[subro]

Hi everyone. I want to build a homemade air conditioner with an ice chest, some freezed water bottles and a fan. I need your opinion.

I have a freezer that can reach -24 ºC. My question is, what is better to cool a room? Regular ice at -24ºC (it will heat until 0ºC and then melt) or brine ice at -24ºC (it will heat until -21ºC and then melt).

I think brine ice will cool the room faster because melting is the slower step (and -23ºC is cooler than 0ºC), but in the end it will exhaust faster. In a thermodynamically way, the amount of heat absorbed by the ice is barely the same, don't it?

Thank you for your help

[Borek]

Compare specific heat capacities of water and brine (and melting heats for both).

In general, you should draw the temperature curve for both systems (assuming constant heat transfer) and compare them, that shall give you a hint about how they perform compared to each other.

[subro]

I have found this graph (http://www.engineeringtoolbox.com/sodium-chloride-water-d_1187.html)



Pure water specific heat is 4.18 J/gK (liquid, doesn't needed) and 2.18 J/gK (solid) and brine at 23% w/w is 3.3 J/gK (liquid) and 2.65 J/gK (solid) (Physics of Snow and Ice : proceedings = 雪氷の物理学 : 論文
集, 1(1): 599-610 Calculated with ecuation of Fig. 1)

Pure water melting heat is 335.55 J/g and brine at 23% w/w is 355 J/g (Physics of Snow and Ice : proceedings = 雪氷の物理学 : 論文
集, 1(1): 599-610 Calculated with ecuation 15)

So:

Regular Ice (-24 ºC ice to 0ºC liquid)

Total heat=specific heat (-24 ºC to -0ºC, ΔT=24) + melting heat (335.55)
Total heat=2.18*24+335.55=387.9 J/g

Brine ice (-24 ºC ice to 0ºC liquid)

Total heat=specific heat (-24ºC to -21ºC, ΔT=3) + melting heat (355) + specific heat (-21ºC to 0ºC, ΔT=21)
Total heat= 2.65*3+355+3.3*21=432.5 J/g

Brine will absorb more heat than regular water, so it is better to use brine.

Is this correct? Am I missing something?

PS: I've also calculated how many degrees will 2 Kg of ice brine decrease in a regular room (5mx4mx3m)

432.5 KJ/Kg*2Kg=1.19 Kg/m^3 (density of air at 25 ºC)*60m^3*1.012 KJ/(Kg*K) (specific heat of air at ordinary conditions)*ΔT

ΔT=12.0 ºC

Wow! 12 ºC is a lot of temperature, even if total heat transfer doesn't occur is enought for me.

However, same calculations determine that regular ice will decrease temperature in 10.7 ºC. I think the difference between regular ice and brine is negligible...

[Borek]

I have not checked the numbers, just skimmed - and nothing caught my attention as being obviously wrong. That's how I would approach the problem in general.

[Enthalpy]

Random thoughts...

The latent heat of fusion seems smaller for brine, not bigger:
http://eprints.lib.hokudai.ac.jp/dspace/bitstream/2115/20328/1/1_p599-610.pdf page 605

If the goal is to ease the heat transfer to air, then you should consider the heat amount up to the fusion point, not up to 0°C.

A lower freezing point makes the freezer's task more difficult. Ice's composition will vary across the depth: the outer parts that freeze first contain purer water, the inner parts concentrate the salt.

You might consider alcohols too, for instance diluted vodka or antifreeze.

Water in a wide bottle takes very long to freeze.

To improve the transfer with air, I wouldn't put the effort on the ice's melting temperature. Thinner container walls (plastic bottle for sparkling water) are more efficient, increased exchange area (long narrow tube) even more.