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### Topic: Basic Phase change problem help  (Read 19495 times)

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#### ARGOS++

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##### Re: Basic Phase change problem help
« Reply #15 on: January 31, 2009, 05:37:26 AM »

#### mr cool

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##### Re: Basic Phase change problem help
« Reply #16 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.

#### ARGOS++

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##### Re: Basic Phase change problem help
« Reply #17 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++

#### mr cool

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##### Re: Basic Phase change problem help
« Reply #18 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?

#### ARGOS++

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##### Re: Basic Phase change problem help
« Reply #19 on: January 31, 2009, 04:45:47 PM »

#### mr cool

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##### Re: Basic Phase change problem help
« Reply #20 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.

#### ARGOS++

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##### Re: Basic Phase change problem help
« Reply #21 on: January 31, 2009, 05:50:48 PM »

#### mr cool

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##### Re: Basic Phase change problem help
« Reply #22 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?

#### ARGOS++

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##### Re: Basic Phase change problem help
« Reply #23 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.
Please calculate once more.

Good Luck!
ARGOS++

#### mr cool

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##### Re: Basic Phase change problem help
« Reply #24 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.
Please calculate once more.

Good Luck!
ARGOS++

i got 103.5 not 1.035

#### ARGOS++

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##### Re: Basic Phase change problem help
« Reply #25 on: January 31, 2009, 06:49:38 PM »

#### mr cool

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##### Re: Basic Phase change problem help
« Reply #26 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.

#### ARGOS++

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##### Re: Basic Phase change problem help
« Reply #27 on: January 31, 2009, 07:17:31 PM »
« Last Edit: January 31, 2009, 07:28:57 PM by ARGOS++ »

#### mr cool

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##### Re: Basic Phase change problem help
« Reply #28 on: January 31, 2009, 07:28:11 PM »
Do the sig figs come out to 3 or 4?

#### ARGOS++

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##### Re: Basic Phase change problem help
« Reply #29 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++