# Ideal Gas Carnot engine refrigeration

• supermanii
In summary, the conversation discusses the relationship between heat absorbed and rejected by reservoirs at different temperatures, and the rate at which heat is added to the hot reservoir using a heat pump. The equation Qh = Wθh/(θh-θc) is derived and confirmed to be correct, with further explanation provided through the concept of coefficient of performance.
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.

supermanii said:
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.

supermanii said:
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 $\Delta Q = |Q_h|-|Q_c| = \Delta U + W = 0 + W = W$.

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

AM

Thanks :)

I would like to clarify that the question is asking about the relationship between heat absorbed and rejected by the reservoirs and their temperatures in an Ideal Gas Carnot engine refrigeration system. This system follows the Carnot cycle, which is a reversible thermodynamic cycle consisting of two isothermal and two adiabatic processes. The relationship between heat and temperature in this system is described by the Carnot efficiency, which is given by the equation:

η = (T_hot - T_cold) / T_hot

where T_hot and T_cold are the temperatures of the hot and cold reservoirs, respectively. This equation shows that the efficiency of the Carnot cycle is dependent on the temperature difference between the two reservoirs.

In terms of a heat pump, the rate at which heat is added to the hot reservoir can be calculated using the equation:

Q_hot = W * (T_hot / (T_hot - T_cold))

where W is the work done by the pump. This equation shows that the rate at which heat is added to the hot reservoir is directly proportional to the temperature of the hot reservoir and inversely proportional to the temperature difference between the two reservoirs. Therefore, the higher the temperature of the hot reservoir and the smaller the temperature difference, the more heat can be added to the hot reservoir by the heat pump.

I hope this clarifies the relationship between heat absorbed and rejected by the reservoirs and their temperatures in an Ideal Gas Carnot engine refrigeration system.

## What is an Ideal Gas Carnot engine refrigeration?

An Ideal Gas Carnot engine refrigeration is a theoretical refrigeration system that uses an ideal gas as the working fluid to transfer heat from a cold reservoir to a hot reservoir. It follows the Carnot cycle, which consists of four reversible processes: isothermal compression, adiabatic expansion, isothermal expansion, and adiabatic compression.

## How does an Ideal Gas Carnot engine refrigeration work?

The refrigeration process starts with the compression of the ideal gas at a constant temperature, which causes an increase in its pressure. Then, the gas is cooled at a constant pressure through the process of heat rejection to the cold reservoir. Next, the gas expands adiabatically, which causes a decrease in its temperature. Finally, the gas is heated at a constant pressure through the process of heat absorption from the hot reservoir. This cycle can be repeated to continuously transfer heat from the cold reservoir to the hot reservoir.

## What are the advantages of using an Ideal Gas Carnot engine refrigeration?

One advantage of using an Ideal Gas Carnot engine refrigeration is that it is a theoretical model, which means it is highly efficient and has no practical limitations. It also follows a reversible process, which means it can be run in reverse to act as a heat pump, providing both heating and cooling capabilities. Additionally, it uses a single working fluid, making it a simple and efficient system.

## What are the limitations of an Ideal Gas Carnot engine refrigeration?

One limitation of an Ideal Gas Carnot engine refrigeration is that it is a theoretical model and cannot be implemented in real-life systems. It also requires perfect insulation and reversible processes, which are not achievable in practical applications. Furthermore, it can only operate in a narrow range of temperatures, making it unsuitable for most refrigeration needs.

## How is the performance of an Ideal Gas Carnot engine refrigeration measured?

The performance of an Ideal Gas Carnot engine refrigeration is measured by its coefficient of performance (COP), which is the ratio of the heat transferred from the cold reservoir to the work done by the engine. The higher the COP, the more efficient the refrigeration system is. In an ideal scenario, the COP would be infinite, but in reality, it is limited by practical considerations.

• Introductory Physics Homework Help
Replies
1
Views
773
• Introductory Physics Homework Help
Replies
1
Views
959
• Introductory Physics Homework Help
Replies
3
Views
928
• Thermodynamics
Replies
1
Views
764
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
765
• Introductory Physics Homework Help
Replies
14
Views
8K
• Introductory Physics Homework Help
Replies
33
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K