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Topic: General Change of State Queston  (Read 7363 times)

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Offline Borek

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Re: General Change of State Queston
« Reply #15 on: November 24, 2013, 05:01:50 PM »
The change in temp for the ice:

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

No, so far exactly 2092 J were used, 7175-2092 is what is "left in the ice".
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #16 on: November 24, 2013, 05:13:55 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:

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

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

I have fixed the bold part. Now, in the mixture, there is some ice at -24.8 Celsius and some water at 0 Celsius.

The final temp:

Offline Borek

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Re: General Change of State Queston
« Reply #17 on: November 24, 2013, 05:37:11 PM »
Think what will happen now. Some water will freeze. All of it? What is the temperature of water left? Up to what temperature does the ice heat up during freezing of the water?
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #18 on: November 24, 2013, 05:42:10 PM »
Think what will happen now. Some water will freeze. All of it? What is the temperature of water left? Up to what temperature does the ice heat up during freezing of the water?

So, there are 34.78 grams of water at 0 Celsius and 115.22 grams of ice at -24.8 Celsius.


Offline Borek

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Re: General Change of State Queston
« Reply #19 on: November 24, 2013, 06:18:11 PM »
115.22 grams of ice at -24.8 Celsius.

You are probably right about mass (again: I am not checking numbers). But why at -24.8°C? Ice from freezing water has exactly the same temperature water had (its temperature doesn't change, it just solidifies), ice that was initially much colder warmed up during water freezing. To what temperature dead it heat up?
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #20 on: November 24, 2013, 06:22:13 PM »
115.22 grams of ice at -24.8 Celsius.

You are probably right about mass (again: I am not checking numbers). But why at -24.8°C? Ice from freezing water has exactly the same temperature water had (its temperature doesn't change, it just solidifies), ice that was initially much colder warmed up during water freezing. To what temperature dead it heat up?

Isn't that what I calculated in the "change in temp calculation"? Or am I missing something?

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Re: General Change of State Queston
« Reply #21 on: November 25, 2013, 03:44:35 AM »
I am not sure which part of your calculations you refer to, but yes, it still seems as if you were missing a thing.

You were perfectly OK up to the "100g of ice at -24.8°C and 50 g of water at 0°C", but this is where you got confused.

Ice will be warming up and the water will freeze. You have already checked that there is not enough ice to absorb all the heat water is capable of loosing while freezing. You have already correctly calculated mass of water that will freeze.

For some reason you are still not seeing that after the freezing ends you have some water left (water at 0°C), some freshly created ice (at 0°C), and 100 g of old ice, that was heat up to 0°C. As everything now has identical temperature no further changes are possible, and the system is in the thermal equilibrium.
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Offline ThatGuy

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Re: General Change of State Queston
« Reply #22 on: November 25, 2013, 07:00:53 AM »
I am not sure which part of your calculations you refer to, but yes, it still seems as if you were missing a thing.

You were perfectly OK up to the "100g of ice at -24.8°C and 50 g of water at 0°C", but this is where you got confused.

Ice will be warming up and the water will freeze. You have already checked that there is not enough ice to absorb all the heat water is capable of loosing while freezing. You have already correctly calculated mass of water that will freeze.

For some reason you are still not seeing that after the freezing ends you have some water left (water at 0°C), some freshly created ice (at 0°C), and 100 g of old ice, that was heat up to 0°C. As everything now has identical temperature no further changes are possible, and the system is in the thermal equilibrium.

Thank you. I greatly appreciate you taking the time to assist me.

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