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. ---------------------------------------------------------- 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*(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...
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).
Yes sir. Google it [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.
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.