# Work during melting

1. Jan 7, 2008

### Distr0

if ice is heated to say 20C so it melts (done so slowly) is work done during the melting process equal to the latent heat? (I'm assuming there are no volume or pressure changes)

just a small thing i cant recall or find a quick answer for right now.

2. Jan 7, 2008

### rohanprabhu

If you heat it from $0^o \to 20^o$, what u do is basically, melt it first and then heat the liquid ice [or.. water] to $20^oC$. so basically, the work done would be the latent heat (required to melt it) + the heat required take it's temp. to $20^oC$...

$$W = mL_m + mC\delta t$$

Where..
$L_m$ is the latent heat of fusion,
$m$ is the mass of the liquid,
$C$ is the Specific heat of water,
$\delta t$ is the change in temperature

Note however that we assumed that 'C' remains constant over $\delta t$. This, in practical sense is not true. 'C' changes with temperature, but for our purpose, we can neglect that change.

If you take that ice from say, $-10^o \to 20^o$, the work done will be the work done previously + the work done to increase temperature of the ice from $-10^o \to 0^o$.

3. Jan 7, 2008

### Distr0

ok cheers and yes i've approached the problem in 3 stages for this.

work from -15 to 0
work for melting
work from 0 to 20

thanks for the quick reply, im sure i'll be back

4. Jan 7, 2008

### Staff: Mentor

BTW, that other heat - the heat required to raise the temp from 0 to 20 C (and from -15 to 0, if you start that cold) is called "sensible heat".

5. Jan 8, 2008

### pkleinod

The original post asked if there were any WORK done. The only work done would be due
to the expansion of the water against the ambient pressure, and this is truly negligible. Therefore, if volume changes are neglected (as DistrO assumes), then there is no work done at all.

6. Jan 8, 2008

### rohanprabhu

Since water is converted from one state to another, the average velocity of the molecules has increased. so has the $v_{rms}$. By any formula, the energy of water has increased. You cannot explain this increase in energy if you say that no work is done.

What you are referring to is:

$$W = P\Delta V$$

But this is true only for a system where no external heat is applied. But in this case, we are clearly applying external heat in order to increase temperature..

7. Jan 8, 2008

### Distr0

yeh thats the thing that got me rohanprabhu, thermal physics really isnt my thing right now. exam in 12 days too

thanks for the help anyway, hate being asked questions which we havent been taught

8. Jan 8, 2008

### stewartcs

This is probably the point of confusion...This does not equal the work done by the system. This is just the amount of heat (Q) required to melt the ice.

Think First Law of Thermodynamics

$$\Delta E = Q - W$$

The change in energy is due to the heat added (or lost) and the work done by (or on) the system.

If you consider the block of ice by itself as the system, the internal energy change is due to the heat added.

If you consider the block of ice and the heat source as the system, the work done by the heat source causes the change in energy.

Hence, depending on how you define the system, there is either no work done by the ice (assuming the OP's given assumptions) and only the heat added is responsible for the change in energy; or the work done by the ice/heat-source system is equal to the work of the heat source, and is responsible for the change in energy.

CS

Last edited: Jan 8, 2008
9. Jan 8, 2008

### Staff: Mentor

Be careful with your definitions here. Energy is the capacity to do work, it is not work. Remember the definition of work W = f.d = PΔV. So in this case the work is negligible since the ΔV is negligible. So heat (Q) is added to the ice, work is not done (W=0) by the ice, and the increased energy (ΔE = Q-W = Q-0 = Q) is accompanied by a temperature increase and a state change without any change in volume.

The word "work" always refers specifically to mechanical work as described above.

Last edited: Jan 8, 2008
10. Jan 8, 2008

### stewartcs

Work does not always specifically refer to mechanical work. In Thermodynamics, the Mechanic's concept of work is generalized.

In Thermodynamic problems such as this, any means for changing the energy of a system, other than heat, is called work. You can have a push-pull work (e.g. in a piston-cylinder, lifting a weight), electric and magnetic work (e.g. an electric motor), chemical work, surface tension work, elastic work, etc...

In defining work, we focus on the effects that the defined system has on its surroundings, which I believe is your point (as was mine).

It is essential that one first clearly specifies the system being analyzed.

CS

11. Jan 8, 2008

### Staff: Mentor

In thermodynamics, "work" and "energy" are the same thing unless some qualifier is added to the word (ie: "mechanical work"). So I assumed that when the OP talked about "work", he meant the heat energy added to the system. Note the definition (first sentence) and the qualifier (second sentence):
http://en.wikipedia.org/wiki/Work_(thermodynamics)

12. Jan 8, 2008

### rohanprabhu

that cleared a lot.. so basically you meant to say that by 'heat' we mean that the heat we supplied and by 'work' we mean that the mechanical work we have done and then these two components add together to give you the change in U so that the energy is conserved?

13. Jan 8, 2008

### pkleinod

To Russ Watters: Sorry, but I am sticking to my previous statement. The first law equates the change in the internal
energy of the system (here the ice) to the heat transferred from the surroundings plus
the work done on the system by the surroundings. It is non-specific about the kind of
work. We are concerned here with melting some ice by heating, so the only kind of work that comes into question is mechanical work (the change of volume at a constant pressure), and this is negligible. Note too, that there IS an accompanying transfer of entropy, namely Q/T, where Q is the amount of heat required to melt the ice at the
temperature T, plus the entropy transferred on heating the water from 0 to 20 deg. C.
The definition of work you give refers to the change in the internal energy of the system: dE = TdS - PdV. In the system under discussion here, the work term PdV can be neglected and the change in internal energy is then by your own definition due almost solely to the change in entropy (i.e. to the heat added to the system). Calling heat energy "work" is the first step on a royal road to confusion.

14. Jan 8, 2008

### Staff: Mentor

That's fine. I don't know for sure, I just suspect that that is what the OP was looking for.

15. Jan 12, 2008

### TVP45

Near the end of his life (probably late 1990s), Arnold Arons wrote a paper on the possibility of confusion when equating thermal energy and work. I can't lay my hands on it at the moment, but think it was in AJP. And, if I recall correctly (a real challenge anymore), Bridgeman also did some thinking about the same topic, though I couldn't begin to guess where that is published.

16. Jan 13, 2008

### rohanprabhu

From wikipedia,

So, if that is the case, then it is not only the mechanical work we are concerned with.

17. Jan 13, 2008

### TVP45

I found the Arons reference. It's in AJP, 67, Dec 199, p. 1067. However, I don't think it's a lot of help except to further point out the need for clarification.

I would argue we have to say "mechanical work" when we mean Fd; otherwise work will generally mean the gain or loss of energy (and, increasingly, we have to think of how to describe internal energy changes as well).

18. Jan 13, 2008

### stewartcs

There is another AJP article by Claude M. Penchina that I remember on the Pseudowork-Energy Principle regarding this matter. I found it helpful.

AJP Volume 46, No. 3, March 1978, p. 295-296.

CS

Last edited: Jan 13, 2008
19. Jan 13, 2008

### rohanprabhu

what is AJP?

20. Jan 13, 2008

### stewartcs

American Journal of Physics.