Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: mr cool on January 30, 2009, 06:08:26 PM

alright so im doing phase change problems and this is the question... How many calories must be released to convert 100.0 grams of summer lake water at 65.0 C to winter ice at 10.0?
im really stuck on where to start then pretty much everything after that. i hate phase change problems
also can someone tell me if my work on this problem is correct??
How many calories are required to convert 45.00 grams of liquid water at 30.0*C to steam at 110.0*C?
here is my work/answer...
(80cal/g)(45.00g)+(80cal/g*C)(30.0*C)+(45.00g)(540cal/g) and i got 30300 calories.
ps.. the * is my makeshift degrees symbol for now.
thanks


this is impossible i have clue what numbers to start with


Would it be (.5cal/g)(100.0g)(65.0 C 10.0 C)? It came out too 3240


Would i use heat of formation to do this? Such as 80 cal/g? If so do i have to make it like 80 cal/g or something?
Does this work?.
(80.0 cal/g)(100.0g) = 8000 cal.?
This is for b.) by the way

Dear reezyfbaby;
Bravo!  Correct!
Now whats the final value for c.)?
Good Luck!
ARGOS^{++}

I wound up with 17740 calories...
Thanks for the help man.


Ok now i got ya... It is a lot easier if you just take it step by step. I got 15000 cal


Oki doki, here is a shot at my other problem,
A.)(1.0 cal/g)(45.00)(100*C  30*C)=3150 I did 100  30 to get up too 100*C
B.)(540 cal/g)(45.00) = 24300 im not sure if i did B.) correct
C.)(.5 cal/g)(45.00)(0+10) = 225
Total.) 27675 calories.
Im not positive but i think i did alrght other than step number 2.
Anyone feel like seeing if i did it correct? Thanks.

Dear reezyfbaby;
Congratulation!,  Exact result!
Good Luck!
ARGOS^{++}

Thanks for the help man, i'm going to hit the hay and maybe put up some more tries at my problems tomorrow. Thanks again


Here is a new problem that i just tried..
Melting ice in your mouth requires a lot of heat energy, this is why it cools you down. How many calories will your body burn if you melt 6.0 grams of 25.0^{o}C ice in your mouth to body temperature of 37.0^{o}C?
Here is my stab at it....
(1.0 cal/g)(6.0g)(37^{o}C  0^{o}C) = 2220 cal.
(80 cal/g)(6.0g) = 480
(.5 cal/g)(6.0g)(0^{o}C  ^{}25^{o}C) = 75 cal
Total = 2775 calories needed
The wording of this problem threw me off a bit so i didn't know if it was an exothermic or endothermic process. i think "this is why it cools you down" gives it away though, yes? I think i did it as exothermic.

Dear reezyfbaby;
How much is 1 * 6 * 37?
And think also about that you have to spend the heat, so the process is consuming heat.
Good Luck!
ARGOS^{++}

oh my bad, you are right it should be 222 not 2220, i must have hit some bad keys on my calculator.
was the problem correct though?
and isn't it exothermic because the ice is taking energy (heat) from the mouth?


Alright i think this is the last problem i have....
A serving of Doritos contains 160.0 Kilo calories (big C's). Your mouth is at body temperature 38.0^{o}C. What mass of 25.00^{o}C ice do you have to eat to burn all the calories in one serving of Doritos.
Do i have to convert the 160.0 Kilo calories into regular calories. 160.0 Kilo calories is 160000 calories.
so far i have..
160000 calories = (.5 cal/g)(25.0^{o}C)(M)+(80 cal/g)(M)
for this problem i used M as a place holder for mass.


here is my second try...
160000 calories = (.5 cal/g)(25.0oC)(M)+(80 cal/g)(M) + (1.0 cal/g)(M)(38.0^{o}C  0^{o}C) = 103.5 g of ice required?

Dear reezyfbaby;
The values in the equation are correct, but I get more than 1.035 kg Ice.
Please calculate once more.
Good Luck!
ARGOS^{++}

Dear reezyfbaby;
The values in the equation are correct, but I get more than 1.035 kg Ice.
Please calculate once more.
Good Luck!
ARGOS^{++}
i got 103.5 not 1.035


Yea i got 130.5 too. I wrote the wrong number off the calculator lol. I wound up with 1.22 g of ice needed.


Do the sig figs come out to 3 or 4?

Dear reezyfbaby;
Of course NOT!, because the limited precision.
But you wrote: gram ice instead of kg ice.
Good Luck!
ARGOS^{++}

What was the (6) for in 1.22(6)
