Formula for temperature changes?

[NuclearRich]

I am going to be donig my largest batch this weekend, and my kettle wont support my strike water's volume. Googling is not helping me at all. I am looking for a formula to tell me what different temperatures of two different volumes of water will produce.
The easiest thing will be for me to take 3 gallons of boiling water and add it to 8 gallons of water at the same time into the MLT. What temperature should the 8 gallons be if I want the whole 11 gallons to be 162.7F?
Thanks for any help!

 

[Nateo]

http://www.tastybrew.com/calculators/infusion.html

something like that?

 

[pbarning]

I use the formula referenced on www.picobrewery.com/askarchive/mashtemp.html to determine my temperatures. It is pretty spot on.

If you can heat both strike water pots at the same time, then just figure out what your strike water temp should be collectively and add the water when they both reach that temp.

 

[menschmaschine]

The answer is 144.21°F. Here's the long way:

The formula is:

(B grams of water)(Btemp°C - Final temp°C)(4.184 J/g°C) = (A grams of water)(Final temp°C - Atemp°C)(4.184 J/g°C)

Where:

B grams of water is 8 gallons, or 30283 grams (1 gram = 1 mL)
Btemp = X, this is what we're solving for
Final temp = 167.2°F or 72.611°C
4.184 J/g°C is a constant- it's the specific heat capacity of water
A grams of water is 3 gallons, or 11356 grams
A temp = 212°F or 100°C

So, X is 62.34°C or 144.21°F

Double check my math because I only did the calculation once.

 

[jmf143]

The answer is 144.21°F. Here's the long way:

The formula is:

(B grams of water)(Btemp°C - Final temp°C)(4.184 J/g°C) = (A grams of water)(Final temp°C - Atemp°C)(4.184 J/g°C)

Where:

B grams of water is 8 gallons, or 30283 grams (1 gram = 1 mL)
Btemp = X, this is what we're solving for
Final temp = 167.2°F or 72.611°C
4.184 J/g°C is a constant- it's the specific heat capacity of water
A grams of water is 3 gallons, or 11356 grams
A temp = 212°F or 100°C

So, X is 62.34°C or 144.21°F

Double check my math because I only did the calculation once.

Why is the 4.184 J/g°C constant needed? Its on both sides of the equation so (I think) algebraically superfulous.

 

[NuclearRich]

The answer is 144.21°F. Here's the long way:

The formula is:

(B grams of water)(Btemp°C - Final temp°C)(4.184 J/g°C) = (A grams of water)(Final temp°C - Atemp°C)(4.184 J/g°C)

Where:

B grams of water is 8 gallons, or 30283 grams (1 gram = 1 mL)
Btemp = X, this is what we're solving for
Final temp = 167.2°F or 72.611°C
4.184 J/g°C is a constant- it's the specific heat capacity of water
A grams of water is 3 gallons, or 11356 grams
A temp = 212°F or 100°C

So, X is 62.34°C or 144.21°F

Double check my math because I only did the calculation once.

This is EXACTLY what I was looking for, thanks so much.

Thanks for trying from the other replies, but a simple infusion calculator isnt quite sufficient for my self inflicted complications.

 

[menschmaschine]

Why is the 4.184 J/g°C constant needed? Its on both sides of the equation so (I think) algebraically superfulous.

It's an equation for combining any two liquids (with different specific heat capacities, for example), so keeping them in the equation would be good practice, but in the case of combining water with water, you're right. So, scientifically, the equation is right, but algebraically, they're not necessary.