Work During Melting of Ice at 20C: Is Latent Heat Equal?

[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 can't recall or find a quick answer for right now.

 

[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 $-10^o \to 0^o$.

 

[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, I am sure i'll be back

 

[russ_watters]

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".

 

[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.

 

[rohanprabhu]

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.

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..

 

[Distr0]

yeh that's the thing that got me rohanprabhu, thermal physics really isn't my thing right now. exam in 12 days too

thanks for the help anyway, hate being asked questions which we haven't been taught

 

[stewartcs]

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

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

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

 

[Dale]

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.

 

[stewartcs]

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.

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

 

[russ_watters]

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.

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):

In thermodynamics, work is the quantity of energy transferred from one system to another without an accompanying transfer of entropy. It is a generalization of the concept of mechanical work in mechanics. In the SI system of measurement, work is measured in joules (symbol: J).

http://en.wikipedia.org/wiki/Work_(thermodynamics )

 

[rohanprabhu]

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.

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?

 

[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.

 

[russ_watters]

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

 

[TVP45]

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.

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.

 

[rohanprabhu]

From wikipedia,

In thermodynamics, work is the quantity of energy transferred from one system to another without an accompanying transfer of entropy.

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

 

[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).

 

[stewartcs]

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).

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

 

[rohanprabhu]

what is AJP?

 

[stewartcs]

American Journal of Physics.

 

[Dale]

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

Correct. I definitely overstated it above. In this case there is no electrical, chemical, or other work, only mechanical work. But in general the proper thermo definition of work is more general than F.d as given above.

 

[pkleinod]

In one regard, a very important one, DaleSpam's first post did NOT overstate
matters: in thermodynamics all forms of work are considered to be
inter-convertible and can thus be expressed as an equivalent amount of
mechanical work. e.g. if a process could raise a heavy piston through
expansion (of a gas, say), then mechanical work is done (raising a weight
against the force of gravity). This piston could then be allowed to fall
to its original position and the energy expended could be used to generate
electricity. Conversely, a process could produce electrical energy that
then could be used to raise a weight. This is why the "surroundings" are
often described in terms of heat reservoirs, weights and pulleys; only
the equivalent mechanical work is discussed. This could be a point of
confusion to one who is not aware of this.

The great leap forward, with regard to heat, was the recognition by Joseph
Black, Count Rumford, and Joule that heat too has a mechanical equivalent.
i.e. a system can increase its internal energy by absorbing heat or by
absorbing work (any kind of work). This was not always obvious. Thermodynamics
expresses this with the first law: Delta(E) = Q + W,
where Delta(E) is the change in internal energy of the system, Q is the
amount of heat absorbed by the system, and W is the work absorbed by the
system. The conservation of energy is expressed by
Delta(E)_system + Delta(E)_surroundings = 0.

The reason we do not have to specify the kind of work in most cases is that, as pointed
out, all forms of work can be considered to be inter-convertible in
principle. The reason
why heat has to be considered separately is because of the second law:
there are limits in principle to the amount of heat that can be converted
to work during a particular process. Thermodynamics concerns itself with
this basic fact of experience.

The molecular view of things brings a bit of insight into this dichotomy
between heat and work. When work is done, all the molecules making up a
body move partially in unison (e.g. a piston moving up or down). When a body is heated,
the molecules absorb energy but their movement is uncoordinated. (Of course,
there is some uncoordinated motion of the molecules about their lattice positions
in all bodies, moving or not.)

For a clear exposition of these matters at an elementary level, I can recommend
"Four Laws that Drive the Universe" by Peter Atkins. (Oxford, 2007).
ps. I am not Peter Atkins.

 

[Dale]

Oh no! Did I overstate my above statement about my original overstatement too?

Some days you just can't win.