How can I find the composition of a mixture between ice and water when an equilibrium is reached?

Jun 2017
345
6
Lima, Peru
The problem is as follows:

A flask whose heat capacity is negligible is filled with $20$ grams of ice at $-80^{\circ}C$ and $7$ grams of water at $80^{\circ}C$. Find the composition of the mixture in the flask when the equilibrium is reached.

The alternatives given are as follows:

$\begin{array}{ll}
1.&\textrm{2 g. of water and 25 g. of ice}\\
2.&\textrm{5 g. of water and 22 g. of ice}\\
3.&\textrm{7 g. of water and 20 g. of ice}\\
4.&\textrm{3 g. of water and 21 g. of ice}\\
5.&\textrm{4 g. of water and 23 g. of ice}\\
\end{array}$

For this problem I'm not sure how to proceed. What I've attempted to do was to assume that the warmer water will lose heat and give it to the ice so it will melt. But I don't know exactly how to translate this into an equation.

So what I did was to add the heat to warm up the ice and melt the ice with that of the cooling of the water.

$mL_{f}=q_{water}$

$80m=7\left(1\frac{cal}{g^{\circ}C}\right)\times 80$

$m=7$

But I don't know exactly if this is the right way to approach this problem. Can someone help me with what would be the right concept?
 
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mathman

Forum Staff
May 2007
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762
Do you know the heat capacity of ice?
 
Jul 2008
5,233
52
Western Canada
My thermodynamics is a bit rusty, but I think the solution of the problem will need to include the latent heat of fusion of water, which is 334 J/g (79.77 c/g). So, for every gram of ice to be converted to water you must transfer 334 J from the liquid water to the ice, in addition to the heat that is transferred to equalize temperature. This continues until the temperatures of both the solid and liquid reach the same temperature at which point no further energy transfer occurs. Obviously, if the final equilibrium temperature turns out not to be 0° then you'll have either all ice or all water.

(Also, one of the answer choices can be ruled out immediately.)
 
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Jun 2017
345
6
Lima, Peru
My thermodynamics is a bit rusty, but I think the solution of the problem will need to include the latent heat of fusion of water, which is 334 J/g (79.77 c/g). So, for every gram of ice to be converted to water you must transfer 334 J from the liquid water to the ice, in addition to the heat that is transferred to equalize temperature. This continues until the temperatures of both the solid and liquid reach the same temperature at which point no further energy transfer occurs. Obviously, if the final equilibrium temperature turns out not to be 0° then you'll have either all ice or all water.

(Also, one of the answer choices can be ruled out immediately.)
Can you develop the heat balance from those words into an equation?.

So far I'm getting that the equation for this process would be

$mc\Delta T + mL{f}=mc\Delta T$

But I don't know exactly what mass should be put into those equations. Help, please?
 
Jul 2008
5,233
52
Western Canada
The heat capacity of ice, is not the same as the heat capacity of liquid water. It is almost exactly half the value of liquid water, or 0.5 cal/°C/g.

Assume for now that the final equilibrium temperature is 0°C. That assumption will be verified later. So, to reach 0°, the liquid water must give up the following amount of heat:
$Q_W = 7\; g \times 80^\circ C \times 1 \; cal/^\circ C/g = 560\; cal$
And the ice must acquire this amount of heat:
$Q_I = 20 \;g \times 80^\circ C \times 0.5 \; cal/^\circ C/g = 800\; cal$
That's a difference of 240 cal. If the liquid water didn't freeze, then the final temperature would be below 0°. However, the liquid water will continue to release heat causing some of it to freeze. Since the latent heat of fusion is 79.77 cal/g, the amount of water that will freeze is equal to the heat difference divided by the heat of fusion. So that the mass of water converted to ice is
$m = \dfrac {240 \; cal }{ 79.77 \; cal/g} = 3.01\; g$
This is less than the total amount of available liquid water, so we can conclude that the final temperature does not go below 0°C, and the assumption that the final temperature is 0°C is valid. (If all of the ice was converted to water, or all of the water was converted to ice, then the final temperature would not be 0°C.)
 
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Jul 2008
5,233
52
Western Canada
And of course, I should have used the word "enthalpy" instead of "heat" (I did say that my Thermodynamics is a bit rusty).
 
Jun 2017
345
6
Lima, Peru
The heat capacity of ice, is not the same as the heat capacity of liquid water. It is almost exactly half the value of liquid water, or 0.5 cal/°C/g.

Assume for now that the final equilibrium temperature is 0°C. That assumption will be verified later. So, to reach 0°, the liquid water must give up the following amount of heat:
$Q_W = 7\; g \times 80^\circ C \times 1 \; cal/^\circ C/g = 560\; cal$
And the ice must acquire this amount of heat:
$Q_I = 20 \;g \times 80^\circ C \times 0.5 \; cal/^\circ C/g = 800\; cal$
That's a difference of 240 cal. If the liquid water didn't freeze, then the final temperature would be below 0°. However, the liquid water will continue to release heat causing some of it to freeze. Since the latent heat of fusion is 79.77 cal/g, the amount of water that will freeze is equal to the heat difference divided by the heat of fusion. So that the mass of water converted to ice is
$m = \dfrac {240 \; cal }{ 79.77 \; cal/g} = 3.01\; g$
This is less than the total amount of available liquid water, so we can conclude that the final temperature does not go below 0°C, and the assumption that the final temperature is 0°C is valid. (If all of the ice was converted to water, or all of the water was converted to ice, then the final temperature would not be 0°C.)
I was almost there to your assumption regarding the difference between those heats but what it doesn't seem yet clear to me is this sentence.

the amount of water that will freeze is equal to the heat difference divided by the heat of fusion
Why is that?. How can I justify this?. Can this be seen in a graph or anything?. Because I'm stuck there.

I believe that if the amount of water that is converted to ice is $3\,g$ then this should be added to the total mass of ice which is at the beginning so:

$20+3=23\,g$

and because this is lost by the water then:

$7-3=4\,g$

Therefore the answer would be the fifth option which it checks with my answers sheet. But I'm still not very convinced about the statement which you mentioned regarding the difference between those heats. Can you explain this with more details?. Btw yes I understood that formaly enthaly would be more accurate.
 
Jul 2008
5,233
52
Western Canada
Why is that?. How can I justify this?. Can this be seen in a graph or anything?. Because I'm stuck there.
If you draw a graph of temperature vs. enthalpy for water, with temperature along the vertical axis and where the temperature covers a range from negative to positive values, you'll get an inclined straight line with a positive slope from the starting negative value until it hits zero degrees. At that point the line will become horizontal as the water continues to absorb heat and converts from solid to liquid, but remains at zero degrees while this is happening. Once all of the water is converted to liquid, then the graph again changes to an inclined line with positive slope as the enthalpy increases further. The horizontal section on the graph represents the latent heat of fusion. It explains why water can exist indefinitely as either a solid or a liquid at exactly the same temperature.TempEnthalpy.png
 
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