Calorimetry of a non-isolated system

In summary, the conversation explores the idea of whether it is possible to keep ice from melting at room temperature or other temperatures above freezing by controlling the volume, concentration of water, and concentration of ice in a container. The speaker initially considers using a calorimetry approach but realizes that the system's interaction with the environment makes it more complex. They also discuss the effects of using a rigid container and observe that the ice does not create a vacuum despite taking up more space when melted. The conversation concludes with the speaker questioning whether it is possible to slow down the melting rate of the ice and how initial conditions may affect this. Ultimately, it is suggested that the heat conductivity of the container and other factors such as convection may make it difficult to accurately
  • #1
ThomGunn
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I had a bottle half full of ice and filled it with water and placed it in my fridge. I noticed that the ice was taking a significant time to melt. In fact after half a day it looked like it hadn’t melted at all.

That got me to thinking. Say I did this with a rigid container and filled up the entire volume of the container with ice and water. Say 50-50. I’m curious if there is some combination of volume, concentration of water, and concentration of ice, that would allow for the ice to remain ice at say room temperature or any other temperature above freezing for a significant amount of time.
Normally this would be a simple calorimetry problem, and I’d use the conservation of energy and find an equilibrium temperature. However, allowing the system to interact with the environment is causing me some confusion. I obviously can’t have the mass of all the air in a room, but I would then treat the environment like a reservoir. So then the equilibrium temperature would have to be the temperature of the reservoir. Which would ensure that the ice would melt.

After thinking of that I got to thinking about how the volume of the container might play some role. I took my bottle out of the fridge and after a short while the ice melted and the sides of the bottle contracted as to indicate that there was some kind of vacuum present. This makes sense, Ice is less dense than water and thus must occupy a larger volume. I didn’t expect a vacuum to develop in such a way that the sides of the bottle would contract. So I got to thinking what would happen if the container was perfectly rigid. This is where I got into things I do not know about. I assume the ice would still melt and the container just wouldn’t crush. Since the volume in the container is fixed, would it be possible to set up the initial concentrations so that there is no space available for all of the ice to melt? Or would this always just lead to an increase in pressure, with the work being done by the reservoir? Allowing the ice to always melt? Lastly, could I calculate the rate at which the ice melts and how the initial conditions affect that rate? To determine if perhaps there is a way to slow the rate at which it melts, because I believe I can now conclude that the ice must melt….although I have no idea how to calculate at what speed this occurs.
Any help or direction would be appreciated. I believe my reasoning up to this point is sound, but if I’ve overlooked something I would very much appreciate you pointing it out
 
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  • #2
ThomGunn said:
I’m curious if there is some combination of volume, concentration of water, and concentration of ice, that would allow for the ice to remain ice at say room temperature or any other temperature above freezing for a significant amount of time.
Concentration is not the right word here, but think about it: for a given amount of water, what proportion should be solid over liquid to have it solid as long as possible?

ThomGunn said:
Normally this would be a simple calorimetry problem, and I’d use the conservation of energy and find an equilibrium temperature. However, allowing the system to interact with the environment is causing me some confusion. I obviously can’t have the mass of all the air in a room, but I would then treat the environment like a reservoir. So then the equilibrium temperature would have to be the temperature of the reservoir. Which would ensure that the ice would melt.
This is not an equilibrium situation. You have to use the Fourier heat conduction law, and you will see that the time required to melt the ice will depend mostly at the temperature of the air and at how well the heat can be transferred from the water to the air, which will depend on the characteristics of the container.

ThomGunn said:
I took my bottle out of the fridge and after a short while the ice melted and the sides of the bottle contracted as to indicate that there was some kind of vacuum present. This makes sense, Ice is less dense than water and thus must occupy a larger volume. I didn’t expect a vacuum to develop in such a way that the sides of the bottle would contract. So I got to thinking what would happen if the container was perfectly rigid. This is where I got into things I do not know about. I assume the ice would still melt and the container just wouldn’t crush.
It's the other way around: the bottle contracted, preventing a vacuum to form. Even in a rigid container, you would not have formation of a vacuum as the water vapour will occupy the space where no liquid water is present, but it would lead to a lower pressure inside the container.

ThomGunn said:
Since the volume in the container is fixed, would it be possible to set up the initial concentrations so that there is no space available for all of the ice to melt?
Does melting ice occupy more or less space?

ThomGunn said:
Lastly, could I calculate the rate at which the ice melts and how the initial conditions affect that rate? To determine if perhaps there is a way to slow the rate at which it melts, because I believe I can now conclude that the ice must melt….although I have no idea how to calculate at what speed this occurs.
See above. But it will be difficult to calculate in real world situations if you do not know the heat conductivity of the bottle, its surface area, and if you have to take into account convection inside the room. You will often get a layer of colder air surrounding the bottle, reducing the effective transfer of heat from the room to the bottle.

If you want to keep the ice for longer, you simply need good insolation. See for instance http://en.wikipedia.org/wiki/Vacuum_flask
 
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1. What is calorimetry?

Calorimetry is the scientific study of heat transfer and the measurement of heat energy. In the context of a non-isolated system, it refers to the measurement of heat energy exchanged between the system and its surroundings.

2. How is calorimetry of a non-isolated system different from an isolated system?

In an isolated system, there is no exchange of heat energy with the surroundings, whereas in a non-isolated system, there is heat transfer between the system and its surroundings. This makes the measurement of heat energy more complex in a non-isolated system.

3. What is a calorimeter and how is it used in calorimetry of a non-isolated system?

A calorimeter is a scientific instrument used to measure the heat energy exchanged between a system and its surroundings. In calorimetry of a non-isolated system, the calorimeter is used to contain the system and measure the temperature changes that occur due to heat transfer.

4. What factors affect the accuracy of calorimetry in a non-isolated system?

Several factors can affect the accuracy of calorimetry in a non-isolated system, including the heat capacity of the system and its surroundings, any heat loss or gain to the surroundings, and the duration of the experiment. It is important to control these factors to ensure accurate measurements.

5. How is the heat energy exchanged in a non-isolated system calculated using calorimetry?

The heat energy exchanged in a non-isolated system can be calculated using the formula Q = mcΔT, where Q is the heat energy, m is the mass of the system, c is the specific heat capacity, and ΔT is the change in temperature. This formula takes into account the temperature changes of both the system and its surroundings.

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