# Thermodynamics, Check my work.

1. Dec 16, 2009

### Dan Feerst

I've never had to do one quite like this. Just wanted to check my work.

1. The problem statement, all variables and given/known data
150g of water and 50g of ice are in a container made from 20.0g of aluminum. The system consisting of the water, ice and the container are at thermal equilibrium at 0C. The container is wrapped in a thermal insulation that isolates it from the environment. A water warmer rated at 200 watts is placed in the container and switched on. The contents of the container are gently stirred to keep the temperature of the system uniform.

a. Determine the amount of heat(in joules) required to raise the temperature of this system to 100C (Lf=333J/g, Cw=4.186J/gK, CAl=0.900J/gK)

b. How long will it take the warmer to heat the mixture to 100C. Show Calculation

c. After the system has been heated to 100C, and additional 50.0g of ice at 0C are added. What is the final temperature of the mixture? Show work.

2. Relevant equations
Q=mCdT

3. The attempt at a solution
Here is what i did. Sorry I used d in place of lambda, and omitted degree symbols. not sure how to work this thing.

a.
mwCwdTw+miCwdTi+MAlCAldTAl+miLf
=> 100(4.186(50+150)+20*0.9)+50*333=Qs=102170J

b.
w=j/s =>
102170J=200t
t=511s

c.

I'm really unsure of this part. I tried to treat all three components of the original system as one.
msCsdTs=miCidTi+miLf
(20+150+50)(4.186+0.900)(300-3Tf)=45Tf+16650
595302=Tf(45+6119.52) =>Tf=97.6C

Well, am I at least close?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 16, 2009

### kuruman

Parts (a) and (b) look OK.

For part (c), you need to say that the sum of all the heat losses or gains add up to zero. You may consider the 150 g of water at 100 oC as one entity, but you cannot add the specific heat of aluminum to the specific of water as you have done. You need to separate the expressions for the heat loss of the 150 g of water and of the aluminum.

3. Dec 16, 2009

### Dan Feerst

It seems like that should be mathematically equivalent. I suppose I just don't completely follow.
If all of my Q values from the system are in equilibrium at 100C, I would think I could set it equal to the heat of the final cube to get the final temperature. If this is correct, I can't see why combining terms wouldn't work. If you don't mind could you calculate C and tell me what answer your get. perhaps your answer will give me a better idea of what I'm doing wrong.

Thanks

4. Dec 16, 2009

### kuruman

I will not calculate your answer but I will show what you are doing wrong.

You have 150 g of water and 20 g of Al going from 100 oC to equilibrium temperature Tf. The heat transferred out of these is

Q = 150*4.186*(100 - Tf) + 20*0.9*(100 - Tf)

You can pull out a common factor of (100 - Tf) to get

Q = (150*4.186 + 20*0.9)*(100 - Tf)

but there is no mathematically correct way to combine the left factor into something like

Q = (150 + 20)*(4.186 + 0.9)*(100 - Tf)

Also, that (300 - 3Tf) is completely nonsensical.

It is seems that your algebra skills need some improvement.

5. Dec 16, 2009

### Dan Feerst

Alright, I admitted I didn't know what I was doing, you don't have to be a jerk about it.

anyway,
I tried it again, and I think I see where I screwed up, I skipped too many steps. This time I got an answer of 64.7C which makes a bit more sense.
question... well two actually.
I used 200g of water in the calculation not 150, since the first 50 grams of ice would be approximately 50g of water, is this correct?
Assuming thats correct, should I be correcting for the difference in densities?
I would think that is something the instructor would specifically ask for if he wanted it done.

6. Dec 16, 2009

### kuruman

If you are seeking help, it is wise and appropriate not to antagonize the people who might provide it, even if you think they are jerks.
Yes, my mistake, 200 g is the correct number.
What "correction for the difference in densities" are you thinking of?

7. Dec 16, 2009

### Dan Feerst

No, believe me, I do appreciate your help. but I would rather go elsewhere than deal with rude comments. If you feel that I should set your comments aside then maybe you shouldn't be making them in the first place

8. Dec 16, 2009

### ideasrule

4.5 cm^3 is not negligible, but I don't understand why the density difference should be corrected for when density was never used in the solution. Look at the question:

"150g of water...50g of ice... 20.0g aluminum"

Where was either volume or density ever mentioned?