Chemistry

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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3. 👁 72
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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posted by Mike
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If you don't remember to make the sign + or - depending upon heat being added or withdrawn, you may want to write the formulas as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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