Find Coefficient of Performance of a refrigerator

In summary, the problem at hand involves finding the coefficient of a refrigerator that uses a monatomic gas and only has two steps. The known parameters are the initial pressure, initial temperature, initial volume, and final pressure. The challenge in this problem is that most formulas require four steps or knowledge of the outside and inside temperature, which is not provided. The only formula that can potentially be used is ##COP = \frac{Q_C}{W}##, but it may not be accurate due to the lack of information. The student has attempted to use this formula and got a COP of 1, but later realized that it may not be the right approach. They are now trying to compare their result with the one from the Carnot
  • #1
grandpa2390
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Homework Statement


I need to find the coefficient of a refrigerator that uses a monatomic gas and only has two steps.
I know the:
initial pressure of the refrigerant
initial temperature
initial volume
and final pressure

Every formula I know requires 4 steps or a knowledge of the outside and inside temperature. This question is unlike anything I have seen to date. The only other problem I have seen like this was for a real refrigerator. it had four steps, and I was given a table for the saturate refrigerant.

Homework Equations


##COP = \frac{Q_C}{W}##
##COP = \frac{Q_C}{Q_H-Q_C}##
##COP = \frac{T_C}{T_H-T_C}##

The Attempt at a Solution



Right now, the only thing I can think of is that I could use the equation ##COP = \frac{Q_C}{W}## by finding Q and W of the process the way I did in earlier chapters. And assume that the change in Q is Q_C .
With this attempt, I could take advantage of the formulas that give me these values based on whether or not the gas is monatomic.

Is this the right direction?

edit: I did the problem this way and got a COP of 1...
I assuming that the change in internal energy is 0. the first process is isothermal, and the second process is linear bringing it back to the same initial point. I get that Q=-W so Q/W is 1
 
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  • #2
edit: I see. I think I see my current mistake. It isn't Q of the entire process. Just the Q_in. So I got a number for the coefficient and I'm onto the next part. I need to compare it into the result I would get from the carnot cycle at the same temperature difference. What I don't get is what the temperature difference is.
All I am given is the temperature of the gas and that the first step is isothermal.
 
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Related to Find Coefficient of Performance of a refrigerator

1. What is the coefficient of performance (COP) of a refrigerator?

The coefficient of performance (COP) of a refrigerator is a measure of its efficiency in removing heat from a space. It is calculated by dividing the heat removed from the space by the work required to remove that heat. A higher COP indicates a more efficient refrigerator.

2. How is the COP of a refrigerator determined?

The COP of a refrigerator is determined by measuring the amount of heat removed from the space and the amount of work or energy input required to remove that heat. This can be done through experimental testing or by using known values for the heat and work.

3. What factors affect the COP of a refrigerator?

The COP of a refrigerator can be affected by various factors such as the design and size of the refrigerator, the type of refrigerant used, the temperature difference between the inside and outside of the refrigerator, and the efficiency of the compressor and other components.

4. How can the COP of a refrigerator be improved?

The COP of a refrigerator can be improved by using more efficient components, such as a better compressor or a more advanced refrigerant. Proper maintenance and regular cleaning of the refrigerator can also help to improve its efficiency.

5. How does the COP of a refrigerator compare to other cooling systems?

The COP of a refrigerator is typically higher than other cooling systems, such as air conditioners, because it is designed specifically for removing heat from a small space. However, the COP can vary depending on the specific design and efficiency of each system.

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