Help me understand how much heat sink possible with frozen H2O

  • Thread starter Thread starter maximiliano
  • Start date Start date
  • Tags Tags
    Heat Heat sink
AI Thread Summary
The discussion centers on the feasibility of using frozen water as a personal cooling system for outdoor activities in high temperatures. A 128-ounce block of ice, weighing 8 pounds, has limited heat sink capacity, likely insufficient for cooling a person during a three-hour climb in 90°F heat. Calculations suggest that the ice could provide cooling for approximately 1 hour and 20 minutes under optimal conditions, with additional cooling from evaporative methods. There is also a consideration of starting with ice at lower temperatures, such as -40°F, to enhance cooling efficiency, although this raises concerns about potential frostbite. Overall, the practicality of using frozen water for cooling in strenuous conditions remains questionable.
maximiliano
Messages
43
Reaction score
0
So many questions on this topic...but I guess the starting point would be to explain the reason for the question. That way, maybe someone can help me work backwards to answer the question...and thus I can apply that to other situations.

I have a friend who thinks he can make a personal cooling system, using frozen water (0c) as the coolant. I say that the amount of heat sink capacity of say 128oz of frozen water, is probably way too small to matter, in terms of cooling a person off while climbing a mountain, in 90 F degree heat, for 3 hours. Add to that the weight of the water itself is 8 pounds. It seems to me the ice would be melted and assume the ambient temp so quickly, that little would have happened in terms of cooling the person off...over a 3 hour period of exertion.

So, I'm trying to understand (maybe there's a solid formula for this) how much heat sink is possible, using ice as the sink, and an ambient temp. or 90F...or 99F I guess, since the purpose is to sink energy from the person, which is surely 99F.

I have another question, which is related...but I'll ask it separately.

Thanks in advance as always!
 
Last edited:
Science news on Phys.org
8 lb (3.6kg) ice takes 3.6*334 = 1200 kJ to melt it.
Walking slowly generates 150W. For mountain walking I see a figure of 300W, but I'm not sure whether that is heat generated or power consumed; the difference is that a lot of the energy goes into raising you up the mountain, so does not end up as heat. Let's take 250W.
In practice, some ice will melt as a result of heat from the environment, but suppose we could prevent that; or, at least, suppose that it roughly balances the heat lost by the walker to the environment. 1200kJ/250W = 4800 seconds = 1 hour 20 minutes.
The melted ice would still be cold, so you could get another 20 minutes from that. Better still, use it to wet a bandana, so now you get evaporative cooling too, and the weight reduces.
 
And the climber has something to drink.

Why start at 0c? Why not start at -40c or -50c? I realize most of the heat is absorbed during melting but starting colder certainly can't hurt.
 
mrspeedybob said:
Why start at 0c? Why not start at -40c or -50c? I realize most of the heat is absorbed during melting but starting colder certainly can't hurt.
No harm in replacing heat stroke by frostbite...
 
Thread 'Thermo Hydrodynamic Effect'
Vídeo: The footage was filmed in real time. The rotor takes advantage of the thermal agitation of the water. The agitation is uniform, so the resultant is zero. When the aluminum cylinders containing frozen water are immersed in the water, about 30% of their surface is in contact with the water, and the rest is thermally insulated by styrofoam. This creates an imbalance in the agitation: the cold side of the water "shrinks," so that the hot side pushes the cylinders toward the cold...
Back
Top