# Thermodynamics Problems

1. ### Norngpinky

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/(TH-TL)

3. 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.

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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

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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*(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...

Last edited: May 7, 2009
2. ### diazona

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. ### Norngpinky

13
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

4. ### diazona

2,155
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. ### djeitnstine

615
Yes sir.

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

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

6. ### Norngpinky

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.