Help me understand how much heat sink possible with frozen H2O

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SUMMARY

The discussion centers on the feasibility of using frozen water (ice) as a personal cooling system during strenuous activities, specifically mountain climbing in 90°F heat. The analysis reveals that 128 ounces (8 pounds) of ice can absorb approximately 1200 kJ of heat, providing cooling for about 1 hour and 20 minutes under ideal conditions. However, the effectiveness diminishes due to ambient temperature and the heat generated by physical exertion, estimated at 250W for mountain walking. The conversation also suggests that starting with ice at temperatures below 0°C could enhance cooling efficiency, although it raises concerns about the risk of frostbite.

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  • Understanding of thermodynamics, specifically heat transfer principles.
  • Knowledge of energy units, particularly kilojoules (kJ) and watts (W).
  • Familiarity with the concept of phase changes, particularly melting ice.
  • Basic principles of evaporative cooling techniques.
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  • Research the thermodynamic properties of water and ice, focusing on heat capacity and phase change.
  • Explore the calculations for heat transfer in outdoor environments, particularly during physical exertion.
  • Investigate advanced cooling techniques, including evaporative cooling and the use of phase change materials.
  • Examine the physiological effects of heat stress and cooling strategies for athletes in extreme conditions.
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Outdoor enthusiasts, climbers, athletes, and anyone interested in personal cooling solutions during physical exertion in high-temperature environments.

maximiliano
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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!
 
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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...
 

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