Thread: reaching a low temperature of water

Using probably more than 10mL of water and as much ice as I need, how can I achieve the lowest temperature possible if I'm allowed to use (at most 20g) of any one substance?

for example: using 10mL of water, a ton of ice, and Sodium Chloride(salt) The temperature yields about -15 degrees Celsius.

I think because NaCl is ionic and thus has a stronger electronegativity it sticks to the H2O molecules causing them to melt?


What would be the best substance to use for reaching the lowest temp?
Magnesium sulfate (hepta/hexa/etc..)hydrate
or
Magnesium carbonate
or
glycerol (I'm guessing this is a bad choice because its a mostly non-polar substance)

yeh NaCl is pretty god

what it does is make it harder for the water to form the lattice structure that it exists in, when in the solid state.
something to do with the physical and chemical effect of NaCL.

Originally Posted by organic god

yeh NaCl is pretty god

what it does is make it harder for the water to form the lattice structure that it exists in, when in the solid state.
something to do with the physical and chemical effect of NaCL.

[EDIT]

I lowered it down to three substances:
1. Magnesium sulfate heptahydrate
2. sodium carbonate
3. calcium chloride dihydrate
Out of these three, which will lower the temperature of water+ice the lowest?

Adding a solute to water lowers the freezing point (also raises the boiling point) of that solution because its allows for less contact between H2O molecules. Freezing point depression is a colligative property which means that it depends on the number of particles that dissociate rather than the mass or nature of the particle.

Change in freezing temp = Kf x molality of solute x i

Kf is a constant for the solvent
i is the Van't Hoff factor or how many particles the solute dissolves into (sugar would be 1 NaCl would be 2)

So if you're comparing equimolal concentrations of the compounds you listed and since Kf is constant, the more particles the solute dissolves into, the lower the freezing temperature.

Originally Posted by rancidchickn

Adding a solute to water lowers the freezing point (also raises the boiling point) of that solution because its allows for less contact between H2O molecules. Freezing point depression is a colligative property which means that it depends on the number of particles that dissociate rather than the mass or nature of the particle.

Change in freezing temp = Kf x molality of solute x i

Kf is a constant for the solvent
i is the Van't Hoff factor or how many particles the solute dissolves into (sugar would be 1 NaCl would be 2)

So if you're comparing equimolal concentrations of the compounds you listed and since Kf is constant, the more particles the solute dissolves into, the lower the freezing temperature.

So according to that, magnesium sulfate heptahydrate should have a lower freezing point than calcium chloride dihydrate because MgSO4•7H2O has more molecules than CaCl2•2H2O. Am I right?
OR
Does the hydrate part even matter? Does CaCl2 have a lower melting point than MgSo4 because CaCl2 has more ions?

Which is correct?

I think the number of water molecules attached to the ionic solid wouldn't matter. When they dissociate they wouldn't interfere with freezing like a non-water particle would. I'd say CaCl2 and Na2CO3 would lower it about equally, because they both dissociate into 3 ions. Maybe someone who knows more could give you a definite answer.

Originally Posted by rancidchickn

I think the number of water molecules attached to the ionic solid wouldn't matter. When they dissociate they wouldn't interfere with freezing like a non-water particle would. I'd say CaCl2 and Na2CO3 would lower it about equally, because they both dissociate into 3 ions. Maybe someone who knows more could give you a definite answer.

Ok right. Thats what I was thinking because the hydrates should not affect dissociation.

So the substances left for me to choose are:
1. CaCl2 • 2H20
2. Na2Co3

Gahhh which one will lower the temperature of water the most?

I don't know if this helps:
The melting point of CaCl2 • 2H20 is (I think) 772 degrees Celsius.
The melting point of Na2Co3 is 851 degrees Celsius.

Well the idea behind colligative properties is that the composition of the solute doesn't matter. All that matters is how many particles it dissolves into. Plugging both of those compounds into the equation above will give you the same result.

The melting point of CaCl2 • 2H20 is (I think) 772 degrees Celsius.
The melting point of Na2Co3 is 851 degrees Celsius.

You're not melting the ionic compound. The melting/freezing point of water is what's being lowered.

Another thing to recognize might be the enthalpy change that's associated with dissolving the solute. I think Na2CO3 might release heat (exothermic) when it's dissolved (strong bases do). If MgCl2 is like NaCl it will absorb heat and lower the temperature when dissolved. The freezing point depression is still equal for the two though.

Originally Posted by rancidchickn

I think Na2CO3 might release heat (exothermic) when it's dissolved (strong bases do). If MgCl2 is like NaCl it will absorb heat and lower the temperature when dissolved. The freezing point depression is still equal for the two though.

What makes you think that Na2CO3 might release heat during the process but not CaCl2 ?

The chart here.
http://en.wikipedia.org/wiki/Enthalpy_of_solution

I'm sure you could pull up actual values for the two compounds in a reference book or some place on the internet. I don't know if this has any real significant effect on the temperature. Once you figure out the enthalpy of solution for the two you could multiply it by the number of moles you have and, plug it into the equation ΔQ = mcΔT. That would solve for the temperature change experienced.

I think I have reached a conclusion:

solute = CaCl2
molarity = 20g/10mL = 0.18moles/0.010L = 18M
produces 3 moles of particles
3*18M = 54M
KfM = (1.86)(54M) = 100.44
Freezing point = -100.44

solute = Na2CO3
molarity = 20g/10mL = 0.19moles/0.010L = 19M
produces 3 moles of particles
3*19M = 57M
KfM = (1.86)(57M) = 106.02
Freezing point = -106.02

solute = MgSO4
molarity = 20g/10mL = 0.166moles/0.010L = 16.6M
produces 2 particles
2*16.6M = 33.2M
KfM = (1.86)(33.2M) = 61.75
Freezing point = -61.75

Therefore Na2CO3 must have the lowest freezing point!!!!!!!!
correct?!?

Please reply!!

You need to be using molality (mol/kg) not molarity. This is because molarity (since it relies on volume) changes with temperature.
Kf for water is 1.858 K·kg/mol

I guess it doesn't make a difference if you're just looking for a qualitative comparison, but you're not going to get 20g of those solutes to dissolve in only 10 mL of water.

Originally Posted by rancidchickn

You need to be using molality (mol/kg) not molarity. This is because molarity (since it relies on volume) changes with temperature.
Kf for water is 1.858 K·kg/mol

I guess it doesn't make a difference if you're just looking for a qualitative comparison, but you're not going to get 20g of those solutes to dissolve in only 10 mL of water.

Shouldn't molarity and molality not matter if the solvent is water?
So I am correct that Na2CO3 will freeze the lowest?

Does the water have to remain a liquid? Or can you win by making very very cold ice?

Look into the KCal/mole of each molecule. Remember the first law of thermodynamics comes into play. The group with the higher temperature will give energy to the lower energy group.

It's been brought up that this is a chemistry forum. Maybe if you studied Morrison and Boyd, but if you study G. Marc Loudon you'll find he introduces you to the physics of chemistry, especially with sp orbitals and such. It has everything to do with the energy of activation required to control the heat of a molecule. Remember physics is the substrate of chemistry as chemistry is the substrate of biology. Personally I think physics is a piece of cake compared to chemistry, but having been a practical chemist in the field of plastics engineering for over 20 years, the need to understand the molecules has reared its ugly head.

Have fun, but this is just becoming a good discussion...