Iron and Ice in Aluminum bucket of water

  • Thread starter LiorSh
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  • #1
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Homework Statement


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.

Homework Equations


Delta Q = 0 in equilibrium

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.

Thank you for your help!
 
Last edited:

Answers and Replies

  • #3
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I think I wrote it by mistake - the given number is 448 for the Cp of the iron
 
  • #4
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the given number is 448 for the Cp
Okay. Transcription errors happen to all of us.
since I ignored the Iron and the Aluminum in my last equation.
What's the heat capacity of water + bucket + iron? That's the system you're cooling with the ice.
 
  • #5
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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
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Cp=2090, Lf=3.33e5, Ti=-5 Celsius
These numbers aren't just along for the ride.
Do I need to divide it by 333 now?
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
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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
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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
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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
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Yes, and is there that much heat available in the bucket-water-iron?
 
  • #11
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Nope, we have only 82655 left in the bucket
 
  • #12
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And the final T is what?
 
  • #13
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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
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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
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It was a typo. In my notebook I used the right number. Thank you so much!
 

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