Physics or Chem knowledgable Help

Teromous said:

I can't muster the energy to find my books, but I know it's not as simple as doing a ratio of .75 to .25...it has to consider entropy and all that garbage of unused energy that I don't have the patience or brainpower to figure out. I'd much rather spend my Friday drinking a beer and keeping this stuff out of my head damn you!

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brewit2it said:

I have a simple thermodynamics question but can't muster the energy to search the net to figure out how to solve it so hope someone who can do it off the top of their head can help out.

Assuming a closed system and all temps in F:

3 gallons wort (you can just assume H20 which will be close enough) at 205 degrees. Add 1 gallon Ice at 6 degree. What is the equilibrium temp.

Thanks for any help

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BendBrewer said:

Why don't you just dump the ice in and take a temp reading?

Side not - I don't recommend adding ice to your wort.

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petep1980 said:

M1 = mass of your wort. I would say this is your volume * density * gravity. H1 is the enthalpy of water at that temp and pressure.

M2H2 = mass and enthalpy of your ice

M3H3 = Well conservation of mass means M3 = M1+M2. Then when you cross reference H, use liquid water.

I did an open system btw, only because there is a model for it in my book.

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petep1980 said:

Only because I have a thermo exam on Monday I did the work.

You're looking at 162F.

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brewit2it said:

Okay, looks like I'm going to have to take a crack at it myself. This may be close (or maybe not) but here's my simple solution. Again I'm making assumptions for simplicity which I know are inaccurate but close enough. For instance it isn't a closed system, and the wort isn't water. But hey, we aren't curing cancer here, we're making beer.

So we already calculated what the answer would be ignoring the effect of latent heat absorption from one gallon of water going from solid to liquid state and that was 155 F. So if we can calculate the amount of energy in joules absorbed during this change of state then calculate how much that amount of joules absorbed from the whole volume of 4 gallons will drop the temp we should get an approximate answer.

So as can be easily looked up, the latent heat for the fusion of water is 334 kj/kg:

1 gallon x 3.8 kg/gallon x 334kj/kg = 1269 kj absorbed energy from melting 1 gallon water.

So now that we have the amount of energy absorbed but need to calculate how much this will drop the temp of the entire closed system. To solve this we need to use the specific heat of water which happens to be 4.18 j per gram per degree C. 4 gallons of water weighs about 15 kg so:

Delta (X) degrees C = 1,269,000 J/(15,000g x 4.18 j/g/degree):

Solving for X we get a drop of 20 degrees C of the entire system from the change of ice to water.

155 F = 68 C
68-20= 48 C

Convert back to degrees F and you get about 118 F

So actually this looks like it could work pretty well since I have been doing a 90 minute 4 gallon boil and with my pot, and boiling outside I've been getting about a gallon boil off. By adding 1 gallon RO ice plus using a water bath I should easily be able to get down to pitching temps in 1/2 hour or less.

Not sure what the hangup on adding Ice to wort is but it's just water

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trigger said:

The highschool answer is 122.9 F. This is assuming that you have 1 gallon of water that you froze, rather than one gallon of ice. I did it with units mass, since the ratio is what's important, not the specific volumes, neglecting changes in density. Basically, this is a gross approximation.

Also, I did this really quickly since my quantum mech homework is calling my name, so I was pretty cavalier with the formulas and units and such. No guarantee on accuracy, but it should be somewhere in the ballpark of 130 F.

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trigger said:

You're neglecting that the specific heat of ice is about half that of liquid water, so that move from 6 F to 32F of the ice has approximately half the dH that it would as water. Your starting temperature of 155 is off because of this.

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trigger said:

Yea, that's actually a much more complex problem to solve. It's one of those things where taking the density into account gives you a guess, but without a lot more information it's just that, a guess. Another thing to take into a aacount is the loss of heat in the form of escaping steam. That will tend to drive the temperature down in an open system.

My guess of 130 was more for a closed system of boiling wort with a gallon of ice, since this is what I figured you were getting at. Either way, this is one of those cases where genchem thermodynamics is going to give you an estimate. For an accurate answer you'd need to know the masses of wort and ice used, the termal mass and thermal conductivity of the vessel, the ambient temperature and humidity, the [solutes] in both the wort and ice, and reactions occurring between the disparate solutions, and a whole bunch of other stuff.

In a case like this your best bet is to monitor the temperature next time you brew, noting the volume ice added, volume wort, temperatures of each and temperature when the ice melts completely. That'll give you an idea of what to expect in the future.

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brewit2it said:

You probably need to do a lot more studying and a lot less drinking before monday:cross:.

Even if you ignore the change of state from ice to water you would get a simple ratio of (205 x 3) + (6 x 1) /4 = 155 degrees which is already less than your answer of 162F, but we know it is something significantly less than 155 F due the the effect of latent heat during change from ice to water in a thermodynamic system.

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