ok i know i posted this before but it got totally confusing and its late and its due tom.
When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 14.0 fluid ounce drink from 75°F to 35°F, if the heat capacity of the drink is 4.18 J/g·°C? (Assume that the heat transfer is 100% efficient.)
i got 212.56 and rounded it to 213 but it said it was wrong. i need help!
9 answers
Show your work and I'll find the error.
ok.. i got
(miX330j/g)+(396.893X4.148X+40)+0
70145.28+330j/g(mi)
70145.28=330j/g(mi)
70145.28/330
mi=212.56
(miX330j/g)+(396.893X4.148X+40)+0
70145.28+330j/g(mi)
70145.28=330j/g(mi)
70145.28/330
mi=212.56
Three errors and two of them compensated for each other.
First,
(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.
First,
(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.
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
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.