# Homework Help: Entropy and internal energy

1. Dec 7, 2008

### JoshMG

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!

2. Dec 7, 2008

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

3. Dec 7, 2008

### JoshMG

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!

4. Dec 8, 2008

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 $\Delta U_{sys}=0$

5. Dec 9, 2008

### JoshMG

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.

6. Dec 9, 2008

### Andrew Mason

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: Dec 9, 2008
7. Dec 9, 2008