Thermodynamic refrigerator Problems

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Homework Help Overview

The discussion revolves around thermodynamic problems related to refrigerators, entropy changes in phase transitions, and heat exchange in insulated systems. Participants are exploring concepts in thermodynamics, specifically focusing on the coefficient of performance, entropy calculations, and the principles of heat transfer.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply the coefficient of performance formula to determine the lowest temperature in a refrigerator setup. Questions arise about the correct temperature units and the implications of using Celsius versus Kelvin.
  • In the entropy change problem, there is confusion regarding the conversion of volume to mass and the correct application of the entropy formula, with participants questioning their assumptions about the density of water.
  • For the insulated cup and water system, participants are calculating changes in entropy but are unsure about the signs of heat transfer and the changing temperature during the process.

Discussion Status

Some participants have provided guidance on unit conversions and the need to consider density in calculations. There is an ongoing exploration of the correct methods for calculating entropy changes, and participants are actively questioning their approaches without reaching a consensus on the correct answers.

Contextual Notes

Participants are working within the constraints of homework assignments, which may impose specific requirements for unit usage and problem-solving approaches. There is a noted lack of clarity regarding the application of temperature in calculations and the assumptions about material properties.

Norngpinky
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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...

totally 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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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.
 
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
 
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).
 
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

\Delta{S} = \int_1^2 \frac{Q}{T}dT

So you have to perform the change in entropy for both the water and the aluminum cup Then add the resulting entropies.
 
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.
 

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