How many joules are needed to heat 8.50 grams of ice from -10.0 degrees to 25.0 degrees

I came up with 2124 joules

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  1. I believe I worked that problem yesterday. Perhaps not for you.
    heat needed to move T of ice from -10 to 0
    q1 = mass x specific heat ice x delta T

    heat needed to melt the ice.
    q2 = mass x heat of fusion.

    heat needed to move T from 0 to 25.
    q3 = mass x specific heat water x delta T.

    Total heat required is q1+q2+q3.
    Post your work if you get stuck.

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  2. For the specific heat of ice I looked it up and found it to be 2.05 and the heat fusion to be 334

    q1 = 8.5 x 2.05 x 10 = 174.25

    q2 = 8.5 x 334 = 2839

    q2 = 8.5 x 4.184 x -25

    and that gave me a total of 2124 joules...also for my 2nd question from yesterday...

    Caculate the hydronium ion concentration and the hydroxide ion concentration in blood, the pH of which is 7.3 (slightly alkaline)

    [hydronium ion]= 5.0 x 10^-8
    [hydroxide ion]= 2 x 10^-7

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  3. q1 is ok.
    q2 is ok.
    q3 (which you typed in as a second q2) is not correct because you made it a negative number. Since heat is being added it must be a positive number. You should have +889.1 J and the total should have been 3902 J but check my arithmetic.
    If you don't remember to make the sign + or - depending upon heat being added or withdrawn, you may want to write the formulas as
    mass x specific heat x (Tf-Ti) where Tf is final T and Ti is initial T.
    For 0 to 25 that will be
    8.5 x 4.184 x (25-0) = a + number
    It works for the -10 to 0 also.
    8.5 x 2.05 x [0-(-10)] = 8.5 x 2.05 x 10 = a + number.
    I don't remember seeing the blood problem.

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