View Full Version : If water didn't freeze, how much colder would it have to be to match the cooling power of ice?
04-23-2012, 02:27 PM
Ice at 32 degrees will keep things cold much, much longer than water at 32 degrees. But what if water didn't freeze? Say it always stayed liquid no matter how cold it got. If you wanted to match the cooling power of ice, how much colder would the water have to be?
For example, I put 1 kg of ice in an insulated box. As long as the ice remains, the box stays at 32 degrees. After X minutes the ice has fully melted and the temperature in the box will start to rise. So now assume that water doesn't freeze. I want the box to remain at or below 32 degrees for the same time X. I put 1 kg of very cold water at temperature T inside the box. After X minutes the box finally rises above 32 degrees. What temperature is T?
04-23-2012, 02:37 PM
According to the graph here, http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html it looks like in celsius, the point you're looking for is -79.7C
Conversion to your 'Merkin F degrees is up to you. ;)
NOTE - that is the temperature at which the 'liquid cold water' will absorb as much heat to get to 0C as it does for ice to melt completely. HOWEVER, a substance at -79.7C may receive more heat from the environment, because the rate of heat transfer is based on the difference between the two temperatures. However, I don't see any easy way to model that difference based on the information you're providing. I think that -79.7C is what you were interested in anyway.
04-23-2012, 02:52 PM
You're asking about the enthalpy of fusion (http://en.wikipedia.org/wiki/Enthalpy_of_fusion), or the latent heat, of water. It's 334 kJ/kg. So a kilo of ice at 32 degrees has 334 kJ less energy than a kilo of water at 32 degrees. A slushy mix has something in between.
So what water temp would have 334 kJ less energy if it wasn't frozen at 32 degrees? For that, you need the specific heat, or heat capacity (http://en.wikipedia.org/wiki/Heat_capacity#Table_of_specific_heat_capacities). In Joules per gram per Kelvin, it's 4.1813 for liquid water. I assume it'd be the same for our magic water.
So I get 334 kJ/(1kg*K) = 4.1813 J/(g*K) = 79.88 degrees difference.
Since real water freezes at 0C, we can assume our magic, supercool water would have to start at -79.88 C to match real-world ice.
04-23-2012, 03:08 PM
Note that the heat capacity of water changes with temperature (http://en.wikipedia.org/wiki/Properties_of_water#Heat_capacity_and_heats_of_vaporization_and_fusion), so the above answers are really only an approximation. To get a more accurate answer would require more detailed info about this special "non-freezing water". Probably -80°C is as many "significant figures" as I'd be comfortable guaranteeing.
04-23-2012, 05:15 PM
That's really cold. It's about the temperature of dry ice! Good thing ice works the way it does, otherwise packing for a picnic could be pretty dangerous.
04-23-2012, 05:48 PM
And the heat of vaporization of water is even higher, so steam can burn things much worse than liquid water could if it were at the same temperature.
04-23-2012, 08:21 PM
1 g of ice when it melts releases 80 calories
1 g of water needs 1 cal to change temperature by 1 deg C
So you need 1 g of cold water at -80 C (melting point of ice is 0 C)
But this comparison is misleading since water at -80C will cool things down faster due to the temperature difference. So X in the second case will be shorter since the heat loss to the environment will be higher/faster.
04-24-2012, 01:40 AM
Also misleading is that on the way to above freezing temperatures, the contents of your cooler will go via a lower temperature. How much lower depends on how big your cooler is and how full it is, but with sufficient 193 K No-Freeze water and a couple of beers, drinking could be hazardous.
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