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

AI Thread Summary
The discussion centers on whether the internal energy and entropy of a system comprising warm water and melting ice change when the process occurs in an insulated box. Participants debate the implications of the first law of thermodynamics, concluding that since there is no heat transfer (Q=0) and minimal work done, the internal energy of the combined system remains unchanged. They also discuss the entropy changes, with some asserting that the melting ice indicates a change in entropy despite the overall energy balance. The consensus leans towards the understanding that while the ice melts, the sealed nature of the box limits significant volume change and work done, ultimately leading to no change in internal energy for the entire system. The complexities of analyzing the ice and water as separate systems are also acknowledged, but the main conclusion remains focused on the overall system's energy and entropy balance.
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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