1. ## water calculations

How much lbs of 85F and 45F do you need to get 52lbs of 65F water? What is the formula to get it?

2. Originally Posted by gumby
How much lbs of 85F and 45F do you need to get 52lbs of 65F water? What is the formula to get it?
That is totally incomprehensible, please post the question in English

Ok lets guess that you mean:

How much water (in pounds) at 85 degrees Fahrenheit and at 45 degrees Fahrenheit to you need to mix to obtain 52 pounds of water at 65 degrees Fahrenheit

Let the two masses be $\displaystyle $$x and \displaystyle$$ y$, then:

$\displaystyle x+y=52$

and:

$\displaystyle 85x+45y=65\times 52$

CB

3. Use the formula:

$\displaystyle Q = mc\Delta \theta$

Q is the heat energy,
m the mass
c the specific heat capacity of water
Delta theta the change in temperature

Energy is conserved, so, Q_1 = Q_2.

$\displaystyle m_1c\Delta \theta_1 = m_2 c \Delta \theta_2$

$\displaystyle m_1(85 - 65) = (52) (65-45)$

EDIT: Oh, and this is a physics question by the way, nothing to do with University Business Math

4. ## water calculations

Hi gumby,
This is not an algebra problem but anyway here is another solution

Let m = lbs ofwater @45 F
Then 52-m = lbs of water @85 F
Datum temp = 45 F
Heat content of m = 0
Heat content 0f 52-m =(52-m)x 40xCp
Heat content of mix = 52x20xCp
Heat added must equal heat absorbed

(52-m) x40 xCp = 52x20xCp Cp's cancel out

m=26 lbs and 52-m= 26

bjh