Calculating Heat Energy in a Refrigerator - Solving for Qh and Qc

In summary, the conversation discusses how to find the amount of heat energy exhausted into the room as 100 mL of water at 288 K is changed to ice at 258 K in a refrigerator with a coefficient of performance of 4.0. The conversation mentions using the Brayton-cycle refrigerator as a model, but it is determined that it is not necessary. The solution is to find the heat lost by the water through temperature change and latent heat, then using the relationship between heat lost and coefficient of performance to find the heat exhausted into the room. The specific heat and latent heat values for water and ice are given for reference.
  • #1
dekoi
100 mL of water at 288 K is placed in the freezer compartment of a refrigerator with a coefficient of performance of 4.0. How much heat energy is exhausted into the room as the water is changed to ice at 258 K?

(The answer is 5.34 x 10^4 J)

I tried using the Brayton-cycle refrigerator to understand this problem.
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I am assuming that the lower isobar is the part of the graph that I need to analyze, where the low-temperature heat exchanger warms the gas and cools the refrigerator. With this method, I know that [tex]Q = nC_v\Delta T[/tex]. However, I cannot think of a way to find the number of moles (n).

I also don't know if my reasoning is correct (that is, only using the lower isobar of the graph to answer my question).

I am also assuming that my lower temperature (258 K) is equivalent to Tc, and my higher temperature (288 K) is equivalent to Th.
I am not sure if this is correct, however.

I have realized that I need to use work (in) to answer the question, because K = Q / W.

However, the work for an isobar is W = pV, and I do not know the pressure, and am unable to find it since I do not know the number of moles.

Any advice will be greatly appreciated, thank you.
 
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  • #2
Seems like you are overcomplicating things. How much "heat" must be removed from the water to (1) cool it to freezing, (2) freeze it, and (3) cool it to its final temperature?
 
  • #3
From the other thread (double posted):
https://www.physicsforums.com/showthread.php?t=101662

You don't know, or have to know, the details of the cycle. All you have to do is find [itex]Q_H = W + Q_C[/itex] where:
[tex]COP = Q_C/W[/tex]

[tex]Q_C[/tex] is the heat lost by the water (you have to take into account temperature change + heat of fusion).

dekoi said:
What is COP?

And I don't think we are required to know heat of fusion, so there must be some other way to solve the problem.

COP = Coefficient of performance = Qc/W (for a refrigerator)

Of course you have to know the heat of fusion! How else do you determine how much heat is removed from the water? If you use 33.4 KJ/100g and add the heat lost due to temperature decrease you will get the right answer.

AM
 
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  • #4
Doc Al said:
Seems like you are overcomplicating things. How much "heat" must be removed from the water to (1) cool it to freezing, (2) freeze it, and (3) cool it to its final temperature?
Am I finding Q for each step of the reaction, and then finding the overall Qh?Also, am I able to use the reversed Brayton-cycle as my model for the refrigerator, because that is for an ideal gas refrigerator, and my refrigerator has water (which is not an ideal gas).
 
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  • #5
dekoi said:
Am I finding Q for each step of the reaction, and then finding the overall Qh?
You are simply finding how much heat must be removed from the water in order to take it from one temperature to another. This is Qc, not Qh!


Also, am I able to use the reversed Brayton-cycle as my model for the refrigerator, because that is for an ideal gas refrigerator, and my refrigerator has water (which is not an ideal gas).
Forget about the details of the cycle. All you need to know is the COP, which is given. (I think you are confusing the water in this refrigerator with the working fluid in a Brayton-cycle refrigerator. This water is just something being cooled by the refrigerator; it's not the working fluid!)
 
  • #6
Okay, so I understand that I am not using the Brayton-cycle refrigerator.

So, how do I find Qc from the information given?
What is Qc equal to?
nCpT?
or something else?

What is Cp for water?
 
  • #7
[tex]K = Qc / W_{in}[/tex]

therefore,
[tex]Qc = (K)(W_{in})[/tex]

But what is [tex]W_{in}[/tex]?
[tex](P)(\Delta V)[/tex]?
But I do not know the pressure... I only know the two volumes.

I think I can find moles by taking the density of the water to be 1000 kg/L...
but then i get n = 5555.56 mol, which I thought was rather large.
 
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  • #8
dekoi said:
[tex]K = Qc / W_{in}[/tex]
therefore,
[tex]Qc = (K)(W_{in})[/tex]
But what is [tex]W_{in}[/tex]?
[tex](P)(\Delta V)[/tex]?
But I do not know the pressure... I only know the two volumes.
I think I can find moles by taking the density of the water to be 1000 kg/L...
but then i get n = 5555.56 mol, which I thought was rather large.
Read the above responses again. We have given you all you need to know.

Use [itex]Q = Cm\Delta T[/itex] and [itex]Q_f = mH_f[/itex] where C is the specific heat (water = 4186 J/Kg*deg.K, ice = 2020 J/Kg*deg.K, and Hf = latent heat = 3340 J/Kg. Add all that up. That gives you Qc. Then use the relationship between Qc and COP to find Qh.

AM
 
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  • #9
Andrew Mason said:
Read the above responses again. We have given you all you need to know.
Use [itex]Q = Cm\Delta T[/itex] and [itex]Q_f = mH_f[/itex] where C is the specific heat (water = 4186 J/Kg*deg.K, ice = 2020 J/Kg*deg.K, and Hf = latent heat = 3340 J/Kg. Add all that up. That gives you Qc. Then use the relationship between Qc and COP to find Qh.
AM

I think that's what I will have to do...
I was unsure at first because I don't recall learning that, but I looked back in my textbook and it's here...

Well I'm going to see my professor today anyway, just in case.

Thanks :)
 

1. How does a refrigerator use heat energy to cool food?

A refrigerator uses a process called vapor compression to transfer heat energy from inside the refrigerator to the outside, thus cooling the inside. This process involves compressing a refrigerant gas, which increases its temperature. The hot gas then travels through coils on the back of the refrigerator, where it releases its heat energy and condenses into a liquid. The liquid then flows through an expansion valve, which causes it to evaporate and absorb heat from inside the refrigerator, thus cooling the air inside.

2. Does a refrigerator consume heat energy to cool food?

Yes, a refrigerator consumes heat energy to cool food. This is because the process of transferring heat from inside the refrigerator to the outside requires energy input, typically in the form of electricity. The refrigerator's compressor and other components use this energy to compress and circulate the refrigerant gas, thus creating the cooling effect.

3. How does a refrigerator maintain a constant temperature?

A refrigerator maintains a constant temperature through a thermostat, which senses the temperature inside the refrigerator and turns the compressor on and off as needed to maintain a set temperature. The thermostat also controls the flow of the refrigerant gas, allowing it to circulate and cool the inside of the refrigerator until the desired temperature is reached.

4. Can a refrigerator produce heat energy?

Yes, a refrigerator can produce heat energy. This is due to the fact that the process of transferring heat from inside the refrigerator to the outside results in the release of heat energy into the surrounding environment. Additionally, the compressor and other components of the refrigerator also generate heat as they operate. This is why refrigerators can feel warm to the touch on the outside even as they cool the inside.

5. How does the size and location of a refrigerator affect its use of heat energy?

The size and location of a refrigerator can affect its use of heat energy. A larger refrigerator will typically require more energy to cool the larger space, while a smaller refrigerator will require less energy. The location of the refrigerator is also important, as placing it in a warm or poorly ventilated area can cause it to use more energy to maintain a constant temperature. Additionally, keeping the refrigerator away from sources of heat, such as direct sunlight or a stove, can also help reduce its energy consumption.

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