Newton's Law of Cooling and Brewing

[MisterTipsy]

Anyone know what the constant (k) value for 5 gallons of beer in a plastic bucket is ?


Newton's Law of Cooling equation is:

T2 = T0 + (T1 - T0) * e(-k * Δt)

where:

T2: Final Temperature (C)
T1: Initial Temperature (C)
T0: Constant Temperature of the surroundings (C)
Δt: Time difference of T2 and T1 (hours)
k: Constant to be found


I reverse engineered a rough value of 0.1. This was based on observed temperatures and times. I read a dead human body has a K value of 1.7. Therefore, 5 gallons of beer =/= dead body!

Is this consistent with your observations?

http://www.endmemo.com/physics/coollaw.php

cooling rate calculator ^

 

[MisterTipsy]

Sir Isaac Newton probably brewed beer.

 

[MisterTipsy]

How long does it take to cool an 80F five gallon bucket down to 65F in a 32F fridge?

 

[rustbucket]

im way to lazy to get my thermodynamics textbook out.. i was not so lazy..

 

[kwingert]

Anyone know what the constant (k) value for 5 gallons of beer in a plastic bucket is ?


Newton's Law of Cooling equation is:

T2 = T0 + (T1 - T0) * e(-k * Δt)

where:

T2: Final Temperature (C)
T1: Initial Temperature (C)
T0: Constant Temperature of the surroundings (C)
Δt: Time difference of T2 and T1 (hours)
k: Constant to be found


I reverse engineered a rough value of 0.1. This was based on observed temperatures and times. I read a dead human body has a K value of 1.7. Therefore, 5 gallons of beer =/= dead body!

Is this consistent with your observations?

http://www.endmemo.com/physics/coollaw.php

cooling rate calculator ^

My head hurts

 

[dlaramie08]

You'll need to know the specific heat of water. However that is for pure water. So you'll need to know the specific heat of the specific concentration of sugar concentration as I'm sure it's different than water alone.

 

[MisterTipsy]

If all the temperature and time values are known, you can use the cooling calculator to determine the constant coefficient. I did that. That's how I got a k value of 0.1.

If more people tell me exactly how long it took them to cool 5 gallons of beer from one temperature to another with a known refrigerator temperature I can validate the constant.

 

[mike20793]

Just solve for the constant using the boundary conditions. That is just the solution to a simple differential equation. Maybe this site will help with it. I would do it, but I just came back from the emergency room and I'm a little doped up.

http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html

 

[MisterTipsy]

I'll have to wait until you aren't doped up.

 

[mike20793]

You just need the initial temperature of what you are cooling, the temperature around what you are cooling, and the time it takes to get to a specific temperature (say 10 minutes to get to 120, along those lines. It doesn't have to be the final temperature). Then plug back into the solution (the cooling equation) to get an equation in terms of T with respect to time, T(t). Then use the time it took to reach the temperature and you will have all knows except k. Then just a little algebra will get you the constant. That website I posted earlier does a better job explaining it. Hope that helps.

 

[mike20793]

Okay, I'm not doped up anymore, so I should be able to explain in more detail. The T2 in the equation is actually T(t). The equation should actually look like this: T(t) = Ta + (To - Ta)e^-kt , where Ta is the temperature surrounding what is being cooled, To is the temperature of what is being cooled, and T(t) is the temperature after a specific time, t. Note the changes to the original equation that was posted. If the initial temperature of the liquid is 212 degrees, and the ambient temperature is 70 degrees, the equation becomes: T(t) = 70 + (212 - 70)e^-kt. So, if your liquid is cooled to 120 degrees after 10 minutes, then T(10) = 120 and t = 10. The equation becomes: 120 = 70 + (212 - 70)e^-10k. Now you can solve for k using the properties of natural logs.

120 - 70 = 142e^-10k
50/142 = e^-10k
ln(0.352) = -10k
-1.044 = -10k
k = 0.1044

Note that these are just values that I came up with randomly. Hopefully this helps, but you definitely need to know those boundary conditions to solve for an exact solution (Ta, To, T(t), and t), rather than a particular solution (a solution with unknown constants).

Finally that math minor paid off!!! One more reason to love homebrewing .

 

[MisterTipsy]

I had plenty of math in college, but haven't needed to use it for a long time.

Determining the K value with arbitrary temperatures and times doesn't determine the actual K value of 5 gallons of beer unless you correctly guessed the time and temps. This is why I asked for some input on how long it takes other peoples' 5 gallons of beer to go from one temp to another in a fridge.

The time part of the equation is measured in hours, not minutes, and the temperature is measured in Celsius so your K value is valid if it took the object 10 hours to cool instead of 10 minutes and the temperatures were 212C, 120C and 70C.

I'll observe the time and temperatures the next time I have to chill beer. I'll use the online calculator to determine K again to see if my observations are consistent.

 

[mike20793]

It must have been a long time because you're wrong; the time is either minutes or hours. It's completely arbitrary. That just determines the units of the constant. Minutes gives it in minutes and hours gives it in hours; it's up to you which you prefer. As long as your consistent, the solution is the same. And Yes, you need to have specific observations for the equation to work. I made up values to explain how to solve the equation you posted. You don't need to use the calculator if you know anything about differential equations. That calculator probably uses the same equation I just explained how to solve.

Not to brag, but I know quite a bit about differential equations. I have a masters in structural engineering and much more complex differential equations were the basis of all my graduate level classes. This one is a very, simple first order differential equation solved in second year undergrad math classes. Not trying to flame you, but I am definitely not wrong.

 

[sieglere]

The k value you generate will probably be a combination of convective and conductive transfer. Basically it'll only be "accurate" for where you set the bucket that specific time since there will be different air flow patterns elsewhere.

What I'd suggest doing is skip the theory, attach a thermometer, take readings every minute, fit a curve and use that as a guide for next time. The k values are often empirically generated and how better do experiments then in a way that generates yummy beer?

 

[mike20793]

That k value from the equation is based on Newton's Law of Cooling, which doesn't take those types of heat transfer into account. There are more complex laws that do. It's just a simple differential equation that is really easy to solve; but yes, it's easier to just use a thermometer and measure. I was just trying to explain how to solve the equation that was originally posted.

 

[MisterTipsy]

It must have been a long time because you're wrong; the time is either minutes or hours. It's completely arbitrary. That just determines the units of the constant. Minutes gives it in minutes and hours gives it in hours; it's up to you which you prefer. As long as your consistent, the solution is the same. And Yes, you need to have specific observations for the equation to work. I made up values to explain how to solve the equation you posted. You don't need to use the calculator if you know anything about differential equations. That calculator probably uses the same equation I just explained how to solve.

Not to brag, but I know quite a bit about differential equations. I have a masters in structural engineering and much more complex differential equations were the basis of all my graduate level classes. This one is a very, simple first order differential equation solved in second year undergrad math classes. Not trying to flame you, but I am definitely not wrong.

My point is if you aren't using real world measurements, you are determining the k value of something that is most likely not 5 gallons of beer in a plastic bucket.

 

[mike20793]

Well I misunderstood the thread I guess. I thought you just needed help solving the equation. That's my bad. Sorry, I don't have any real world measurements and sorry for the confusion. I guess I wasn't any help after all!

 

[MisterTipsy]

Well I misunderstood the thread I guess. I thought you just needed help solving the equation. That's my bad. Sorry, I don't have any real world measurements and sorry for the confusion. I guess I wasn't any help after all!

It's ok.

I was merely looking for more data to support my observation of the K value of 5 gallons of beer in a plastic bucket, but the math lesson has been fun

or, if someone already knew what the K value was, I was ready to hear it.


Now, we will have a pop quiz:


If I put 5 gallons of 80F beer in a 35F fridge, how long will it take to cool to 65F ????


Anyone? Anyone?

 

[sieglere]

It's ok.

I was merely looking for more data to support my observation of the K value of 5 gallons of beer in a plastic bucket, but the math lesson has been fun

or, if someone already knew what the K value was, I was ready to hear it.

Now, we will have a pop quiz:


If I put 5 gallons of 80F beer in a 35F fridge, how long will it take to cool to 65F ????

Anyone? Anyone?

About 8!

 

[MisterTipsy]

About 8!

How did you get 8 ?

 

[sieglere]

How did you get 8 ?

There's no units...so just being a bit sarcastic

 

[MisterTipsy]

Oh, I see.

 

[mike20793]

You need one more boundary condition to solve that problem. Otherwise, it's not possible. You need to know how long it takes the beer to cool to a specific temperature, not necessarily the temperature you mentioned. You need that to solve for k. If you assume k equals 0.1 (in hours) from your observations then it would be 4.05 hours. Without the k, you can't get the exact solution, which I guess is your problem to begin with. You can measure the time it takes to drop one degree. That would be enough to calculate the constant, k.

 

[MisterTipsy]

I measured the temperature drop after 4 hours and determined K was roughly 0.1 using the calculator.

It took 4 hours to go from 80F to 65F with a fridge temp of 35F.

 

[mike20793]

Awesome. That was about what I got assuming that value of k (I got 4.05 hours). Looks like your initial judgement was correct. Good thinking. It's nice when you can verify you were right all along.

 

[MisterTipsy]

Now I can chill beer precisely on a tight schedule!