# Carnot refrigerator and work done

• sportsrules
In summary: ENDMENT: In summary, a heat pump is like a refrigerator engine that uses the inside and outside of a building as hot reservoirs. For a Carnot process between 25 C and 40 C, the coefficient of performance is 20.86. For each Joule of heat transferred from the cooler reservoir, the Carnot refrigerator engine requires 0.048 Joules of work input.
sportsrules
The question reads...
A heat pump is essentially a refrigerator engine that uses the inside of a building as a hot reservoir in the winter and the outside of the building as the hot reservoir in the summer. A) For a carnot refrigerator engine operating between 25 C and 40 C in the summer, what is the coefficient of performance? B) For each joule of heat transfer form the cooler reservoir per cycle, how many joules of work are done by the carnot refrigerator engine?

Ok, I figured out part A, which was simply plugging the numbers into the equation

K=1/[(Thot/Tcold)-1]...and I got that K=19.9

I can't really figure out part B. I know that there is another equation

K=Qc/W
and
K=1/[(Qh/Qc)-1]

but I cannot figure out how to rearrange get an answer for the work b/c I don't know what Qc is or how to go about finding it. Any help would be great! Thanks!

sportsrules said:
The question reads...
A heat pump is essentially a refrigerator engine that uses the inside of a building as a hot reservoir in the winter and the outside of the building as the hot reservoir in the summer. A) For a carnot refrigerator engine operating between 25 C and 40 C in the summer, what is the coefficient of performance? B) For each joule of heat transfer form the cooler reservoir per cycle, how many joules of work are done by the carnot refrigerator engine?
For a heat pump, the coeff. of perf. is the ratio of heat output (heat transferred from the cold reservoir to the hot reservoir) to the work input.

$$COP = \frac{Q_H}{W}$$

So:

$$W = Q_H/COP$$

$$COP = Q_H/W = Q_H/Q_H-Q_C = \frac{1}{1-\frac{Q_C}{Q_H}}$$

For the Carnot process (reversible so $\Delta S = 0$), the heat is exchanged at constant temperature, so:

$$\Delta S = Q_h/T_h - Q_c/T_c = 0$$

$$Q_h/T_h = Q_c/T_c$$

$$Q_h/Q_c = T_h/T_c$$

So: $$COP = \frac{1}{1-\frac{T_C}{T_H}}$$

COP = 1/(1-298/313) = 20.86

So for each Joule of heat delivered, W = 1/20.86 = .048 Joules of work input are needed

AM

Last edited:

A) The coefficient of performance (K) for a Carnot refrigerator engine operating between 25 C and 40 C in the summer is calculated using the equation K=1/[(Thot/Tcold)-1]. Plugging in the values, we get K=19.9. This means that for every joule of work input, the refrigerator can transfer 19.9 joules of heat from the cold reservoir to the hot reservoir.

B) To find the work done by the refrigerator engine, we can use the equation K=Qc/W, where Qc is the heat transferred from the cold reservoir and W is the work done. Rearranging this equation, we get W=Qc/K. We already know that K=19.9, so now we just need to find Qc.

To find Qc, we can use the equation K=1/[(Qh/Qc)-1]. Rearranging this equation, we get Qc=Qh/(K+1). We know that Qh is the heat transferred from the hot reservoir, which is equal to the work input. So, Qh=W. Plugging this into the equation for Qc, we get Qc=W/(K+1).

Now, we can plug in the values for W and K to find Qc. W is equal to 1 joule (since we are looking at the work done per joule of heat transfer from the cold reservoir) and K is 19.9. Therefore, Qc=1/(19.9+1)=0.0499 joules.

So, for every joule of heat transfer from the cold reservoir, the Carnot refrigerator engine does 0.0499 joules of work.

## What is a Carnot refrigerator?

A Carnot refrigerator is a type of refrigeration system that uses a reversible thermodynamic cycle to transfer heat from a cold source to a hot source, thereby cooling the cold source. It is based on the principles of thermodynamics and is known for its maximum theoretical efficiency.

## How does a Carnot refrigerator work?

A Carnot refrigerator works by using a reversible thermodynamic cycle, which involves four processes: isothermal compression, adiabatic expansion, isothermal expansion, and adiabatic compression. During these processes, the refrigerant (usually a gas) absorbs heat from the cold source, cools it down, and then releases the heat to the hot source, thus cooling the cold source further.

## What is the efficiency of a Carnot refrigerator?

The efficiency of a Carnot refrigerator is given by the Carnot efficiency formula, which is the ratio of the absolute temperature difference between the hot and cold sources to the absolute temperature of the hot source. This means that the efficiency of a Carnot refrigerator can only be 100% if the hot source is at an infinite temperature, which is not possible in practical applications.

## What is the work done by a Carnot refrigerator?

The work done by a Carnot refrigerator is equal to the difference between the heat absorbed from the cold source and the heat released to the hot source. This work done is used to drive the refrigeration cycle and is dependent on the efficiency of the refrigerator.

## What are the limitations of a Carnot refrigerator?

One of the main limitations of a Carnot refrigerator is its theoretical nature, as it assumes perfect reversibility and no energy losses. In practical applications, these assumptions are not possible, and the efficiency of a Carnot refrigerator is much lower than its theoretical maximum. Additionally, the efficiency of a Carnot refrigerator decreases with an increase in the temperature difference between the hot and cold sources, making it less efficient for cooling large temperature differences.

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