OUNT OF WORK INPUT FOR A CARNOT REFRIGERATOR

• swain1
In summary: Thanks for that. I understand that now. I got the answer 8.3E6 J which I hope is right. For part b) would the amount of power used just be (8.3E6)/0.65=1.3E7J or am I missing something?For part b) the work input would be 1.3e7 J.
swain1

Homework Statement

a) Assume that a Carnot cycle is used to achieve that 250kg of water at 0 degrees celsius is frozen into ice using an ideal refigerator The room temperature is 27 degrees. What is the minimum amount of work input to the refrigerator to achieve this?

b)Assume the refigerator works with 65% efficiency (compare to a carnot engine), how much work input does it require.

Homework Equations

Latent heat of melting ice =3.33E5 J/kg

The Attempt at a Solution

For part a) I don't see why the amount of work input is not just that required to turn the water to ice, which would be 3.33E5*250. I think I might be missing something as the room temperature is given. I haven't yet started the second part as I think I need to understand the first part. Any help would be appreciated.

swain1 said:

Homework Statement

a) Assume that a Carnot cycle is used to achieve that 250kg of water at 0 degrees celsius is frozen into ice using an ideal refigerator The room temperature is 27 degrees. What is the minimum amount of work input to the refrigerator to achieve this?

b)Assume the refigerator works with 65% efficiency (compare to a carnot engine), how much work input does it require.

Homework Equations

Latent heat of melting ice =3.33E5 J/kg

The Attempt at a Solution

For part a) I don't see why the amount of work input is not just that required to turn the water to ice, which would be 3.33E5*250. I think I might be missing something as the room temperature is given. I haven't yet started the second part as I think I need to understand the first part. Any help would be appreciated.
The amount of HEAT removed from the water has to be this amount. This is not the amount of work input, however.

The refrigerator has to remove heat from a 0 degree reservoir and deliver it as heat at 27 degrees. In order to do that, work must be input. If it was a Carnot cycle, the coefficient of performance would be:

$$COP = Q_c/W = Q_c/(Q_h-Q_c) = \frac{T_c/T_h}{1 - T_c/T_h}$$

In this case, what is the COP?

That will give you Qh and that will, in turn, give you W.

AM

Ok I see that puting in the temps into the formula give the COP. Which I have found to be 10.1. I am not sure where this comes from though.
How am I meant to find the value of Qc so I can work out the value of W.

So for the first part I don't need the latent heat of melting ice? Thanks

swain1 said:
Ok I see that puting in the temps into the formula give the COP. Which I have found to be 10.1. I am not sure where this comes from though.
COP is Qc/W (the ratio of "cooling" or heat removal to work input). For a Carnot cycle heat flow is at constant temperature and $\Delta S = Q_h/T_h - Q_c/T_c = 0$ so Qc/Qh = Tc/Th
How am I meant to find the value of Qc so I can work out the value of W.

So for the first part I don't need the latent heat of melting ice? Thanks
Qc is the heat removed from the water which you have correctly stated as the mass x latent heat/mass

AM

Thanks for that. I understand that now. I got the answer 8.3E6 J which I hope is right. For part b) would the amount of power used just be (8.3E6)/0.65=1.3E7J or am I missing something?

swain1 said:
Thanks for that. I understand that now. I got the answer 8.3E6 J which I hope is right. For part b) would the amount of power used just be (8.3E6)/0.65=1.3E7J or am I missing something?
For part a) I get 8.23E6 J:

$$W = Q_c/COP = 3.33e6*250/(273/27) = 8.23e6 J$$You have the right approach to b). If the real refrigerator (b) was 65% as efficient as the Carnot, it would require 1/.65 of the work needed in the Carnot cycle (COP would be .65 * COP of the Carnot). So the work would be 1.27e7 J.

AM

1. What is a Carnot Ideal Refrigerator?

A Carnot Ideal Refrigerator is a theoretical refrigeration system that operates on the Carnot cycle, which is a reversible thermodynamic cycle. It is considered an ideal refrigeration system because it operates at maximum efficiency and has no internal losses.

2. How does a Carnot Ideal Refrigerator work?

A Carnot Ideal Refrigerator works by transferring heat from a low-temperature reservoir to a high-temperature reservoir using a refrigerant. It follows the Carnot cycle, which consists of four processes: isothermal compression, adiabatic compression, isothermal expansion, and adiabatic expansion.

3. What makes the Carnot Ideal Refrigerator more efficient than other refrigeration systems?

The Carnot Ideal Refrigerator is more efficient because it operates at maximum efficiency, meaning it has the highest possible coefficient of performance (COP). This is due to the reversible nature of the Carnot cycle, which minimizes internal losses and maximizes the efficiency of heat transfer.

4. What are some real-life applications of the Carnot Ideal Refrigerator?

The Carnot Ideal Refrigerator is a theoretical concept and is not used in practical applications. However, it serves as a benchmark for measuring the performance of real refrigeration systems and can be used to optimize their efficiency.

5. Are there any limitations to the Carnot Ideal Refrigerator?

One limitation of the Carnot Ideal Refrigerator is that it is a theoretical concept and cannot be achieved in practice. It also assumes ideal conditions and does not take into account factors such as friction, heat transfer through walls, and non-ideal gas behavior. Additionally, the refrigerant used in the Carnot cycle would need to have a very high boiling point, which is not practical for real-world applications.

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