Ice Water Heat Transfer
Some ice is floating in a glass and the temperature of the ice, water, and the surroundings are all equal to 0 ∘ C .
In which direction will heat flow between the ice and water?
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8 solutions
What abut the energy needed for conversion of ice to water?
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That is why it will forever remain as a mixture of ice and water (provided the environment is always at 0C). Both need more energy to change state but neither are allowed to transfer.
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That is incorrect. Water has additional heat (energy) bound up in it compared to the ice because of the energy required to phase change between ice and water. Just because they are the same temperature, that doesn't mean that they have the same energy -- a temperature measurement is insufficient to calculate all of the energy bound up in the water and ice. Of course they're allowed to transfer energy. Surface molecules of the ice will eventually collide with molecules in the surrounding water and energy will be transferred from the water to the ice.
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@Bart Humphries – Thought experiment: if water and ice are in thermal contact within a closed system at 0C, does the mixture eventually become purely water or purely ice?
Hint: the energy needed to melt ice at 0C is the same as the energy that must be lost to freeze water at 0C.
The water system systems need more energy to convert the ice to water without it freezing but that would mean it is at more than 0C. Similarly ice would have to have be at a lower energy and temperature to force the water to freeze without it melting.
For a discussion on the rest I refer you to my exchange with Cody in the Zacharys comment thread.
Disagree. Due to ambient air temp, unless also at zero, heat will flow from glass to water and from there to ice.
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The ice and the water may absorb heat from the surroundings, but as long as they are both at zero degrees, heat can't flow between them.
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That is incorrect. Water has additional heat (energy) bound up in it compared to the ice because of the energy required to phase change between ice and water. Just because they are the same temperature, that doesn't mean that they have the same energy -- a temperature measurement is insufficient to calculate all of the energy bound up in the water and ice.
The two can exchange heat at thermal equilibrium. But, as this happens, the water becomes ice and vice versa. The net exchange of heat is zero, but it will always be from water to ice because the water is always freezing and the ice is always melting. If there was no exchange, the cubes would always remain cubes. But, in fact, you will end up with a slushy mess in the end because heat is always flowing from water to ice.
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That is incorrect- the state of equilibrium means that once both the ice and water are at 0C, there is no change in the average amount of either ice or water. Whatever amount of ice melts is compensated by an equal amount of water freezing.
Furthermore the statement that there is heat transfer only from water to ice is incorrect- the amount of heat that leaves the water for it to be converted to ice is restored by the amount of heat gained by the ice melting elsewhere.
To see this more clearly, if heat could only flow from water to ice, how do they remain in thermal equilibrium?
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Sorry for the late reply. When the heat flows from water to ice, there is a change of state which uses up that heat without changing the temperature.
Would the water not still transfer kinetic energy to the ice? Would that not be "heat" transfer?
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It would not if they had the same temperature. What is commonly called heat transfer is a macroscopic (large average) phenomenon. At the microscopic level it boils down to kinetic energy but notice it is the average kinetic energy.
Some water particles could have large energies while others could have relatively lower energies, and the same applies for the ice. Will there be a *net exchange? No! The average effect, which we see, is encoded in the laws of thermodynamics. This allows us to ignore the horrible details of what is actually happening and arrive at correct predictions.
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But as I studied molecular physics, heat is literally defined as "the average random molecular kinetic energy." Also stated that as the phases of matter, that each stage up by default has more kinetic energy than those below it.
If the water didn't have more molecular kinetic energy than the ice, then wouldn't it already be ice?
Also it's a glass of water, not a thermos. It never states anywhere about this being a closed system or to assume the ambient temperature is the same as the glass. This problem is just flawed. It need to Be More Specific.
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@Cody Elkins – "Heat is literally defined as "the average random molecular kinetic energy."."
It is not: this is a grossly oversimplified definition of heat, which is why it is best to think in terms of God given macroscopic laws. In this case it particularly fails because there are non-trivial interactions that arise, namely, the formation of hydrogen bonds. It also fails because the systems (ice and water) are free to exchange particles.
"f the water didn't have more molecular kinetic energy than the ice, then wouldn't it already be ice?"
No and here's why: let's say that both have the same kinetic energy and let's ignore what I said in the previous paragraph. Ice is different from water in that there are strong hydrogen bonds between the water molecules holding the crystalline structure together whereas these bonds spontaneously form and break in liquid water. Therefore, energy is needed to break these hydrogen bonds for the ice to become water.
In other words, both have the same amount of kinetic energy but ice has a lower amount of potential energy. Is either one allowed to transfer heat to the other to overcome this potential barrier? No! Heat transfer is only allowed across gradients and they both have the same temperature and thus kinetic energy!
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@A Former Brilliant Member – You never addressed the open system. Yes it matters. A poorly worded question, even if it is the "basic" section, can mean allot of over thinking.
And my answer wasn't a "gross oversimplification" of the definition of heat, I looked the actual definition up and it says "The motion of atoms and molecules creates a form of energy called heat or thermal energy which is present in all matter."
Even IF this were a closed system, wouldn't the energy transfer continue until you got a slushy water instead of ice cubes floating in it?
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@Cody Elkins – That is a dictionary definition- there is a difference between the proper definition of a quantity i.e. the definition you use to make calculations, and descriptive definitions.
The zeroth law of thermodynamics address systems in thermal contact only and serves to define the notion of temperature as the quantity which determines if a set of systems are in thermal equilibrium.
Once you allow for particle exchange and the matter becomes more complex and indeed becomes the subject of transport theory. All of this is completely unnecessary as we are dealing with a macroscopic phenomenon.
If it were a closed system, it would eventually become a slushy mixture but crucially you would have the same amount of ice. The reason it becomes slushy and does not remain in cube form is because of the second law of thermodynamics.
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@A Former Brilliant Member – Now that I am actually awake, I see what you are saying. You are saying that, even though there is an exchange of heat on a Microscopic scale, there isn't an exchange of heat on a Macroscopic scale. This problem is poorly worded, and should be rewritten.
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@Cody Elkins – Here's the reason why I like this problem- heat exchange, just like temperature, is a macroscopic phenomenon. It is ill defined on the microscopic scale. To make this more concrete consider the following:
Suppose a bit of water gives enough heat to melt a bit of ice and becomes frozen itself. Has there been a heat exchange between the system which was that bit of water and the system which was that bit of ice? Yes!
(This can never happen- If there is such a heat exchange then the water must be warmer on account of some temperature fluctuation perhaps- and cannot therefore lose enough heat to become ice-and there will correspondingly be some water elsewhere which will be slightly cooler and will in fact become ice to compensate for the ice that was melted.)
However has there been a heat exchange between the system defined as ice in our closed system and water in our the closed system? No: there was no (net) thermal heat exchange and no (net) particle exchange.
The bit of water that lost heat joined the ice system and the heat that went to the ice system melted it thereby restoring that heat to the water system.
There is an equilibrium in exchange of particles between both systems and an equilibrium in heat exchange- that is where the subtlety lies. Equilibrium does not mean there is no local heat exchange, it means there is no global heat exchange.
This is by no means a basic problem, it brings to light many subtle considerations of thermodynamics. More to the point, it forces you to think more carefully about what happens, and how things are defined.
If the question asked is there an exchange of heat between the sytem of particles in the ice cubes and water in the glass then we say yes on the basis of the second law of thermodynamics. We know that there must have been local heat exchanges since we end up with that slushy mess so that the zeroth law is not violated either.
But then why can't we just appeal to straight away to the zeroth law? Because in the case of just ice cubes and water, there is more than just thermal contact- there is particle exchange whose equilibrium we cannot speak for:
Some or even all of the particles in the cube system may remain in the ice state or join the water state.
But in the case of system of ice in our closed system and system of water in our closed system, we know that we must have an equilibrium in particle exchange on account of the state of thermal equilibrium between the two.
I am not certain, but given this hypothetical state and the latent heat of the fusion of water, the second law suggests a heat flow of 334 joules per gram of ice. All still zero degrees. Entropy prohibits the water becoming the more organised ice structure without work being done. Energy must flow from the water to the ice. I guess it depends if you define latent heat as heat flow. As "heat flow" is really a tautology as flow is implicit in the word "heat". ONLY if it is a "flow" is it heat, a bit like saying ATM machine, THEN "latent heat" must mean the proverbial "heat flow"
Ok, that's my argument.
Heat is:
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as a technical term in thermodynamics, a spontaneous transfer of thermal energy
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only in non-technical parlance, thermal energy in general
to be clear, if it is not flowing, it is NOT heat (in thermo physics).
Technically, by this definition, a microwave oven does not 'heat' food, it's radiation performs work on food, Reference: Prof Richard Wolfson ("The Great Courses"), not a transfer of energy to a cooler body. The sun DOES heat the earth, transfer of energy by radiation due to temperature gradient. Not so for the microwave oven. No temperature gradient from the water to ice, so maybe the water does work on the ice?
relevant wiki: https://en.wikipedia.org/wiki/Heat
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Yes, you are right about the technical definition of heat. We are in agreement with it. There is no exchange of thermal energy in this case.
What about the excess amount of energy present as Latent heat of fusion. It is extra energy so to equalise emergy has to flow from ice to water
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Heat only flows between two objects if there is a temperature difference between them. The temperature gradient determines how heat will flow. So even though water has more latent heat than ice, if they are both at 0 ∘ C , heat won't flow between them.
I somewhat disagree with the answer. Sure on a macroscopic level, if the temperature is the same in both systems, no energy should flow from a system to the other. But if we are at some equilibrium limit, very small changes can create a snowball effect and change the macro state.
We know the water and the ice are at 0°, but given that a non-null quantity of energy is needed to go from 0° ice to 0° water, we don’t know the real quantity of energy in the water and in the ice. The system can be in a stable state : A "lot" of energy is lacking in the ice for the ice to become water, and there is way too much energy in the water, for it to become ice. This imply the probability for a state change in a given part of the ice/water, given the microscopic internal temperature fluctuations, is very very small.
Now if the water is "near" to become ice, or the ice is "near" to become water, we should remember that even if there is a very big probability for the temperature of the system to stay "almost" homogeneous, there is also a very big probability for it to never stay "perfectly" homogeneous. The very small internal temperature fluctuations will imply that some water will become ice and some ice will become water. Now because the volume from ice to water change, the very small and random volumes of water becoming ice, will become colder (<0°), and energy will start to flow from the adjacent water to these new volumes of ices, so the adjacent water will become ice, etc… At the end all the water will become ice, with a temperature below 0°, and energy will flow from the 0° original ice to the rest of the system.
Please explain me if and where I am wrong.
Before the ice became ice, it was water at the same temperature. In order to become ice, it had to lose more heat. So the ice actually has less heat in it than the water, and this will create a gradient that will cause heat to flow from the water to the ice.
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Heat only flows between two objects if there is a temperature difference between them. The temperature gradient determines how heat will flow. So even though water has more latent heat than ice, if they are both at 0 ∘ C , heat won't flow between them.
This is sort of bunk though because the water would be transferring energy to the ambient environment thereby heating up which would cause the ice to transfer heat to the water.
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Thanks, the clarification that the system is also at 0 ∘ C has been added to the problem.
The problem does not say that the system is isolated, and in everyday experience a glass of water and ice is a room temperature (why put some ice in water at all if there's already 0 celsius in the enivronment?) I think that the "Isolated System" hypothesis should be clearly stated in the problem text
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Thanks, the clarification that the system is also at 0 ∘ C has been added.
I guessed and checked this problem.How does the heat flow work? Why doesn't the heat flow, in this problem? I give this problem a 10 because I didn't know the context of it.I decided to try it out, even thought I don't know a lot about physics.
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Heat only flows from high temperature to low temperature. Since water and ice are both at 0 ∘ C , there is no temperature gradient, and heat won't flow.
I encourage everyone to get into the habit of channelling their feedback into the Reports section of a problem (via the [ ∘ ∘ ∘ ] menu under the available answers) when they either; think the problem is poorly stated, or think that the answer to the problem is actually incorrect, rather than commenting on a solution.
The problem author and Brilliant staff won't necessarily read your comments against a solution, or subsequently address the issues you raise.
If you think that a solution to the problem author's chosen answer is incomplete, or incorrectly explains their chosen answer, by all means comment on the solution detailing what is missing or needs changing.
I've only seen comment saying that the problem is wrong thus far, so I'm just going to wait for something conclusive from Brilliant on this problem.
There is heat in everything
Heat must flow as latent heat of frosting
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Heat flows only if there is a temperature difference. In this case, both temperatures are equal, so there is no net heat flow. The heat moving from water to ice is equal to the heat moving from ice to water.
Relevant wiki: Zeroth law of thermodynamics
This is a rather subtle question so I will provide a detailed discussion for it: On the face of it we have the Zeroth Law of Thermodynamics which tells us that heat will flow between two systems brought into thermal contact until both systems are in thermal equilibrium.
The purpose of this law is to tell us that there is a property that all systems have which tells us if there is a net heat flow between them and that property is temperature.
In this instance we have both the ice and the water at precisely 0C. Assuming we are dealing with standard conditions (since the freezing and boiling point will change if the external pressure changes etc) then that will be the end of it.
"But surely the water will "want" to melt the ice or the ice will "want to freeze the water because of some mumbo jumbo in the second law of thermodynamics about order and disorder and what have you"
This is the bit where there is misconception. If the ice were colder when the water reaches zero, then some water will freeze. If the water is warmer when the ice reaches zero, then some ice will melt. If they both reach zero at the same time then neither one is allowed to supply the extra energy needed to change the state of the other.
The second law of thermodynamics has no role in this problem because, quite simply, it only tells us if it is possible for a process to take place and not whether it will actually take place.
Here's the reason why I like this problem- heat exchange, just like temperature, is a macroscopic phenomenon. It is ill defined on the microscopic scale. To make this more concrete consider the following:
Suppose a bit of water gives enough heat to melt a bit of ice and becomes frozen itself. Has there been a heat exchange between the system which was that bit of water and the system which was that bit of ice? Yes!
(This can never happen- If there is such a heat exchange then the water must be warmer on account of some temperature fluctuation perhaps- and cannot therefore lose enough heat to become ice-and there will correspondingly be some water elsewhere which will be slightly cooler and will in fact become ice to compensate for the ice that was melted.)
However has there been a heat exchange between the system defined as ice in our closed system and water in our the closed system? No: there was no (net) thermal heat exchange and no (net) particle exchange.
The bit of water that lost heat joined the ice system and the heat that went to the ice system melted it thereby restoring that heat to the water system.
There is an equilibrium in exchange of particles between both systems and an equilibrium in heat exchange- that is where the subtlety lies. Equilibrium does not mean there is no local heat exchange, it means there is no global heat exchange.
This is by no means a basic problem, it brings to light many subtle considerations of thermodynamics. More to the point, it forces you to think more carefully about what happens, and how things are defined.
If the question asked is there an exchange of heat between the sytem of particles in the ice cubes and water in the glass then we say yes on the basis of the second law of thermodynamics. We know that there must have been local heat exchanges since we end up with that slushy mess so that the zeroth law is not violated either.
But then why can't we just appeal to straight away to the zeroth law? Because in the case of just ice cubes and water, there is more than just thermal contact- there is particle exchange whose equilibrium we cannot speak for:
Some or even all of the particles in the cube system may remain in the ice state or join the water state.
But in the case of system of ice in our closed system and system of water in our closed system, we know that we must have an equilibrium in particle exchange on account of the state of thermal equilibrium between the two
Posts on stack exchange support your arguments: https://physics.stackexchange.com/questions/287367/why-can-a-block-of-ice-at-0-circ-rm-c-and-water-at-0-circ-rm-c-coexist-in/287371
transformation from a phase to another is a thermic reation. From ice to liquid, it's an exothermic reaction. We assume that the ice melts, so there should be a heat transfert from ice to liquid. I'm confused w/ the explaination
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First off, Ice to liquid is an endothermic reaction, so heat is taken from the water to melt some ice. But the heat that is taken from the water causes some of it to freeze. You have an equal amount of water freezing and ice melting- this is the state of equilibrium.
Nothing is said about ambient temperature or where the glass is. At room temperature, water to ice.
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No- even if the ambient were at room temperature, the water would have to be warmer in order for heat transfer to occur i.e. the temperature of the water must be greater than 0C if heat is to be exchanged between it and ice.
No heat is exchanged at thermal equilibrium which is in turn determined by temperature .
nothing is said about the environment that the glass is in. if the glass is not sitting in a vacuum chamber of sorts, then it will absorb heat from, or radiate heat to, the environment, dependent upon the ambient temperature in which it sits. thus heat will be conducted to, or from, the water which will then conduct to, or from, the ice.
Relevant wiki: Zeroth law of thermodynamics
Both ice and water can exist a zero degrees Celsius. You have to add about 80 calories a gram to ice (the heat of fusion) to cause it to convert to water at zero degrees
Celsius. After this point each calorie added will raise the temperature of one gram of water one degree Celsius.
Same for liquid water. You have to remove the same amount of heat from zero degree water to form a gram of ice at zero degrees Celsius. Continue removing heat and the
temperature of the ice will drop.
as,no external force is there to change the condition,they will be in thermal equilibrium condition .so, no heat will flow.
thank you sir
No temperature difference→no heat flow
This is correct, neglecting the effects of the temperature of the room. If the room is also at a uniform 0 degrees, then no heat transfer occurs. If however, there is a differential temperature between the glass and the room, heat transfer will occur, ultimately either to or from the suspended ice.
This is true - b this transfer would be a transfer of heat from air to ice, and from air to water - not between water and ice, which is what the question asks. Exception: If the air transfers its heat more quickly to one than to the other, which could occur if, e.g., a greater proportion of the water than the ice were exposed to the air (which is likely to be the case, as we should include not just the surface area of the water at the top of the glass, but all the water against the sides of the glass, which are engaged in heat transfer with the room air). In this case, then some of the heat transferred from the room to the water would then be transferred from the water to the ice.
First of all, the water will freeze
I recall being taught that thermal equilibrium is part of the definition of temperature.
If they are both at the same temperature. 1. How will there be any heat!! 2. I they have the same heat their is no place for the heat to go or be produced!! 3. How is it even possible to produce heat if you're in a state when water freezes and wouldn't the glass become colder if left in a room with no sunlight/ heat!! Anyway that's what I think!! <(^-^)>
Relevant wiki: Zeroth law of thermodynamics
Simplifying the process somewhat: Transference (flow) of heat energy within an isolated system occurs until the bodies of the system reach Thermal Equilibrium . With the ice and the water in this problem already at equilibrium, the net heat transference within the system becomes zero - No heat will flow
I use the term 'net' because, even at 0 ∘ C , both the ice and the water do emit heat energy - Objects emit heat energy unless they are at 0 ∘ K ( − 2 7 3 . 1 5 ∘ C ). So it is still conceivable that the ice and the water continually exchange an equivalent amount of heat energy with each other, such that the net change over time that each experiences is zero.