# Carnot refrigerator cost problem

• RockJockey
In summary: That should help you get started. In summary, in order to make 2 kg of ice at 0° C using a Carnot refrigerator with a freezer compartment temperature of -13.5° C and a kitchen temperature of 24.5° C, the COP needs to be calculated using the formula COP=Qc/W. The cost of electrical energy is ten cents per kilowatt · hour and the conversion factor for kilowatt · hours to Joules is 3,600,000 = 1 kilowatt hour.
RockJockey

## Homework Statement

Three kilograms of liquid water at 0° C is put into the freezer compartment of a Carnot refrigerator. The temperature of the compartment is -13.5° C, and the temperature of the kitchen is 24.5° C. If the cost of electrical energy is ten cents per kilowatt · hour, how much does it cost to make two kilograms of ice at 0° C?

## Homework Equations

COP= Qc/W , Qc/Qh=TL/Th , Q=mL L=3.35 x 10^5 j/kg , Qh= W+ Qc, 3,600,000 = 1 kilowatt hour

## The Attempt at a Solution

I used the formula Q=mL to solve for Qc, because this is the heat that needs to be moved to the hot reservoir, got an answer of Qc=1005000J. I then used Qc/Qh = TL/Th to solve for Qh came out to be 1149885.584j. I then used the Qh=W+Qc to solve for W, which equaled 144885.584j. I then solved for the COP = 6.937. I did convert temp values to K. I'm lost after this point I need to know how to come up with the cost.

RockJockey said:

## Homework Statement

Three kilograms of liquid water at 0° C is put into the freezer compartment of a Carnot refrigerator. The temperature of the compartment is -13.5° C, and the temperature of the kitchen is 24.5° C. If the cost of electrical energy is ten cents per kilowatt · hour, how much does it cost to make two kilograms of ice at 0° C?

## Homework Equations

COP= Qc/W , Qc/Qh=TL/Th , Q=mL L=3.35 x 10^5 j/kg , Qh= W+ Qc, 3,600,000 = 1 kilowatt hour

## The Attempt at a Solution

I used the formula Q=mL to solve for Qc, because this is the heat that needs to be moved to the hot reservoir, got an answer of Qc=1005000J.
The problem statement says that you only need to make 2 kg of ice. Even though you put 3 kg of water in the freezer, you only need to make 2 kg of ice (the other 1 kg can remain as water). So you'll have to redo your Qc calculation.
I then used Qc/Qh = TL/Th to solve for Qh came out to be 1149885.584j. I then used the Qh=W+Qc to solve for W, which equaled 144885.584j. I then solved for the COP = 6.937.
What I would have done is first solved for the COP based on Thot and Tcold. Once you have the COP, you can easily solve for the work W using your COP= Qc/W formula. My way is a little easier, but your method works just fine too. But either way, you'll have to redo the calculations using only 2 kg of ice (instead of 3) this time.
I'm lost after this point I need to know how to come up with the cost.
Once you find the work in Joules, you have to convert that to kilowatt · hours.

Joules are a measure of energy. kilowatt · hours are also a measure of energy (just different units).

Here is a hint. 1 Watt = 1 Joule/second.

## What is the Carnot refrigerator cost problem?

The Carnot refrigerator cost problem is a theoretical problem in thermodynamics that aims to determine the minimum cost of operating a refrigerator. It is based on the Carnot cycle, which is a theoretical thermodynamic cycle that represents the most efficient way to convert heat into work.

## How does the Carnot refrigerator cost problem relate to real-life refrigerators?

The Carnot refrigerator cost problem is a theoretical concept that is used to understand the thermodynamic principles behind refrigeration. It does not directly relate to real-life refrigerators, but it serves as a benchmark for evaluating the efficiency and cost-effectiveness of refrigeration systems.

## What factors affect the cost of a Carnot refrigerator?

The cost of a Carnot refrigerator is affected by the temperature difference between the hot and cold reservoirs, the efficiency of the refrigerator, and the cost of the materials and components used in its construction. Additionally, the operating conditions and maintenance costs can also impact the overall cost of a Carnot refrigerator.

## What is the minimum cost of a Carnot refrigerator?

The minimum cost of a Carnot refrigerator is determined by the Carnot efficiency, which is the maximum theoretical efficiency of a refrigerator. This efficiency is calculated by dividing the absolute temperature of the hot reservoir by the difference between the absolute temperatures of the hot and cold reservoirs. The lower the temperature difference, the higher the Carnot efficiency and the lower the minimum cost of a Carnot refrigerator.

## How can the cost of a Carnot refrigerator be reduced?

The cost of a Carnot refrigerator can be reduced by improving its efficiency, reducing the temperature difference between the hot and cold reservoirs, and using cost-effective materials and components. Additionally, regular maintenance and proper operation can also help to reduce the overall cost of a Carnot refrigerator. Advances in technology and materials may also lead to more cost-effective refrigeration systems in the future.

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