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Topic: Specific Heat - Why no temperature change {for phase change}?  (Read 4978 times)

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Offline addem

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So I'm reading that, when water is solid at 0-degrees C, if you add heat to make it liquid, the temperature doesn't change.  This seems paradoxical to me because heat was added--isn't that just about the definition of temperature changing?  Is it that the heat is converted to some other form of energy as the material passes from solid to liquid?

*MOD Edit -- more descriptive title*
« Last Edit: March 04, 2015, 02:33:56 PM by Arkcon »

Offline sjb

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Re: Specific Heat - Why no temperature change?
« Reply #1 on: March 04, 2015, 02:47:26 AM »
So I'm reading that, when water is solid at 0-degrees C, if you add heat to make it liquid, the temperature doesn't change.  This seems paradoxical to me because heat was added--isn't that just about the definition of temperature changing?  Is it that the heat is converted to some other form of energy as the material passes from solid to liquid?


How does the structure of a solid compare to that of a liquid?

Offline addem

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Re: Specific Heat - Why no temperature change?
« Reply #2 on: March 04, 2015, 12:24:34 PM »
How does the structure of a solid compare to that of a liquid?

A solid is tightly packed while a liquid is dispersed.  For water in particular it forms some kind of a rigid structure like crystals or something.  Not seeing what this has to do with it, though, because I would think being liquid means the particles are moving around faster, which is a higher temperature, right?  So after heat is added, it turns liquid, which is moving faster at a molecular level, which means the temperature is higher. 

Offline Ciubba

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Re: Specific Heat - Why no temperature change?
« Reply #3 on: March 04, 2015, 01:10:26 PM »
Are you familiar with the idea of the enthalpy of fusion? It is a form of latent heat, meaning that it is heat that is absorbed without changing the temperature of the molecule(s). Physically, this means that the liquid phase of water has a higher internal energy than that of the solid phase.

In other words, all of the energy that is being absorbed by solid water at about 0 C is going into breaking the lattice rather than into the velocity of the molecules. No increase in velocity=no increase in temperature.

Offline Arkcon

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Re: Specific Heat - Why no temperature change?
« Reply #4 on: March 04, 2015, 02:30:15 PM »
Not seeing what this has to do with it, though, because I would think being liquid means the particles are moving around faster, which is a higher temperature, right?  So after heat is added, it turns liquid, which is moving faster at a molecular level, which means the temperature is higher.

OK, right here, is where you're mixing definitions, and it has confused you.  You're conflating the motion of individual molecules, with the molecular vibrations associated with heating.  These two are not the same.  You may not claim that motion=heat: consider a object in space, it doesn't gain temperature when its accelerated.  No more so, individual molecules in a sample.

You're going to have to accept that the thermal energy you're expecting to see has been used up -- that is to say, the breakup of short range bonds that made a liquid a solid are now potential energy that would be released if the solid were to re-form.

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This seems paradoxical to me because heat was added--isn't that just about the definition of temperature changing?

I would bet the formal scientific definitions of these terms are not equivalent, in writing, anywhere, except for your post.  Why not check a textbook and see?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline addem

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Re: Specific Heat - Why no temperature change?
« Reply #5 on: March 04, 2015, 11:28:03 PM »

OK, right here, is where you're mixing definitions, and it has confused you.  You're conflating the motion of individual molecules, with the molecular vibrations associated with heating.  These two are not the same.  You may not claim that motion=heat:

I see--that's news to me!  OK, so if speed at the molecular level is not the definition of heat, then what is?  Wikipedia and other simple webpages don't say anything that I can understand which would clarify this for me.

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I would bet the formal scientific definitions of these terms are not equivalent, in writing, anywhere, except for your post.  Why not check a textbook and see?

My textbook's definition of heat:  "Heat is the energy associated with the motion of particles."  It doesn't really contradict what I said.

Offline Corribus

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Re: Specific Heat - Why no temperature change {for phase change}?
« Reply #6 on: March 05, 2015, 11:25:58 AM »
Some thoughts on relevant terminology

Although we use them every day, energy, temperature and heat are abstract concepts that are difficult to define precisely, even for seasoned physical scientists. Part of the reason for this is that they are not concrete physical properties of specific objects, even though we often treat them as such in casual conversation. Like many things in thermodynamics, they refer more appropriately to dynamical processes and thus are better defined in a relative sense. That is, in thermodynamics, and more broadly in chemistry, we are more concerned with where, when, and why heat and energy travel than quantifying them in their origins and destinations. Chemistry is about processes and change, not the treatment of static systems. Likewise, while we might say, "what is the temperature of that coffee?", temperature really is a quantity that is defined relative to something else. It specifies where heat or energy is likely to spontaneously go rather than defining any absolute property of a system (although, the Kelvin scale is defined, in a thermodynamic sense, absolutely).

The transfer of energy from one place to another can be roughly divided into two categories: work and heat. I like to think of work as energy that is transferred and brings about a deliberate change in a system that otherwise wouldn't happen on its own. But because no transfer of energy is 100% efficient (the laws of thermodynamics and all that), there will be some energy lost, and this is predominantly what heat is. That lost energy isn't lost in the sense that it disappears from the universe, however. It simply becomes distributed without preference (randomly, one might say) throughout the "environment", however you want to define that. Microscopically, that extra energy gets fed into the random motions of atoms and molecules in the form of kinetic energy. This is fundamentally an entropic process. Note that heat CAN be used to do work, but since it does not act discriminately, at least on a microscopic level, it is a fundamentally ineffecient way to get things done.

Onward to temperature!

Temperature is a measure of where heat (energy) will be spontaneously transfered when you bring two objects together. A hot body has more latent heat than a cold body, and so heat will flow from the former to a latter, because this is entropically favorable. (There's a statistical argument that can be applied here to demonstrate why this is, but we'll leave that be.) Quantified, the temperature is a measure of not only the direction of change, but the magnitude of change. When you touch something and say, "Ah, that feels hot!", what you really mean is that there is more heat contained in the hot object (in the form of vibrating, rotating, and translating molecules) than there is in your relatively cold finger. If you say, "that bathwater is warm", it specifies the same vectoral differential, but a difference of scale. When molecules collide with each other, energy is transfered, somewhat like billiard balls smacking into each other. On a macroscopic scale, there is a net transfer of molecular kinetic energy from the hot object to your cold finger, and you perceive this as a "change in temperature". At equilibrium (such that it exists), the temperature of the two bodies is equivalent because this is the entropic endpoint - but notice that the change in temperature between the hot body and the eventual equilibrium point is NOT necessarily equal to the positive change in temperature of the cold body compared to the equilibrium point. That is, if you drop a hot penny at a temperature of 100 °C into a swimming pool at 20 °C, the temperature of the water in the swimming pool is not 60 °C at equilibrium. This fact, which should be obvious to everyone as a matter of intuition, tells you that there is a fundamental distinction between heat (energy) and temperature. For one thing, there's a difference in scale between a penny and swimming pool. But a more important distinction is described below.

Incidentally, a nice introduction to temperature as a concept can be found here: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/temper.html

What happens when you heat something?

When you heat something, it is prudent to think about what happens. Unlike light energy, which can be directed into specific molecular bonds, functional groups, or delocalized molecular transitions, owing to the Bohr condition, heat energy is non-discriminating for the most part. There are a lot of similarities between heat transfer and mass diffusion; in fact they are governed by many of the same equations and are both fundamentally entropic. Which is to say, in a simplistic treatment, heat doesn't preferentially speed up some atoms more than others. Kinetic energy is transferred to every particle equally. (Although, there are statistical mechanical rules for population of molecular states, but we can safely ignore them in this discussion.) Since heat is essentially a macroscopic manifestation of the average kinetic energy of atoms and molecules, when you bring two objects together, there will be a net flow of kinetic energy from the hot body to the cold body until such time as the transfer rate of kinetic energy from body A into body B exactly matches that of body B into body A - i.e., the temperatures of the two bodies are equivalent. All well and good, and this simplistic model works perfectly well when we talk about an "ideal" system, such as two bodies of ideal gas initially held at different temperatures.. or even two metallic solids.

But, molecules are more complicated than ideal monatomic gasses. Not only do molecules have (translational) kinetic energy and internal vibrations/rotations, they also exhibit intermolecular forces, which are electrostatic forces between charged particles and dipoles. Importantly, form much of the basis of the property of heat capacity, which is partially responsible for the effect described above about the penny in the swimming pool. It takes more energy to raise the temperature of water 1 degree than it does an equivalent mass of copper (specific heat capacity of water = 4.17 J g-1K-1, heat capacity of copper = 0.4 J g-1 K-1), because when heat flows into water a smaller fraction of the heat is transferred as kinetic motion of the water molecules compared to the fraction of heat that is transferred as kinetic motion of copper atoms. Water has all kinds of strong intermolecular forces, plus internal molecular vibrational and rotational states, and when kinetic energy is transferred into water, a sizeable fraction of that energy is absorbed by these intermolecular "bonds". Remember, heat energy does not discriminate where it goes. Imagine a billiard scenario like this: if the cue ball smacks into another ball, and we assume an elastic collision, the cue ball will come to a dead stop and the other ball will fly off at the same velocity (assuming equal mass) because all of the kinetic energy has been transferred. On the other hand, imagine if we tether two billiard balls together with a spring and we smack a cue ball into this tethered system. We can imagine that the cue ball will again come to a dead stop and our spring-loaded pair will again fly off in the other direction, but also the two balls will exhibit some vibration, and possibly also rotate around their center of mass - i.e., spin. In other words, while 100% of the energy will be transferred from the cue ball to the spring-loaded pair, all of the energy will not be transferred as translational kinetic energy. Some of it will be stored in the spring in the form of a vibration (which eventually gets distributed as heat via friction... those balls do eventually come to a rest!) and also as rotational energy. The same happens in molecules. If you drop a hot penny into water, versus, say, an equivalent volume of hexanes (specific heat capacity = 2.3 J g-1 K-1), the final temperatures of the solutions will be different. While the amount of energy transfered may be equivalent, the amount of energy that can be stored in water versus hexanes is different by virtue of the types of internal and intramolecular degrees of freedom (vibrations, rotations, intermolecular bonds) that is characteristic of each substance. This is represented by the heat capacity.

Getting back to phase transitions

At a phase transition, intermolecular bonds are broken (or formed, depending on the direction). Ice, or any solid, is characterized by molecules locked into a rigid lattice. MOlecules are bound together by strong intermolecular bonds, which is why solids hold their shape. The amount of energy required to break these bonds is called the lattice energy. When you transfer heat into a solid, this means kinetic energy is transferred. Some of that kinetic energy is transferred of a kind, meaning as translational kinetic energy of the atoms in the solid, but they only have so far they can move because they are locked into a lattice. Some of this energy is also absorbed by the intermolecular bonds that make up the lattice. In order to melt a solid, you have to supply enough energy to break up those intermolecular bonds.. but that energy cannot simultaneously contribute to translational kinetic energy of molecules and atoms in the solid. Energy can't do two things at once. Rather, the energy is lost to the process of breaking the lattice - specifically, it's converted into potential energy through work. Hence the energy that goes into melting a solid does not increase the temperature of the solid, because - again - we've defined temperature as a relative measure of the ability of a body to transfer heat energy to another body. The energy that goes into breaking up a lattice is not available as translational kinetic energy to transfer elsewhere, hence why the temperature is unaffected by heat of a phase change.

I hate to go back to billiards again, but moving balls can be very instructive as crude illustrations of microscopic phenomena. Suppose again you have a cue ball and another ball, but this time, there is an incline and plateau after the other ball. You hit the cue ball such that is has exactly the amount of kinetic energy equivalent to the gravitational potential energy between the resting point of the other ball and the top of the incline. When the cue ball smacks into the other ball, 100% of the kinetic energy is transfered. The other ball rushes up the incline, but slows as it goes, because it is doing work against the gravitational field. When the other ball reaches the top of the incline, it comes to a rest because we've specified the amount of kinetic energy transferred is exactly equivalent to the gravitational potential energy of the incline. Now, we can say that energy has been transferred from one system to the other, but the receptor ball has no kinetic energy once the process is completed. It certainly cannot transfer kinetic energy to another ball at this point. The process of melting is not so different, both microscopically and macroscopically. Kinetic energy is transferred to a lattice, but a large portion of that energy is used to do work against the electrostatic potential holding the lattice together. The atoms making up the lattice do not obtain translational kinetic energy from this process, and hence they cannot transfer kinetic energy to another nearby body, hence the temperature of the macroscopic system does not change. Once the melting proceses is over, though, additional kinetic energy transferred into the system goes, at least partially, into translational modes of the melted atoms and molecules, and hence the temperature rises again... at least until the boiling point, which is another phase change that obeys the same fundamental principles.

Anyway, I tried to keep this written in rather casual language, so there are places where I've made simplifications and certain fictional allowances, but maybe it helps you understand what's going on.
« Last Edit: March 05, 2015, 12:41:49 PM by Corribus »
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline addem

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Re: Specific Heat - Why no temperature change {for phase change}?
« Reply #7 on: April 27, 2015, 02:20:18 PM »
Corribus, I think you answer was pitch-perfect and excellent, thank you!  Took me a while to get back to this, but I think I much better understand all of the relevant ideas.  This answer should probably be pinned somewhere to help anyone in the future to understand similar concepts.

Offline pgk

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Re: Specific Heat - Why no temperature change {for phase change}?
« Reply #8 on: April 27, 2015, 05:54:59 PM »
A solid needs an additional amount of energy, in order to destroy the surface energy and pass into the liquid phase and thus, the temperture does not increase during melting. Similary, a liquid needs an additional amount of energy, in order to pass into the gaseous phase (evaporation enthalpy) and thus, the temperture does not increase during boiling.
Question: Which one causes more serious burns on contamination with the hand, boiling water at 100oC or vapor at 100oc, and why?

Offline Corribus

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Re: Specific Heat - Why no temperature change {for phase change}?
« Reply #9 on: April 28, 2015, 10:17:10 AM »
Corribus, I think you answer was pitch-perfect and excellent, thank you!  Took me a while to get back to this, but I think I much better understand all of the relevant ideas.  This answer should probably be pinned somewhere to help anyone in the future to understand similar concepts.
You're welcome, glad it was helpful.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline addem

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Re: Specific Heat - Why no temperature change {for phase change}?
« Reply #10 on: April 29, 2015, 10:59:22 PM »
A solid needs an additional amount of energy, in order to destroy the surface energy and pass into the liquid phase and thus, the temperture does not increase during melting. Similary, a liquid needs an additional amount of energy, in order to pass into the gaseous phase (evaporation enthalpy) and thus, the temperture does not increase during boiling.
Question: Which one causes more serious burns on contamination with the hand, boiling water at 100oC or vapor at 100oc, and why?

I'm going to go with the steam because the water will simply transfer heat until equilibrium.  The steam, on the other hand, will transfer the same amount of heat due to equilibrating temperatures; but it will also condense, which constitutes a release of energy as molecules "lock into place" so to speak.

Offline pgk

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Re: Specific Heat - Why no temperature change {for phase change}?
« Reply #11 on: April 30, 2015, 10:58:39 AM »
Yes. It’s the steam but apart the heat transfer, steam at 100oc, contains an additional amount of heat that is equal to the free energy of evaporation [= Evaporation enthalpy –  (273 + 100)oK x Entropy(liquid → gas)].
So, never put your hand over boiling water.

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