Work required to extract heat from a refrigerator

In summary, there is a discussion about a reversible refrigerator engine and the calculation of work required to extract one calorie from the interior of the refrigerator compartment. The conversation includes a mention of the answer key saying the work should be +.043cal, but the person getting a negative sign. They explain their logic and equations used, but it is pointed out that there may be a mistake in their calculations. The conversation concludes with the person realizing their mistake and thanking the expert for their help.
  • #1
guyvsdcsniper
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Homework Statement
A reversible refrigerator engine extracts heat from
the inside of a refrigerator compartment kept at 8°C and rejects unwanted
heat QH to its 20°C exterior. Find the work required to extract one calorie
from the interior of the refrigerator compartment
Relevant Equations
dQ/T=0
The answer key says the work done should be +.043cal. I am getting a negative sign.

I have posted my work in the attached image.
My logic is that since this a reversible engine we can say the integral of dQ/T=0. Looking at the cycle, Qc isi being absorbed (Qc>0) and Qh is being rejected (Qh<0). Then we can say Qc/Tc-Qh/Th=0.

Solving for Qh, we get 1.0427cal.
We also know W= the integral of dQ. So we can say W=Qc-Qh.

Doing so makes me get a negative sign.

I guess from my perspective the negative indicates the system is doing work, hence W<0.

Am I doing something wrong/looking at the situation from a different perspective by getting this negative sign?
IMG_0347.jpg
 
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  • #2
quittingthecult said:
Homework Statement:: A reversible refrigerator engine extracts heat from
the inside of a refrigerator compartment kept at 8°C and rejects unwanted
heat QH to its 20°C exterior. Find the work required to extract one calorie
from the interior of the refrigerator compartment
Relevant Equations:: dQ/T=0

The answer key says the work done should be +.043cal. I am getting a negative sign.

I have posted my work in the attached image.
My logic is that since this a reversible engine we can say the integral of dQ/T=0. Looking at the cycle, Qc isi being absorbed (Qc>0) and Qh is being rejected (Qh<0). Then we can say Qc/Tc-Qh/Th=0.
If ##Q_c>0## and ##Q_h<0##, then ##\frac{Q_C}{T_C}-\frac{Q_H}{T_H}## will be positive. You should either have a plus sign in between or use absolute values.

quittingthecult said:
Solving for Qh, we get 1.0427cal.
We also know W= the integral of dQ. So we can say W=Qc-Qh.
Look at the arrows in your diagram. The energy coming into the system is ##|W|+Q_c##, and the energy going out is ##|Q_h|##. Conservation of energy requires ##|Q_h| = |W| + Q_c##.

quittingthecult said:
Doing so makes me get a negative sign.

I guess from my perspective the negative indicates the system is doing work, hence W<0.
The usual convention is if the system does work on the environment, ##W## is positive. Are you using a different convention or did you make a mistake?
 
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  • #3
vela said:
If ##Q_c>0## and ##Q_h<0##, then ##\frac{Q_C}{T_C}-\frac{Q_H}{T_H}## will be positive. You should either have a plus sign in between or use absolute values.Look at the arrows in your diagram. The energy coming into the system is ##|W|+Q_c##, and the energy going out is ##|Q_h|##. Conservation of energy requires ##|Q_h| = |W| + Q_c##.The usual convention is if the system does work on the environment, ##W## is positive. Are you using a different convention or did you make a mistake?
Ah your right. I overthought this way too hard. I should have just looked at the arrows.

Thank you, your response has helped me realize the mistake i made.
 
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Related to Work required to extract heat from a refrigerator

1. What is the purpose of extracting heat from a refrigerator?

The purpose of extracting heat from a refrigerator is to maintain a cool temperature inside the refrigerator. This is necessary for preserving food and drinks and preventing them from spoiling.

2. How does the process of extracting heat from a refrigerator work?

The process of extracting heat from a refrigerator involves the use of a refrigerant, which is a chemical substance that can absorb and release heat. The refrigerant is compressed and then expanded, causing it to absorb heat from inside the refrigerator and release it outside.

3. What factors affect the amount of work required to extract heat from a refrigerator?

The amount of work required to extract heat from a refrigerator is affected by the temperature difference between the inside and outside of the refrigerator, the efficiency of the refrigerant, and the design and condition of the refrigerator itself.

4. How is the work required to extract heat from a refrigerator measured?

The work required to extract heat from a refrigerator is measured in units of energy, such as joules or kilowatt-hours. This can be calculated by multiplying the amount of heat extracted by the temperature difference and the efficiency of the refrigerant.

5. Can the work required to extract heat from a refrigerator be reduced?

Yes, the work required to extract heat from a refrigerator can be reduced by using more efficient refrigerants, improving the design and insulation of the refrigerator, and minimizing the temperature difference between the inside and outside of the refrigerator.

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