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If you don't remember to make the sign + or - depending upon heat being added or withdrawn, you may want to write the formulas as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