Do I have enough information to solve this density problem?

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The discussion revolves around a density problem involving the freezing of water in a soft-drink bottle. Participants clarify that the mass of the water remains constant at 1495.5 g, regardless of the phase change to ice. The correct volume of ice can be calculated using the density of ice at -10 degrees Celsius, leading to a final volume of approximately 1630.862 mL. There is confusion regarding a potential typo in the problem statement, particularly concerning the mention of ice at 10 degrees Celsius. Ultimately, the problem emphasizes the need to extract and apply given data effectively to arrive at a solution.
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Homework Statement


"If a soft-drink bottle whose volume is 1.50 L is completely filled with water and then frozen to -10 degrees Celsius, what volume does the ice occupy? Water has a density of 0.997 g/cm^3 at 25 degrees Celsius; ice has a density of 0.917 g/cm^3 at -10 degrees Celsius."

Homework Equations


D = m/V
D = m/1500 mL
m = x(0.997) + y(0.917)
x + y = 1500 mL
x = 1500 - y

The Attempt at a Solution


m = (1500 - y)(0.997) + y(0.917)
m = 1495.5 - y0.997 + y0.917
m = 1495.5 - y0.08

I feel like I would need the mass of the bottle's contents or its density in order to proceed. Am I plugging in my numbers wrong?
 
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Question is rather lousy if you ask me. I guess they want you to assume bottle is either elastic enough to expand, or it simply breaks (and all water is converted to ice).
 
I agree; my chemistry professor doesn't seem to know what she's talking about.

But do I have enough information to solve for the volume of the ice?
 
Yes. Mass of water doesn't change.
 
But I don't know what the mass of the water is...
 
You have enough data to calculate it.
 
Eclair_de_XII said:
But I don't know what the mass of the water is...

Re-read the problem statement carefully:

"If a soft-drink bottle whose volume is 1.50 L is completely filled with water and then frozen to -10 degrees Celsius, what volume does the ice occupy? Water has a density of 0.997 g/cm^3 at 25 degrees Celsius; ice has a density of 0.917 g/cm^3 at 10 degrees Celsius."

Some problems will not spoon-feed you all the data you need for solution. You have to do a little work to find it.

It's not clear how the ice exists at a temperature of 10 degrees Celsius, unless this is a typo which occurred when copying the problem statement.
 
If that's the case, then wouldn't the ice take up the whole bottle since it can't be two different temperatures?
 
Eclair_de_XII said:
If that's the case, then wouldn't the ice take up the whole bottle since it can't be two different temperatures?
That wasn't the point I was trying to make, but now that you asked it ...

What happens to ice when the temperature is greater than 0 degrees Celsius?

(Hint: What is the freezing point of water?)
 
  • #10
The freezing point of water is 0 degrees Celsius, isn't it?

If it's greater, then it starts to melt?
 
  • #11
Eclair_de_XII said:
The freezing point of water is 0 degrees Celsius, isn't it?

If it's greater, then it starts to melt?
Yes. So the density of ice at 10 degrees Celsius seems to cover an impossible situation.
 
  • #12
SteamKing said:
It's not clear how the ice exists at a temperature of 10 degrees Celsius, unless this is a typo which occurred when copying the problem statement.

I'm sorry for not pointing this out earlier, but this is correct.
 
  • #13
Eclair_de_XII said:
If that's the case, then wouldn't the ice take up the whole bottle since it can't be two different temperatures?

Please reread my first post in the thread. Final volume of the ice is not 1.5L.
 
  • #14
Okay, 1.50 L of water (at 25 degrees Celsius) is poured into water. It has a mass of m = 1500(0.997) = 1495.5 g. This mass does not change, even when the water is frozen. However, its volume of the water does change as it turns to ice? Given that the density of ice is 0.917 g/mL, I can calculate the volume from that and the mass.

m = 1495.5
1495.5 = V(0.917)
V = 1495.5/0.917
V = 1630.862 mL
 
  • #15
And that would be my approach to solving the question.

Watch your significant figures.
 
  • #16
Okay, thanks.
 
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