Iron and Ice in Aluminum bucket of water

1. Dec 19, 2014

LiorSh

1. The problem statement, all variables and given/known data
Iron ball Mass = 0.8 kg, Cp=484, Ti=80 Celsius edit : Cp = 448
Aluminum Bucket = 0.5 kg, Cp=900, Ti=20 Celsius
Water = 0.6 kg, Cp=4186, Ti=20 Celsius
Ice cube=0.5 Kg, Cp=2090, Lf=3.33e5, Ti=-5 Celsius

Find the final temperature of the bucket.

2. Relevant equations
Delta Q = 0 in equilibrium

3. The attempt at a solution
At fist I took into consideration only the water,iron and Aluminum to find the temperature in equilibrium
Delta Q = 4186*0.6*(Tf-20)+900*0.5*(Tf-20)+448*0.8(Tf-80)=0
The final temperature I got is 26.47 Celsius

Next I calculate the amount of energy release from 0.6 kg of water = 600*4.182*26.47=66,482 J
Then I divided it by the amount of energy needed to bring this amount of water to 0 Celsius - 66,482/333=199.645 Grams of ice.
Since we have 0.5 kg of ice Tf will be equal to 0 Celsius and 0.3 Kg of the ice will remain in the bucket.

In order to proof it I wrote the following equation = Delta Q = 4186*0.6*(Tf-26.47)+2090*0.199*(0-(-5)) +0.199*3.33e5+4186*(0.199)*(Tf-0)=0

I got Tf= -0.65 Celsius which could be a result of roundoff error so it seems like I got the right answer.

But I think that I did something wrong since I ignored the Iron and the Aluminum in my last equation.

Last edited: Dec 19, 2014
2. Dec 19, 2014

Bystander

Try it again.

3. Dec 19, 2014

LiorSh

I think I wrote it by mistake - the given number is 448 for the Cp of the iron

4. Dec 19, 2014

Bystander

Okay. Transcription errors happen to all of us.
What's the heat capacity of water + bucket + iron? That's the system you're cooling with the ice.

5. Dec 19, 2014

LiorSh

4.186*600g*26.47+0.450*800*26.47+0.900*500*26.47= 87880 The heat capacity of the system. Do I need to divide it by 333 now? then I get 263.905 grams of ice needed to bring the system to 0 Celsius.

6. Dec 19, 2014

Bystander

These numbers aren't just along for the ride.
No. 3.33e5, or 3.33 x 105, or 330,000 comes in after one more step; the ice has in initial temperature less than that of the melting point.

7. Dec 19, 2014

LiorSh

Delta Q =(-87880+3320Tf)+(2090*0.500)*(0-(-5)) +0.500*3.33e5+4186*(0.500)*(Tf-0)=0 (The red part is from the first equation).
than I get -20 Celsius which is impossible.
There must be something I'm not aware of. What am I missing?

8. Dec 19, 2014

Bystander

How much heat from bucket, water, and iron to raise the temperature of the ice to 0 C? How much does that lower the temperature of the bucket, water, and iron?

9. Dec 19, 2014

LiorSh

To raise the temperature of the ice to 0 we need 2.090 x 500 x 5 = 5225 J.
Bringing the Ice to 0 Celsius will lower the temp of the bucket by 1.6 Celsius ------> 87880-5225=3320Tf => 24.90 C

Is that correct?
Than Heat to melt the Ice = 3.33e5x0.5 = 166500 J

10. Dec 19, 2014

Bystander

Yes, and is there that much heat available in the bucket-water-iron?

11. Dec 19, 2014

LiorSh

Nope, we have only 82655 left in the bucket

12. Dec 19, 2014

Bystander

And the final T is what?

13. Dec 19, 2014

LiorSh

that means that we can use only 82655 of heat to melt the ice - which leaves us with 166500-82655= 83845
So it will melt 0.248 kg of the ice and the rest of the ice will remain in the bucket at 0 Celsius.

Is that correct? or there is a different way to prove it?

14. Dec 19, 2014

Bystander

I haven't checked all your arithmetic, but the only bug I've seen was the transcription error in the original post, so you should be good to go.

15. Dec 19, 2014

LiorSh

It was a typo. In my notebook I used the right number. Thank you so much!