Does the internal energy of the combined system change when ice melts?

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Homework Help Overview

The discussion revolves around the internal energy and entropy changes when warm water is mixed with ice in an insulated box, specifically questioning whether the internal energy of the combined system changes during the melting of ice.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the relationship between internal energy, heat transfer, and work in the context of an insulated system. Some express confusion regarding the implications of melting ice on internal energy and volume changes, while others question the assumptions about heat transfer and work done.

Discussion Status

There is active engagement with differing viewpoints on whether the internal energy of the system changes, with some participants asserting that no heat transfer occurs, while others argue that work is done due to volume changes. The discussion reflects a lack of consensus on these points, with participants seeking clarity on the underlying principles.

Contextual Notes

Participants note that the process occurs in an insulated box, which raises questions about heat transfer and the implications for internal energy changes. There is also mention of the need to consider the system as a whole versus separate systems.

JoshMG
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PROBLEM:
Some warm water is mixed with ice and the ice melts. Assume that the entire process happens in an insulated box.

Does the internal energy of the combined (ice+water) system change?
Does the entropy of the combined system change?

Considering that the ice and water as separate system, discuss the change in internal energy and in entropy for each.


ATTEMPT:

I know that internal energy is dependent of change in temperature, there is a change in internal energy.

So in this case, I am assuming, it is an isobaric process so the volume is changing. So there is a change in entropy.



Separate systems: also change in internal energy. Again, change in entropy.


There is something definitely wrong with my answer. WHAT IS IT?! I'm going crazy!
 
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Change in Internal Energy=Work+Heat Transfer

You say for the entire system there is a change... how? Where is the work? Where is the heat transfer?

Also, this should be in Introductory Physics
 
PROBLEM:
Some warm water is mixed with ice and the ice melts. Assume that the entire process happens in an insulated box.

Does the internal energy of the combined (ice+water) system change?
Does the entropy of the combined system change?

Considering that the ice and water as separate system, discuss the change in internal energy and in entropy for each.

ATTEMPT:

I believe this is a conceptual question, so this is what I think:

I know that internal energy is dependent of change in temperature, there is a change in internal energy.

So in this case, I am assuming, it is an isobaric process so the volume is changing. So there is a change in entropy.



Separate systems: also change in internal energy. Again, change in entropy.


There is something definitely wrong with my answer. WHAT IS IT?! I'm going crazy!
 
Did you read my post?!

If it happens in an insulated box there is no heat transfer. The total volume of the system cannot increase It is contained in a box.

\Delta U=Q-W For the entire system, both Q and W are zero, so \Delta U_{sys}=0
 
I don't agree. This is what I think:

I do agree that there is no heat transfer, so Q=0. But the prompt does say that the ice melts, therefore there is a change in volume and since there is a change in volume, we have work done by the system. Because there is work done, we have a change in internal energy.

I just don't understand what difference it would make if we looked at the ice and water as different systems.
 
JoshMG said:
I don't agree. This is what I think:

I do agree that there is no heat transfer, so Q=0. But the prompt does say that the ice melts, therefore there is a change in volume and since there is a change in volume, we have work done by the system. Because there is work done, we have a change in internal energy.

I just don't understand what difference it would make if we looked at the ice and water as different systems.
You are asked whether there is a change in energy of the combined water and ice in the container (we will call that the system). There is no addition of energy from the surroundings to the system. There is no removal of energy from the system to its surroundings.. So Q = 0.

Apply the first law. Unless there is work done on or by the system, there can be no change in internal energy.

I don't see where there is any significant work done here. If it was air-tight, there would be a slight increase in the volume of air/vapour as the ice melted. But it would be small. I don't think the question asks you to take that into account.

AM
 
Last edited:
JoshMG said:
I don't agree. This is what I think:

I do agree that there is no heat transfer, so Q=0. But the prompt does say that the ice melts, therefore there is a change in volume and since there is a change in volume, we have work done by the system. Because there is work done, we have a change in internal energy.

I just don't understand what difference it would make if we looked at the ice and water as different systems.

Just because the Ice melts does not mean there is a volume change. The box is sealed, so for the entire system there cannot be a change in volume.
 

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