Posted by Sarah on Wednesday, September 15, 2010 at 10:23pm.
Show your work and I'll find the error.
ok.. i got
Three errors and two of them compensated for each other.
(mass ice x 330) + (mass water x specific heat water x (Tfinal-Tinitial) = 0
1. You reversed the Tfinal and Tinitial. That should be 35-75 = -40 and not +40.
2. When you moved the 70145.28 to the other side you did not change the sign; however, mistake #2 canceled the #1 error.
3. If I multiply 396.893 x 4.184 x -40 I don't get 70,000 something but more like 66,000 something which goes on to obtain slightly more than 200 g ice.
i got 201.28 and rounded to 201 but it said it was wrong
Two possibilities. The other problem I worked had 12 ounces of water and not 14. Could this be the problem?
Second, you are allowed only two places (because of the 40 (assuming the problem gave T as 75 and 35). If the 14 fluid oz is correct, then the correct answer will be 2.0 x 10^2 g ice required.
14 is correct. i tried the 2.0 X 102 and it was wrong. I have no idea what im doing wrong. This is a though question
I have checked my calculations and I stick by the 2.0 x 10^2 grams. I assume you are typing the answer into a database. If you leave out the caret (the ^) the think the database will not put it in for you. Type in 2.0 x 10^2.
in the database im using we have to put 2.0e2 i have tried that but to no evail.
Actually now that look at the prob. again. I think the 40 degrees F. has to be converted to degrees C. so i tried that which makes it 22.22 and put that instead of 40 and got 31.6 grams but it still said it was wrong
Sarah--I'm sorry I didn't catch that. Actually, I didn't catch two or three things. You are right, the F must be converted to C. Also I didn't catch that the ice, which melts at zero C must ALSO be heated to 35 F (converted to C) AND I didn't convert fluid ounces to grams correctly.
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