[tpederson]

1 gallon vegetable stock (3.5L H2O, 4.8g NaCl aprox.)

1 gallon ice water (3.79L)

1c NaCl  (110g)

1/2c Brown sugar (Does it affect the problem?)

 

This is roughly the recipie for the brine that I intend to brine my turkey in.  I'm curious to find out at what temp my 5 gallon bucket with turkey and brine will turn into a salty turkey-sickle.  Myself and my roommates thank you in advance for any help you can render.

ED: Here's a link to the whole recipie in case it helps: http://www.foodnetwork.com/food/recipes/recipe/0,1977,FOOD_9936_8389,00.html

[tpederson]

After doing some research, I've determined that it's a question of molarity and the freezing point.  I have yet to determine exactly how to calculate this... any help?

[tpederson]

Ok, more progress...

Freezing Point Depression:
Delta T_f = T_f⁰ - T_f= i K_f m

T_f (Freezing Point H₂O) = 0
T_f = ?
i (The van 't Hoff Factor, NaCl solves into 2 ions) = 2
K_f (The Freezing Point Constant of H₂O) = -1.86C/m
m (concentration in moles of solute per kilogram of solvent) = 0.269

T_f - 0 = (2)(-1.86)(0.269) = -1.00068C

So...

0C - 1.00068 = -1.00068C = 30.2F

I havent calculated the sugar.  Working on that.

Could anyone double check this?  I'm working on my BS but its in Networking so its a little light on science...

Freezing Point Depression:
http://members.aol.com/profchm/fpdepres.html

The van 't Hoff Factor:
http://dbhs.wvusd.k12.ca.us/webdocs/Solutions/BP-Elev-and-FP-Lower.html

[jdurg]

Your calculations look sound to me.  For the sugar, you use the same equation that you did for the salt, but you'd only use a factor of '1' for the sugar since it doesn't dissociate upon dissolution.  To get the proper values, however, you'll need to combine the sugar and salt equations into one since they are both in the same solution they will both affect the freezing point depression.

[jdurg]

In order to combine everything into one equation, you need to figure out how many moles of sugar you're adding.  Brown sugar is a homogeneous mixture of various compounds, but we'll just make an assumption that it has the same molecular weight as table sugar does (C12H22O11, 342 g/mole).  Looking online, I found the density of brown sugar to be about 0.721 g/mL.  So your half-cup has 85.29006 grams of sugar.  For the Freezing Point Depression equation, we'll combine the moles of sugar with twice the moles of salt to come up with the overall molarity of the solute.  (We'll use twice the moles of salt since NaCl dissociates into two ions, and then use 1 for the van't Hoff Factor).

Delta Tf = Tf0 - Tf= i Kf m

Tf0 = 0C
Tf=?
i = 1
Kf = -1.86C/m
m=(molesNaClx2 + molesC12H22O11)/7.29L = 0.572589M/L solute.

Tf -0 = (1)(-1.86)(0.572589) = -1.06502 C

Tf = 30.083 Degrees Fahrenheit.

As you can see, the large amount of liquid you have in there causes the dilution factor to be VERY large so you're not going to see a huge depression in the freezing point.  In reality, you can assume that it will freeze at the normal freezing point of water.