So we’ve moved into the mountains–7,200 feet to be exact. My 5 minute Oatmeal takes 10 minutes to cook. I’m sure that’s because the water boils way before it gets to 212. So how much salt do I have to add (per gallon, lets’ say) to raise the boiling point to sea-level normal?
Once I know I’ll decide if I can live with that much in my food.
It’s gonna be a lot more salt than you want to eat. I don’t have all the equations in front of me, but 4 teaspoons of salt per gallon raises the boiling point about 1-2 degrees Fahrenheit. At 7700, the boiling point is about 193. I suspect the relationship isn’t linear between the two, but a back-of-the-envelope estimate tells me its A LOT of salt you’ll need to add. Well over half a cup. (My wide guesstimate is from 1/2 to 1 1/2 cups).
Adding 58 grams of salt to a liter of water will raise the boiling point by about 0.51°C (0.92°F). Adding another 58 grams of salt to the water will raise the boiling point by a bit less than another 0.51°C. The solubility limit for NaCl will prevent you from raising the boiling point by more than ~5°C.
4 teaspoons of salt (as per pulykamell) is about 60 grams, so we’re pretty much in agreement.
Have you considered asking the folks at Quaker Oats for high-altitude cooking instructions? You can’t be the first person in history who eats oatmeal in a mountainous setting.
Sorry, you’re just going to have to learn to live with it. Wait till you try cookies or cakes or bread - then things will get really interesting. Do not bake your first cake at high altitude and expect to serve it to company that night, or even the dog.
Welcome to Colorado!
I’m thinking put a lid on the oatmeal. It’s still gonna take longer.
No such thing as a 3-minute egg, for instance–I’m in Denver and it’s a 4-minute egg. Probably a 5-minute egg at your altitude. Rice is interesting, too. And coffee.
OK. If I’m doing my math right, I calculate you would need to add about 1.6 kilograms of salt to raise the boiling point of 3.63 kg (approx. 1 gal) of water 8 degrees.
The equation is
ΔT[sub]b[/sub] = K[sub]b[/sub] * m
Now, admittedly, it’s been a hell of a long time, so somebody check my math. The boiling point elevation constant of water (K[sub]b[/sub]) is 0.512ºCmolal[sup]-1[/sup]
So we have:
8 = 0.512ºCm[sup]-1[/sup]*xm
Solving for x, we get:
x=15.625m
This’ll be a backwards explanation, but it’s easier for me to go this way:
1.6 kg of salt equals 28.376 moles [1660g*(1mol/58.5g)]
Since salt dissociates into two ions, Na+ and Cl- in water, the molality of 1.6 kg of salt in 3.63 kg of water is:
2*28.376mol/3.63kg=15.364 molal solution, the x we solved for (minus rounding errors)
Anybody for a quick check? It seems like a lot of salt to me.
OK, 1.6 kilograms does not seem right at all. Did I forget a kilogram to gram conversion somewhere or vice versa? This is bugging me. Totally an equation I would have whipped through in sophomore year high school chemistry and now, fifteen years later, it might as well be ancient Aramaic.
Looks like you missed a mole->gram conversion.
58.44 X 15.625 = 913.1 grams.
The equation only applies to ideal solutions anyway, so once you get above a molar (molal) or so, you start getting diminishing returns on the temperature rise.
Use a pressure cooker?
But read the booklet if you do–IIR, some grains (barley? rice? …?) are not to be pressure-cooked, maybe because they would clog the steam valve. Read their book.
Ah, the joys of living in the High Country! I used to live at a similar altitude, and the advice I have for you is: get a pressure cooker and Air Bake baking pans. You should also be able to get a high altitude coffee maker.
Hmm…no, my conversion is there…where I convert 1.6 kg to moles. And that 15.625 is molality (moles of solute per moles of solvent), hence about 56 moles of solute (Na+ and Cl-) per 3.63 kg (approx 1 gallon) of water to reach a molality of almost 15.625, our target from the first equation.
Maybe it’s not so high after all. 1.6 kg of salt to raise 3.63 kg of water 8 degrees celsius doesn’t seem completely implausible to me.
Isn’t molality actually moles of solute per kilogram of solvent? Or is that another term? I know we used moles of kilogram more often in chemistry.
Ohhh, you’ve got the wrong definition for molality.
Molality (m) denotes the number of moles of a given substance per kilogram of solvent.
ΔT[sub]b[/sub] = K[sub]b[/sub] * molality
K[sub]b[/sub] = 0.512ºCmolal-1
ΔT[sub]b[/sub] = 8
8 = 0.512 * molality
15.625 = molality = 15.625 moles NaCl/Kg water = 913.1 grams/Kg water.
First of all your first response gave the boiling point as 193 degrees F which means you need a delta change of (212-193)= 19 Degrees F or 10.6 Degrees C. To achieve a 10.6 Degree C elevation in boiling point you would theoretically need:
m = 10.56/0.512 = 20.62 molal or 20.62 moles per kilogram of water. Since NaCl dissociates into two ions you need 20.62/2 moles of NaCl or 10.31 x 58.5 g/mole or:
603 grams of NaCl per kilogram of water.
The more exact solution would have to take into account the activity coefficient (or the reduction in dissociation) due to the fact that you are no longer working in a dilute solution.
BTW, one gallon of water would be approximately 3.785 kg.
Forget the math, nix the salt, and get a pressure cooker.
Sorry, that was a typo on my part. I meant moles per kilogram solvent. Although it may be hard to follow, I am indeed dividing by kilograms solvent (3.63 kg, which should more accurately be 3.725 kg, apparently). Using you numbers, then, 913.1 g/KG = 3.3 kg per gallon or, since NaCl disocciates into two ions, about 1.65 kg NaCl per gallon to impel an 8C change.
And 8 C is approximately the change required at 7700. I misstated my Fahrenheit figures to begin with. The boiling point at that elevation is about 198, or 92.2 C. I had looked at a chart, saw 198.3F and 92.4C listed at 7500 feet, and somehow remembered 193F and 92C.
I’m agreeing with Broomstick here, and it sounds like something Alton Brown would say.
(We meet again, pulykamell! Your posts make me feel simple, but I like you!)