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Thread: water calculations

  1. #1
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    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?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by gumby View Post
    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 $$ x and $$ y, then:

    x+y=52

    and:

    85x+45y=65\times 52

    CB
    Last edited by CaptainBlack; Aug 7th 2010 at 05:55 AM.
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Use the formula:

    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.

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

    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
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  4. #4
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    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
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