Thermodynamic refrigerator Problems

In summary, the temperature inside a refrigerator could be as low as 23 degrees Celsius if the refrigerator had a coefficient of performance of 4.5.
  • #1
Norngpinky
13
0

Homework Statement


A restaurant refrigerator has a coefficient of performance of 4.5. If the temperature in the kitchen outside the refrigerator is 28^\circ C, what is the lowest temperature that could be obtained inside the refrigerator if it were ideal?

Homework Equations


According to textbook, COP=TL/(TH-TL)

The Attempt at a Solution


So wouldn't it be 4.5=TL/(28-TL)? And I got 23 degrees celsius but that's wrong...So I don't know where I'm wrong. ----------------------------------------------------------

Homework Statement


What is the change in entropy of 3.00 m^3 of water at 0 C when it is frozen to ice at 0 C?

Homework Equations


So I'm thinking the problem is saying what the change in entropy when water changes from solid to liquid...

If so, then change in entropy = Q/T = (m*heat of fusing)/T (kelvin)

The Attempt at a Solution


I'm stuck on on to convert 3.00 m^3 into mass... would that be the same as 3kg? I know 1kg=1L= either 1cm^3 or 1m^3?
----- If it's equal to 3kg...then change in entropy would be 3kg*333000J/kg / (273K) ?
still didn't get the right answer as 3660J/K

------

Homework Statement


A 120 g insulated aluminum cup at 19 C is filled with 150 g of water at 55 C. After a few minutes, equilibrium is reached.

Estimate the total change in entropy.

Homework Equations


I found final temperture to be 50 degrees, but I'm stuck on how to calculate the total change in entropy...

Totaly change in entropy = change in entropy of aluminum cup + change in entropy of the water

The Attempt at a Solution


S for al cup = Q/T= 900*.120kg*(50-19)/(273+50K) = 10.4

S for water = Q/T = 4186*.150*(55-50)/(273+50K) = 9.72

total S = 10.4+9.72=20.1 J/K

But it says that's wrong...
 
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  • #2
First problem: temperatures should be in Kelvins. (Always) It makes no sense to divide a temperature in degrees Celsius by anything.

Second problem: a liter is defined as a cubic decimeter. An easy way to remember this is that one milliliter is equal to one cubic centimeter.

Third problem: pay attention to signs. One of the cup or water is losing heat while the other is gaining heat, so one of the two will have a negative change in entropy.
 
  • #3
diazona said:
First problem: temperatures should be in Kelvins. (Always) It makes no sense to divide a temperature in degrees Celsius by anything.

Second problem: a liter is defined as a cubic decimeter. An easy way to remember this is that one milliliter is equal to one cubic centimeter.

Third problem: pay attention to signs. One of the cup or water is losing heat while the other is gaining heat, so one of the two will have a negative change in entropy.

I did what you said...

In the first problem... if I converted 28 degress celsius to kelvins it would be 301K
so then would we have 4.5= TL/(301-TL) ??

Second problem...I still don't get how to convert 3.00m^3 into cm^3 or L or Kg for that matter...

Third problem... so I did
S of cup = Q/T= m*c*change in temperature/final temp in kelvein = .12*900*(50-19)/(273+50) = 10.4

S of water = 9.72

10.4-9.72 ... still wrong answer
 
  • #4
First problem: try it!

Second problem: think about it like this: what is the mass of a given volume (in your case, 3 cubic meters) of water? You need the density of water to figure it out - do you know what the density of water is? (In any units?)

Third problem: You also need to take into account the fact that the temperature changes as the water and the cup come to equilibrium. So the T in the denominator changes; it's not 323K until the very end of the process. You'll either need to do this with an integral, or use another formula (if you have one that applies).
 
  • #5
Norngpinky said:
I did what you said...

In the first problem... if I converted 28 degress celsius to kelvins it would be 301K
so then would we have 4.5= TL/(301-TL) ??

Yes sir.

Norngpinky said:
Second problem...I still don't get how to convert 3.00m^3 into cm^3 or L or Kg for that matter...

Google it

Norngpinky said:
Third problem... so I did
S of cup = Q/T= m*c*change in temperature/final temp in kelvein = .12*900*(50-19)/(273+50) = 10.4

S of water = 9.72

10.4-9.72 ... still wrong answer

[tex]\Delta{S} = \int_1^2 \frac{Q}{T}dT[/tex]

So you have to perform the change in entropy for both the water and the aluminum cup Then add the resulting entropies.
 
  • #6
I had calulus I last semester so I sort of forgot how to do integral, but I shall go back on that tomorrow =)

As for the 2nd problem, the density would be 1000 kg/m^3 @ 0 degree celsius...so D=M/V...so.. the mass would be D times volume...so the mass would be 3kg? That sounds wrong.

And I got problem one.

Thank you for your help. i will work more on them tomorrow as it is already almost 2:30 in the morning.
 

1. How does a thermodynamic refrigerator work?

A thermodynamic refrigerator works by utilizing the principles of thermodynamics to transfer heat from a cold reservoir to a hot reservoir. This is achieved by using a refrigerant, which is a fluid that can change between liquid and gas states, to absorb and release heat as it moves through the refrigerator's system. The refrigerant is compressed, causing it to increase in temperature, and then passed through a condenser where it releases heat and changes back into a liquid. The liquid then goes through an expansion valve, which causes it to rapidly expand and cool down. This cold refrigerant is then passed through an evaporator, where it absorbs heat from the inside of the refrigerator, cooling it down. The refrigerant then repeats this cycle, continuously transferring heat from the inside of the refrigerator to the outside.

2. What is the Carnot cycle, and how is it related to thermodynamic refrigerators?

The Carnot cycle is a theoretical thermodynamic cycle used to describe the maximum possible efficiency of a heat engine or refrigerator. It consists of four reversible processes: isothermal compression, adiabatic expansion, isothermal expansion, and adiabatic compression. A thermodynamic refrigerator operates using a reversed Carnot cycle, where the roles of the hot and cold reservoirs are switched. This means that a thermodynamic refrigerator is most efficient when it operates using a Carnot cycle, but in reality, some energy is lost due to friction and other inefficiencies.

3. What factors affect the efficiency of a thermodynamic refrigerator?

The efficiency of a thermodynamic refrigerator is affected by several factors, including the temperature difference between the hot and cold reservoirs, the type of refrigerant used, and the design and components of the refrigerator. Additionally, the efficiency of a thermodynamic refrigerator is limited by the second law of thermodynamics, which states that heat cannot spontaneously flow from a colder body to a hotter body. This means that a thermodynamic refrigerator will never be able to cool the inside of the refrigerator to a temperature lower than the temperature of the outside environment.

4. What are the main challenges in designing a thermodynamic refrigerator?

One of the main challenges in designing a thermodynamic refrigerator is achieving high efficiency while keeping costs and energy consumption low. This requires careful selection of materials and components, as well as optimizing the design and operation of the refrigerator. Another challenge is finding a suitable refrigerant that is environmentally friendly and has a high enough boiling point to be effective in a thermodynamic refrigerator.

5. How does a thermodynamic refrigerator differ from a traditional refrigerator?

A traditional refrigerator, also known as a vapor-compression refrigerator, uses electricity to power a compressor that compresses a refrigerant, which then goes through a condenser and an evaporator to transfer heat and cool the inside of the refrigerator. In contrast, a thermodynamic refrigerator uses the principles of thermodynamics to transfer heat without the need for a compressor, making it more energy-efficient and environmentally friendly. However, thermodynamic refrigerators are currently not as widely used as traditional refrigerators due to their higher cost and more complex design.

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