What is the coefficient of performance of this refrigerator?

In summary: OUNT OF HEAT PER CYCLE is Qh - QcIn summary, the conversation involves a student seeking help with physics problems for an upcoming exam. They mention struggling with thermodynamics and provide four specific questions they need help with. They also mention getting unexpected results on a previous assignment and ask for assistance.
  • #1
sltungle
27
0
Okay, this is a long post so buckle up (not sure if this counts as Homework Help or not - technically it WAS homework, but now it's basically exam revision).

I was wondering if anyone would mind helping me with a few physics problems I have? It was originally for a mastering physics assignment due last Monday, but I didn't get around to finishing all of the problems and now I want to understand how to do them for my upcoming physics exam.

Normally I wouldn't require this great a deal of help - I'm normally incredibly good at physics, I even managed to get high distinctions in our physics course last semester which covered optics, quantum mechanics, and special relativity. Unfortunately thermodynamics seems to be tripping me up a bit, though.

Some of the problems I had in Mastering Physics are as follows:

Question 1

At what pressure ratio would a heat engine operating with a Brayton cycle have an efficiency of 80.0? Assume that the gas is diatomic.

I tried going:

Efficiency = 1 - Qc/Qh, or 1 - Tc/Th with efficiency as 0.80, and then subbing in (pV/nR)cold and (pV/nR)hot for Tc/h but I'm not sure I approaching the problem right. I didn't finish this question.

Question 2

A Carnot refrigerator operating between -20.0 deg and 30.0 degrees Celsius extracts heat from the cold reservoir at the rate 100 J/s. What are (a) the coefficient of performance of this refrigerator, (b) the rate at which work is done on the refrigerator and (c) the rate at which heat is exhausted to the hot side?

What is the coefficient of performance of this refrigerator?
5.06
Correct


What is the rate at which work is done on the refrigerator?
Exhausted 3 of 4 attempts for this question. Can't remember my working unfortunately.

What is the rate at which heat is exhausted to the hot side?
Used 1 of 4 attempts.

This entire question is making me angry at myself. I've always been good at physics but just messing this whole last mastering physics up demoralized me greatly. Any help on any of the problems for that question would be greatly appreciated.

Question 3

250 ml of water at 20.0 degrees Celsius is placed in the freezer compartment of a refrigerator with a coefficient of performance of 4.00. How much heat energy is exhausted into the room as the water is changed to ice at -20.0 degrees Celsius?

Got 0% for that problem. Tried using Q = mcΔT with ΔT as both 40 degrees AND 20 degrees and then relating it to COP = Qc/Qh-Qc, but that got me nowhere. Help on working through it would be appreciated.

Question 4

The power output of a car engine running at 3000 rpm is 500 kW.

How much work is done per cycle if the engine's thermal efficiency is 40.0?Give your answer in kJ.
10.0 kJ
Correct


How much heat is exhausted per cycle if the engine's thermal efficiency is 40.0?Give your answer in kJ.

Gave this one an attempt but didn't get it right

Despite having got the first answer correct (10.0 kJ), upon explaining to a friend how to work through the same problem he used the exact same steps as myself (with different values of course) and Mastering Physics said his answer was wrong. I'm a bit concerned that I fluked the answer after that. My steps were:

1) Convert 3000 rpm to cycles per second (3000/60 = 50 cycles/sec or 50 rps if you will).

2) Divide the output (500 kW or 500 kJ/s) by how many cycles there are per second (which when I look at the units should equal kJ/cycle).

Again - I got the question correct with these two steps, my friend got it wrong.

Sorry for the long post. I'd be happy for any help at all, even if its only one one sub question.
 
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  • #2
Anyone? Like I said, even help with one of the sub questions would be really appreciated (especially the last one - I can't see why I got it correct, but my friend didn't).
 
  • #3
sltungle said:
Question 1

Efficiency = 1 - Qc/Qh, or 1 - Tc/Th with efficiency as 0.80, and then subbing in (pV/nR)cold and (pV/nR)hot for Tc/h but I'm not sure I approaching the problem right. I didn't finish this question.
Look up the equation for the efficiency of an ideal Brayton cycle. Set that to .8 and determine what P2/P1 is (using [itex]\gamma=C_p/C_v = ??[/itex] for a diatomic gas such as air).
Question 2
b)Use COP = Qc/W (= Tc/(Th - Tc) = 253/50 = 5.06)

If Qc = 100 J/s what is W?

c) Since W = Qh-Qc, what is Qh? (using the W you determined in b))
Question 3

Got 0% for that problem. Tried using Q = mcΔT with ΔT as both 40 degrees AND 20 degrees and then relating it to COP = Qc/Qh-Qc, but that got me nowhere. Help on working through it would be appreciated.
You have to take into account the latent heat of fusion of water and the heat capacity of ice (it is different than the heat capacity of liquid water).
Question 4

You have found the work per cycle, W. Efficiency = output/input = W/Qh

So if efficiency = .4 what is Qh?

AM
 

What is the coefficient of performance of this refrigerator?

The coefficient of performance (COP) of a refrigerator is the ratio of the amount of heat removed from the refrigerated space to the amount of work required to remove that heat. It is a measure of the efficiency of a refrigerator.

How is the coefficient of performance calculated?

The coefficient of performance is calculated by dividing the amount of heat removed from the refrigerated space by the work input. This can be expressed as COP = Q/W, where Q is the heat removed and W is the work input.

What is a good coefficient of performance for a refrigerator?

A good coefficient of performance for a refrigerator is typically considered to be above 3. This means that for every unit of work put into the refrigerator, it removes at least 3 units of heat from the refrigerated space.

How can the coefficient of performance be improved?

The coefficient of performance can be improved by using more efficient components and technologies in the refrigerator, such as better insulation, compressors, and heat exchangers. Regular maintenance and proper usage can also help improve the COP.

What are some factors that can affect the coefficient of performance of a refrigerator?

The coefficient of performance of a refrigerator can be affected by factors such as the ambient temperature, the temperature of the refrigerated space, the type and quality of components, and the age and condition of the refrigerator. Energy efficiency ratings and proper installation and usage can also impact the COP.

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