A saturated salt water solution should just be slush at -20 and gradually rise in temp as it melts then becomes 10C
Fresh water at -20 will be ice and absorb heat at some rate until it reaches 0, then plateau in temp as it phase changes to water, than rise gradually to 10C
Which 1 liter bottle (saturated salt water, or fresh water) will absorb the most heat during this temperature change?
Specific heat of liquid water = 4.19
Specific heat of ice = 2.09
Specific heat of fusion = 334
(all easy to find)
So if you do the math, for the pure water the energy required to melt the ice is 80% of the total energy input.
Saline solution
Sodium Chloride Water Solutions
Dissolved salt reduces the specific heat capacity of water.
Specific heat of saturated saline = 3.3
But the salt solution remains liquid to a lower temperature, and ice has half the specific heat capacity.
That same link also says that a saturated saline solution remains liquid to -20C, so not slush - no enthalpy of fusion for the saline solution. That’s the clincher.
So my amateur answer is that the fresh water absorbs more heat, but IANAT.
ETA: It looks like the freezing point of saturated saline is right at -20C. So the caveat here would be that there would be a very large difference in the amount of heat required for the saline depending on whether it’s frozen.
But whichever version has to convert solid to liquid water will be the most heat-absorbtive. Saturated saline at -19C is all liquid. Saturated saline at -21C is all ice. Choosing your saline at -20C and slushy makes the problem indeterminate since we don’t have any way to know how much of the slush is ice and how much is water.
You’d have the same problem if you chose 0C and slushy for the freshwater leg of your experiment.
Which leads me to ask what’s the real question the OP is trying to understand? Is it about the relative importance of freeze/thaw versus sensible heat change, or is it about the importance of saline versus non-saline? You’ve (inadvertently?) picked example temps for your experiment that blend the two issues into a muddy mess.
I got much the same answer. There is enthalpy of solution involved as well, but it is relatively minor.
If we are working in mass, the brine contains less water, so that swings things towards pure water even if the brine freezes.
I ask basically to see if it better (only from a storage cooler sense) to use containers of salty water in an ice chest or fresh water.
I’m not interested in other aspects, only if salty water will keep my food below 10C for longer than fresh water/ice.
I’ll note that the idea that salt water is liquid to -20 and turns to ice right at -20 does not coincide with my actual empirical evidence. Salt water stuck in a freezer will get slushy and not turn to solid ice or completely liquid at well defined temperatures. Its slushy over a broad range. At least this is my actual real experience.
If your freezer doesn’t go to a low enough temp to get the saline to freeze completely, then I think the answer is that the pure water is better. But you could use a lower concentration of salt that you should be able to get to freeze.
Comparing frozen pure water to frozen saline, then… given that most of the energy goes to the melting, perhaps the difference that then becomes interesting is not so much the total energy, which will only differ slightly, but the fact that the saline will sit at its lower melting point for most of the time. So there will be a bigger temperature difference between the coolant and what you are trying to cool (and the outside environment). I’m not sure if that’s good or bad.
As the OP wants to keep stuff cool it all about how much energy the cooling material can absorb to prevent the food warming up.
Your cooler and food is at a given starting temp and the cooler will leak in energy at a certain rate , so you need something to absorb that incoming energy , which is your one L bottle of ice or salty water which is starting at -20c
1L of ice is about 919g , ignoring the thermal expansion coefficient from 0 down to -20 which is pretty small.
I am assuming a closed 1l container and when filled space was left for expansion of the ice hence less mass.
So we take 919g of ice up by 20c to melting point
919 x 2.1 x 20 = 38.5kj
306kj to melt the ice (333j/kg x 0.919kg)
919g x 4.2 x 10 to heat the water up to +10c= 38.5kj
So that’s 380kj or energy absorbed that would have warmed up your food .
Salt saturated water is 1.2 g /cc so we have 1200g in that liter. We can probably ignore the thermal expansion coefficient, the solution at -20C isn’t going to be significantly higher than 1.2gcc
Assuming the salt solution is still all liquid 1200x 3.3 x30 = 118kj ( I used the 3.3 j/g/c from upthread)
So the water / ice bottle will absorb more energy, mostly due to the energy to change phase from ice to liquid.
If you started with 1l of water and froze it and allowed for the expansion in the container, then to water case is even better as your started with more mass.
As others mentioned the slush part is an undefined part of the problem, at any temperature there will be a distribution of kinetic energy in the liquid so as we are close to the freezing point there will be some ice in there as you will never have a equal distribution of energy, but my statistical mechanics course was way too long ago to remember the details , but it narrower than the Maxwell Boltzmann distribution in gasses.
For the saltwater mixture to be the same as the ice bottle it would need to absorb another 262k which is about 700g of water warming from -20 c to 0 then melting, that’s most of the water in your 1 l of saturated brine , so for it to be as good as the pure water , it would have to be pretty much entirely frozen.