# Chemical Forums

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

Title: Basic Phase change problem help
Post 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?

(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
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 30, 2009, 07:00:13 PM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 30, 2009, 08:20:39 PM
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 30, 2009, 08:32:01 PM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 30, 2009, 08:56:20 PM
Would it be (.5cal/g)(100.0g)(65.0 C -10.0 C)?  It came out too 3240
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 30, 2009, 09:04:17 PM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 30, 2009, 09:12:44 PM
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
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 30, 2009, 09:16:17 PM

Dear reezyfbaby;

Bravo! -  Correct!
Now whats the final value for c.)?

Good Luck!
ARGOS++

Title: Re: Basic Phase change problem help
Post by: mr cool on January 30, 2009, 09:17:42 PM
I wound up with 17740 calories...

Thanks for the help man.
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 30, 2009, 09:22:22 PM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 30, 2009, 09:28:12 PM
Ok now i got ya... It is a lot easier if you just take it step by step. I got 15000 cal
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 30, 2009, 09:31:13 PM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 30, 2009, 09:52:22 PM
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.
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 30, 2009, 09:55:52 PM

Dear reezyfbaby;

Congratulation!,   -  Exact result!

Good Luck!
ARGOS++

Title: Re: Basic Phase change problem help
Post by: mr cool on January 30, 2009, 09:59:54 PM
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
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 31, 2009, 05:37:26 AM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 31, 2009, 04:02:40 PM
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 -25.0oC ice in your mouth to body temperature of 37.0oC?

Here is my stab at it....

(1.0 cal/g)(6.0g)(37oC - 0oC) = 2220 cal.
(80 cal/g)(6.0g) = 480
(.5 cal/g)(6.0g)(0oC - -25oC) = 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.
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 31, 2009, 04:21:12 PM

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++

Title: Re: Basic Phase change problem help
Post by: mr cool on January 31, 2009, 04:38:17 PM
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?
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 31, 2009, 04:45:47 PM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 31, 2009, 05:41:34 PM
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 38.0oC.  What mass of -25.00oC 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)(25.0oC)(M)+(80 cal/g)(M)

for this problem i used M as a place holder for mass.
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 31, 2009, 05:50:48 PM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 31, 2009, 06:09:14 PM
here is my second try...

160000 calories = (.5 cal/g)(25.0oC)(M)+(80 cal/g)(M) + (1.0 cal/g)(M)(38.0oC - 0oC) = 103.5 g of ice required?
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 31, 2009, 06:21:36 PM

Dear reezyfbaby;

The values in the equation are correct, but I get more than 1.035 kg Ice.

Good Luck!
ARGOS++

Title: Re: Basic Phase change problem help
Post by: mr cool on January 31, 2009, 06:33:55 PM

Dear reezyfbaby;

The values in the equation are correct, but I get more than 1.035 kg Ice.

Good Luck!
ARGOS++

i got 103.5 not 1.035
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 31, 2009, 06:49:38 PM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 31, 2009, 07:13:05 PM
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.
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 31, 2009, 07:17:31 PM
Title: Re: Basic Phase change problem help
Post by: mr cool on January 31, 2009, 07:28:11 PM
Do the sig figs come out to 3 or 4?
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 31, 2009, 07:31:48 PM

Dear reezyfbaby;

Of course NOT!, because the limited precision.
But you wrote:  gram ice instead of kg ice.

Good Luck!
ARGOS++

Title: Re: Basic Phase change problem help
Post by: mr cool on January 31, 2009, 07:37:38 PM
What was the (6) for in 1.22(6)
Title: Re: Basic Phase change problem help
Post by: ARGOS++ on January 31, 2009, 07:43:29 PM