Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: ThatGuy on November 24, 2013, 07:47:02 AM
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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.
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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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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?
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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?
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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).
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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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).
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
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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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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.
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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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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.
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How would I go about continuing from here?
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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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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?
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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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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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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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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:
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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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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.
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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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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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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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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.