Ideal Gas Carnot engine refrigeration

[supermanii]

Homework Statement

What is the relationship between the heat absorbed and rejected by the reservoirs and their temperature. If a heat pump is used to transfer heat what is the rate at which heat is added to the hot reservoir in terms of the temperatures of the reservoir and the work done by the pump?

Homework Equations

Answer to the first part is \frac{Q_{h}}{Q_{c}}=\frac{\theta_{h}}{\theta_{c}}

The Attempt at a Solution

I got a solution of Q_{h}=\frac{W\theta_{h}}{\theta_{h}-\theta_{c}} however I am very unsure of this and think I went about it the wrong way. Not even sure I have answered the question. Without a correct answer I can not do the rest of the question.

 

[collinsmark]

I got a solution of Q_{h}=\frac{W\theta_{h}}{\theta_{h}-\theta_{c}} however I am very unsure of this and think I went about it the wrong way. Not even sure I have answered the question. Without a correct answer I can not do the rest of the question.

Looks fine to me. (I'm assuming you are using the symbol θ to represent temperature.)

But if you'd like a more formal discussion on the subject, look in your textbook or coursework for coefficient of performance (COP). It will explain the concept probably better than what I can do here.

 

[Andrew Mason]

I got a solution of Q_{h}=\frac{W\theta_{h}}{\theta_{h}-\theta_{c}} however I am very unsure of this and think I went about it the wrong way.

Your answer is correct. Here's why: Since the system returns to its initial state after a complete cycle, there is no change internal energy so \Delta Q = |Q_h|-|Q_c| = <br /> \Delta U + W = 0 + W = W.

If you are using a Carnot heat pump, Qh/Qc = Th/Tc. So Qh/W = Qh/(|Qh|-|Qc|) = Th/(Th-Tc)

AM

 

[supermanii]

Thanks :)