Mixing three water quantities at different temperatures

[beer_brewer]

I am trying to figure out the following in my beer brewing. I have been able to find equations for mixing two quantities of water, but not three. Can anyone help me out here?

So, I have 5 liters at 100 C and I want to add 18 liters (x at 20 C and y at 8 C) to arrive at 23 liters at 24 C. This is because most of the remaining 18 liters is at room temperature (20 C), and I want to find out how much water chilled in the fridge (8 C) will bring me to the desired 23 liters at 24 C?

many thanks

 

[Jobrag]

Solve the two simultaneous equations the answer is in litres. This assumes that all the liquid is water but I guess that the hot liquid is wort which probably has a higher specific heat capacity then pure water in which case the final mixture will be hotter then intended, these should get you fairly close though.

X+Y=19

293X+281Y=4966

 

[beer_brewer]

Thanks Jobrag, but I need some clarification. Why is X + Y = 19 and not 18? And how did you get the second equation. When I put X = 19 - Y into the second equation (which I assume you meant this when you said to calculate both equations simultaneously) I ended up with X=50.08.

cheers

 

[gmax137]

I have been able to find equations for mixing two quantities of water...

OK - you want to mix 5 liters at 100C with 18 liters, to get 23 liters at 24C. What must be the mixed temperature of those 18 liters?

Once you get that figured out, you should be able to figure out how to make up the 18 liters. But I will give you a hint, or rather a prediction: you aren't going to be able to do what you want (your cold water isn't cold enough).

 

[beer_brewer]

Ah ha. That explains the 50 liters. Thank you. I was letting my assumptions get in the way of the calculation.