Ice Melts: Examining the ΔU, w and q of 0oC Water

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In summary, the first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the thermal energy supplied (q) plus the work done (w) on the system. In the case of ice melting at 0°C, the separation between molecules decreases, resulting in a positive value for work (w). However, there is no change in temperature, meaning no change in kinetic energy (KE) of the molecules, but a decrease in potential energy (PE). This results in a negative value for ΔU. According to the first law, q should also be negative to balance out the negative ΔU and positive w. However, the answer key says that both ΔU
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songoku
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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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songoku said:
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.

songoku said:
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?
songoku said:
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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  • #3
DrClaude said:
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
songoku said:
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.

songoku said:
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?"

songoku said:
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?).

songoku said:
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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  • #5
DrClaude said:
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
songoku said:
So the proper answer will be all three variables are positive?
Yes.
 
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  • #7
Thank you very much
 

What is the ΔU of 0°C water?

The ΔU, or change in internal energy, of 0°C water is 0. This is because at this temperature, water is in a state of equilibrium and there is no change in internal energy.

What is the w of 0°C water?

The w, or work done, of 0°C water is also 0. This is because there is no change in volume or pressure at this temperature, so no work is being done.

What is the q of 0°C water?

The q, or heat transferred, of 0°C water is also 0. This is because at this temperature, there is no change in energy being transferred into or out of the system.

Why is 0°C used as a reference point for examining ice melts?

0°C is used as a reference point because it is the temperature at which water transitions from a solid to a liquid state. At this temperature, the ice is melting and there is a change in the substance's physical state, making it a useful point for analysis.

How do the values of ΔU, w, and q change as the temperature of water increases from 0°C?

As the temperature of water increases from 0°C, the values of ΔU, w, and q will also increase. This is because the internal energy, work done, and heat transferred are all dependent on temperature. As the temperature increases, the energy and work being transferred will also increase.

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