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Thermodynamics Problemsby Norngpinky
Tags: thermodynamics 
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#1
May709, 01:23 AM

P: 13

1. The problem statement, all variables and given/known data
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? 2. Relevant equations According to textbook, COP=TL/(THTL) 3. The attempt at a solution So wouldn't it be 4.5=TL/(28TL)? And I got 23 degrees celsius but that's wrong...So I don't know where I'm wrong.  1. The problem statement, all variables and given/known data 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? 2. Relevant 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) 3. 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  1. The problem statement, all variables and given/known data 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. 2. Relevant 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 entrophy = change in entrophy of aluminum cup + change in entrophy of the water 3. The attempt at a solution S for al cup = Q/T= 900*.120kg*(5019)/(273+50K) = 10.4 S for water = Q/T = 4186*.150*(5550)/(273+50K) = 9.72 total S = 10.4+9.72=20.1 J/K But it says that's wrong... 


#2
May709, 01:30 AM

HW Helper
P: 2,155

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
May709, 01:47 AM

P: 13

In the first problem... if I converted 28 degress celsius to kelvins it would be 301K so then would we have 4.5= TL/(301TL) ?? 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*(5019)/(273+50) = 10.4 S of water = 9.72 10.49.72 ... still wrong answer 


#4
May709, 01:56 AM

HW Helper
P: 2,155

Thermodynamics Problems
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
May709, 02:00 AM

PF Gold
P: 619

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


#6
May709, 02:20 AM

P: 13

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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