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Topic: General Change of State Queston  (Read 7357 times)

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Offline ThatGuy

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General Change of State Queston
« on: November 24, 2013, 07:47:02 AM »
Suppose that you have x amount of ice at -40 Celsius and mix it with y amount of water at 10 Celsius.

When you try to solve for the final temperature and use -q = +q ; how do you decide if the ice will become liquid water or the 10 Celsius water will become ice? How do you decide whether to use the heat of fusion on the -q side or use the opposite on the +q side?

Thank you for your help.

Offline Borek

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Re: General Change of State Queston
« Reply #1 on: November 24, 2013, 09:39:42 AM »
You have to check step by step. First, compare how much heat will be lost by water cooled down to zero with amount of heat required to heat up ice to zero. See where you are and what is the next possible step, and so on.
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #2 on: November 24, 2013, 09:41:36 AM »
You have to check step by step. First, compare how much heat will be lost by water cooled down to zero with amount of heat required to heat up ice to zero. See where you are and what is the next possible step, and so on.

What exactly do you mean by next possible step?

Offline ThatGuy

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Re: General Change of State Queston
« Reply #3 on: November 24, 2013, 10:01:08 AM »
For example:

100 grams of ice at -35 degrees Celsius is mixed 50 grams of water at 10 degrees Celsius. Determine the final temperature.

Liquid Water:

q=mct
q= (50)(4.184)(0.10) = -2092 Joules

Ice:

q=mct
q= (100)(2.05)(0-(35)) = 3587.5 Joules

What would I do next?

Offline Borek

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Re: General Change of State Queston
« Reply #4 on: November 24, 2013, 10:20:36 AM »
q=mct
q= (50)(4.184)(0.10) = -2092 Joules

OK, cooling water is capable of losing 2092 J (I assume you meant 0-10, not 0.10).

Quote
Ice:

q=mct
q= (100)(2.05)(0-(35)) = 3587.5 Joules

To warm up the ice you need  3587 J.

Is there enough heat from cooling water to heat up the ice to 0°C?
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #5 on: November 24, 2013, 10:37:00 AM »
q=mct
q= (50)(4.184)(0.10) = -2092 Joules

OK, cooling water is capable of losing 2092 J (I assume you meant 0-10, not 0.10).

Quote
Ice:

q=mct
q= (100)(2.05)(0-(35)) = 3587.5 Joules

To warm up the ice you need  3587 J.

Is there enough heat from cooling water to heat up the ice to 0°C?

No, there isn't enough heat from the cooling water to heat up the ice. Therefore, the liquid water will freeze:

-q=+q
-1((50)(4.184)(0-10) + (-334)(50) + (50)(2.05)(Tf - 0)) = (100)(2.05)(Tf-(-35))

-1 ( -2092 - 16700 + 102.5 Tf ) = 205Tf +7175

Tf = 37 degrees Celsius! I'm certain I did something wrong

Offline Borek

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Re: General Change of State Queston
« Reply #6 on: November 24, 2013, 10:54:47 AM »
At this stage you can't yet tell what the final temperature will be. It can be zero, it can be below zero

Calculate, how much heat will the water release while solidifying. Compare that to the amount of heat ice is still capable of absorbing before its temperature gets to zero.

How much water will freeze? All of it?
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #7 on: November 24, 2013, 11:07:35 AM »
At this stage you can't yet tell what the final temperature will be. It can be zero, it can be below zero

Calculate, how much heat will the water release while solidifying. Compare that to the amount of heat ice is still capable of absorbing before its temperature gets to zero.

How much water will freeze? All of it?

The bold part: heat of fusion = (50)(-334) = -16,700 Joules

The amount of heat that the ice is still able to absorb is 1495 Joules.

Offline Borek

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Re: General Change of State Queston
« Reply #8 on: November 24, 2013, 11:24:16 AM »
You are almost there.

Ice will absorb only part of the heat of solidifying (or negative heat of fusion), so only part of water will freeze (you can easily calculate how much - but now you don't have to). If not all water freeze, in the end you will have a mixture of ice and water. It can have only one temperature.
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #9 on: November 24, 2013, 11:33:48 AM »
You are almost there.

Ice will absorb only part of the heat of solidifying (or negative heat of fusion), so only part of water will freeze (you can easily calculate how much - but now you don't have to). If not all water freeze, in the end you will have a mixture of ice and water. It can have only one temperature.

The ice will only be able to absorb 1495 Joules. So the negative heat of fusion (-16700) plus the heat that the ice will absorb is 1495 Joules for a total of -15205 Joules.
« Last Edit: November 24, 2013, 03:31:00 PM by ThatGuy »

Offline ThatGuy

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Re: General Change of State Queston
« Reply #10 on: November 24, 2013, 03:49:55 PM »
How would I go about continuing from here?

Offline Borek

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Re: General Change of State Queston
« Reply #11 on: November 24, 2013, 04:25:10 PM »
You don't have to continue - you already know the answer.

Note: I have not checked your calculations, I was just following the numbers you listed.
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #12 on: November 24, 2013, 04:31:26 PM »
You don't have to continue - you already know the answer.

Note: I have not checked your calculations, I was just following the numbers you listed.

Yes, I noticed a blunder in one of my calculations. However, we still don't know the final temperature of the mixture, correct?

What if they asked how much of it was ice and how much was water?

Offline Borek

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Re: General Change of State Queston
« Reply #13 on: November 24, 2013, 04:39:05 PM »
We know the temperature.

To what temperature was water cooled down?

At the moment water was cooled down, ice temperature was still a little bit lower. It got warmed up freezing the water - but it could not warm up above the freezing point.
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #14 on: November 24, 2013, 04:45:25 PM »
Ok; so these are my calculations:

The Energy in the Ice: to reach melting point

q = (100)(2.05)(0-(-35))
   = 7175 Joules

To Get Water to the melting point:

q= (50)(4.184)(0-10)
  = -2092 Joules

So, the 2092 Joules were used to heat up the ice

The change in temp for the ice:

(7175 - 2092) = (100)(2.05)(change in temp)

Change in temp = 24.8 Celsius --> Temp of ice is -10.2 Celsius

The Q to freeze ice:

q= (334)(50) = 16700 Joules

Not all the water can freeze; since the ice only has 5083 Joules "left"

5083 = 334 * mass ---> mass = 15.22 grams

Am I doing this correctly up until this point? Thank you for your time.

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