Work required to extract heat from a refrigerator

[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?

 

[vela]

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.

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 $|Q_h| = |W| + Q_c$.

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?

 

[guyvsdcsniper]

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.