - 28)°F = 2.4 BTU.
The Ice would absorb:
0.5 BTU / lb °F * 1 lb * (32 - 28)°F = 2.0 BTU
The water would absorb:
1.0 BTU / lb °F * 1 lb * (40 - 32)°F = 8 BTU
and finally, melting the ice to water would absorb:
144 BTU / lb * 1 lb = 144 BTU
or 92% of the energy absorbed by the system.
Apologies for the English units here.
Let's say you had a heat sink that has 1 pound of water contained in a 1 pound steel vessel, and you froze it to 28 F and needed to keep a final temperature of less than 40 F.
The Aluminum would absorb:
0.2 BTU / lb °F * 1 lb * (40 - 28)°F = 2.4 BTU.
The Ice would absorb:
0.5 BTU / lb °F * 1 lb * (32 - 28)°F = 2.0 BTU
The water would absorb:
1.0 BTU / lb °F * 1 lb * (40 - 32)°F = 8 BTU
and finally, melting the ice to water would absorb:
144 BTU / lb * 1 lb = 144 BTU
or 92% of the energy absorbed by the system.
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