First Law of Thermodynamics -- Are ΔU, w and q positive, negative or zero when ice melts at 0oC

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Homework Statement


The first law of thermodynamics may be expressed in the form: ΔU = q + w, where ΔU is the increase in internal energy, q is the thermal energy supplied to the system, w is the work done on the system.

Is ΔU, w and q positive, negative or zero when ice melts at 0oC to give water at 0oC (note: ice is less dense than water)?

Homework Equations


ΔU = q + w

ΔU = KE + PE


The Attempt at a Solution


When ice melts to give water at 0oC, the separation between molecules becomes smaller so the density of water will be bigger than ice. This will result in positive value of w.

No change in temperature means no change in KE of molecules but smaller separation between molecules means that the PE decreases so ΔU will be negative.

Based on first law of thermodynamics, the value of q should be negative because ΔU is negative and w is positive.

But the answer key says that ΔU and q are positive and w = 0.

Where did I go wrong? I am also confused why I got negative value of q. Ice turns to water requires heating so maybe q should be positive but it doesn't match the math.

Thanks
 
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Answers and Replies

  • #2
DrClaude
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No change in temperature means no change in KE of molecules
That would be only true for an ideal monatomic gas. Equating temperature with kinetic energy is only an acceptable when translation is the only way energy is stored.

but smaller separation between molecules means that the PE decreases so ΔU will be negative.
Careful here. You are separating molecules that used to stick together. Does that increase or decrease their potential energy?


But the answer key says that ΔU and q are positive and w = 0.
I am surprised by the ##w=0##, since your analysis of the change of volume was correct, and the problem notes that there is a change in density. But it is true that ##|w| \ll |q|##.
 
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That would be only true for an ideal monatomic gas. Equating temperature with kinetic energy is only an acceptable when translation is the only way energy is stored.
So in this case I have to look at the speed of thr molecules. When it changes from ice to water, the molecules can move more freely so the speed and KE increases?

Careful here. You are separating molecules that used to stick together. Does that increase or decrease their potential energy?
I try to consider the note given from the question: ice is less dense than water, so I think the molecules become closer not further when changing from ice to water.

So the distance between molecules should increase from ice to water and the thing about ice is less dense than water is related to other factor (not because the distance between molecules decreases)?

I am surprised by the ##w=0##, since your analysis of the change of volume was correct, and the problem notes that there is a change in density. But it is true that ##|w| \ll |q|##.
How can we determine |w| << |q|? Is it because the heat supplied will be bigger compared to change in volume that causes the work done?

Other question: can I directly answer that the q will be positive for process involving heating and negative for process involving cooling?

Thanks
 
  • #4
DrClaude
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So in this case I have to look at the speed of thr molecules. When it changes from ice to water, the molecules can move more freely so the speed and KE increases?
Yes.

I try to consider the note given from the question: ice is less dense than water, so I think the molecules become closer not further when changing from ice to water.
Yes, the average distance between molecules is smaller in water, but the interaction between the molecules is different. But that is not necessarily knowledge that you are expected to have.

I think that the question should be answered with what you are expected to know about the phenomenon, like "What do you have to do to make an ice cube melt?"

How can we determine |w| << |q|?
You can only figure that out with actual numbers. Knowing the change in density of water at 0 °C and the enthalpy of fusion, you can figure it out. But this is again not something that you would have been expected to know (I guess?).

Other question: can I directly answer that the q will be positive for process involving heating and negative for process involving cooling?
Yes, as I said above, that's the kind of knowledge you are expected to have.
 
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Yes, the average distance between molecules is smaller in water, but the interaction between the molecules is different. But that is not necessarily knowledge that you are expected to have.

I think that the question should be answered with what you are expected to know about the phenomenon, like "What do you have to do to make an ice cube melt?"
So I can say the note about density is for finding work, not about the distance between molecules will decrease?

To make ice melt, we need to heat it so this means heat supplied into the system (q positive) and the distance between molecules becomes larger, hence PE increases

You can only figure that out with actual numbers. Knowing the change in density of water at 0 °C and the enthalpy of fusion, you can figure it out. But this is again not something that you would have been expected to know (I guess?).
Nah, I don't think so :biggrin:

So the proper answer will be all three variables are positive?

Thanks
 
  • #6
DrClaude
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So the proper answer will be all three variables are positive?
Yes.
 
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Thank you very much
 

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