# Mixing Water & Temperature Calculations



## loftboy (8/4/08)

Hi All,

I need a simple formula that I can put into Excel, to workout the final temperature of 2 different amounts of water. For example, what would the temperature be if I have 25L of 75 deg C & add 17L of 20 deg C water to it.

If I had of known when I was at school, that science would be useful in homebrewing, I would have paid more attention. 

Thanks.

David.


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## ohitsbrad (8/4/08)

Try this.


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## Offline (8/4/08)

Because they are both water it's not to hard

you just need to think of them as percentages

for your example

25L will be 25/(25+17) percent of you final volume &
17L will be 17/(25+17) percent

there for you get 

42L @
25/(25+17)*75+17(25+17)*20=52.7

i think?


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## Gigantor (14/4/08)

The problem is: what is the final temp when 17l of water @ 20 degC is added to 25 l of water @ 75 degC. Right?

Final temp = [(Mass1*Temp1)+(Mass2*Temp2)]/(Mass1+Mass2) = [(17*20) + (25*75)]/(17+25) = 52.74 degC

I have assumed that 1 litre of water is 1Kg - that shouldn't introduce errors greater than the tolerance of your standard household thermometer.

I think that's the same formula as given by Offline (same answer anyhow). Should work out whether in K or C.

Shouldn't be too hard to put the formula into a spreadsheet.

Cheers


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## sathid (14/4/08)

the thermodynamic way of looking at it is:


m1*Cp1*dT1+m2*Cp2*dT2=0

Basically this is saying that all of the energy from the hot fluid, is being transferred to the cold fluid.

This assumes that there is no heat of mixing (i.e. the total thermal energy of the mixture is exactly equal to the sum of the thermal energy of the two fluids)

Where m1 and m2 are the masses of the two fluids being mixed (and if you know volume and density, you can use mass=density*volume) (SI units)
Cp1 and Cp2 are the heat capacities of the two fluids (at constant pressure - SI units)
dT1 and dT2 are the temperature changes of each fluid (final temp - initial temp). Since it is a difference sum, it does not matter if you use degrees kelvin, or celcius.

I've attached an excel spreadsheet. It can be used as is for water, or you can alter the S.G. and Cp (heat capacity) to use it for wort, if you can find Cp data for wort. It wont be exact in that case (as the Cp data will be only approximately match your wort) but probably close enough for our purposes.

Hope it helps. Gigantor and Offline have the same result, only that it assumes two identical fluids. 

View attachment Mixing_Fluids.xls


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## BrissyBrew (16/4/08)

sathid said:


> the thermodynamic way of looking at it is:
> 
> 
> m1*Cp1*dT1+m2*Cp2*dT2=0
> ...


I love the spreadsheet, now just to get it up on AHB as a tool, so I can send all the questions I have recieved why 25L drum of ice water did not cool my 5gal wort to pitching temps a link in reply. 
Why physics requires 100L of 0*C water to chill 23L of wort at 95*C to a pitching temp of 18.5 is just plain rude when trying to chill ones wort.


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## sathid (16/4/08)

Well think about it. The wort will probably have a higher heat capacity than water does. 

Heat capacity is how much energy is required to raise 1 gram of material, 1 degree in temperature (units are Joules per gram per kelvin).

So if your wort has a Cp of 1.5, and water has a Cp of 1, you will have a situation where for every degree your wort cools, your water will increase 1.5 degrees. That is, provided you have equal masses of wort and water.

Let us just assume for a moment, that you are mixing your cooling water, with your wort (which is a way of approximating an ideal heat exchanger)

In addition, you are trying to cool the wort by 70 degrees, and only heat the water by 18.5 degrees. 
Even if they have the same heat capacity, you will need 3.5 times as much water as you have wort. 

I am assuming we aren't talking about a Heat exchanger setup, and that you are just dunking your kettle in the drum of water?

Remember that you will also have heat transfer between the surroundings and both the wort, and water. If you're water is in say, a stainless steel sink, or metal drum, you probably have 3 or 4 times the surface area of the water exposed to air, as you do the wort (this is considering both area of the container, and surface of the liquid). 
So you will be absorbing more heat into your water from the surroundings, than the wort will be losing to the surroundings. So the amount of water required to cool your wort will increase.


There is a few other factors, but thats it in an approximated manner.

I hope I understood your post correctly.


A slightly neater version of the spreadsheet: 

View attachment Mixing_Fluids.xls


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## mobrien (16/4/08)

sathid said:


> Well think about it. The wort will probably have a higher heat capacity than water does.
> 
> Heat capacity is how much energy is required to raise 1 gram of material, 1 degree in temperature (units are Joules per gram per kelvin).
> 
> So if your wort has a Cp of 1.5, and water has a Cp of 1, you will have a situation where for every degree your wort cools, your water will increase 1.5 degrees. That is, provided you have equal masses of wort and water.



I'm sorry - I have to correct a couple of mistakes here.

1. Heat capacity is the amount of heat to raise anything by 1 degree (or K). You are talking about *specific* heat capacity when you raise 1g of a substance by 1 degree (or K).

2. Water has a *specific* heat capacity of 4.184 J/g/K not 1 as you have stated (you may be thinking of calories, that would be 1cal/g/K)

3. The specific heat capacity of sugar (sucrose) is much less than water - around 1.255 J/g/K (or 0.3cal/g/K)

So....

Because sugar dissolved in water is a mixture (the sugar does not change state as its dissolves) you can work out the specific heat capacity from the percentage sugar in the solution. Personally I would stick to percentage mixing calculations - but its easy enough to do using the conversion factor to convert SG to g/L or % sugar.

I have edited your spreadsheet so it is now correct (I think!) - it now calculates the specific heat capacity of the solutions and final mix, as well as the SG of the final mix. Because some people like extra info, I also have made it spit out the % sugar and g/L sugar in the solutions, calculated from the SG.

Finally, I added two other workbooks - the sheets I posted a few months ago that calculate the time to heat a volume of water with an element, and the volume of cold water to add to hot water to get a temperature which you desire.

Of course if I have made any mistakes, please let me know!

Matt 

View attachment Mixing_Fluids_edit_MO.xls


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## sathid (16/4/08)

mobrien said:


> I'm sorry - I have to correct a couple of mistakes here.
> 
> 1. Heat capacity is the amount of heat to raise anything by 1 degree (or K). You are talking about *specific* heat capacity when you raise 1g of a substance by 1 degree (or K).



Oops. lol you are right. Just getting a bit lazy with terminology. Cp also doesn't need the p (for constant pressure), and should be lower case. )



mobrien said:


> 2. Water has a *specific* heat capacity of 4.184 J/g/K not 1 as you have stated (you may be thinking of calories, that would be 1cal/g/K)


Actually, not sure what I was thinking here...Probably nothing! lol

The original spreadsheet will still work provided both fluids have the same specific heat (because it appears on both sides of the equation), otherwise yes, change it to the correct value! 



mobrien said:


> 3. The specific heat capacity of sugar (sucrose) is much less than water - around 1.255 J/g/K (or 0.3cal/g/K)
> 
> So....
> 
> ...


I'm not so sure on this one. I'm sure it will probably work for our purposes, but I think that the true specific heat capacity of the solutions will vary from what you have suggested. However I don't think it really matters. 

Good idea on blocks for the reverse reactions. Probably more practical than the other way around.

Thanks for picking up my errors!


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## mobrien (16/4/08)

> I'm not so sure on this one. I'm sure it will probably work for our purposes, but I think that the true specific heat capacity of the solutions will vary from what you have suggested. However I don't think it really matters.



http://www.sugartech.co.za/heatcapacity/index.php

There we go - on online calculator for specific heat capacity of sugar solutions based on published engineering tables.

As it turns out, at low concentrations, the spreadsheet in its current form is pretty close at getting the Specific Heat Capacity of the solution. As the concentrations of sugar go up, it gets less accurate. It turns out its due to the energy required to break the hydrogen bonds between the water molecules and the sucrose molecules, so indeed the solution turns out to have a different specific heat capacity than just a mixture based on its component parts.

But now were are getting a little too geeky - a straight percentage mix assuming pure water would get us damn close - now we are talking gravity, density, specific heat capacity and hydrogen bonding, all for the sake of a couple of % accuracy!

Back to the normal programming!

M


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