# Work required to extract heat from a refrigerator

• guyvsdcsniper
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.
guyvsdcsniper
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?

Delta2
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?

BvU and guyvsdcsniper
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.

BvU

## 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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