Calculating final temperature in insulated container

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The discussion revolves around calculating the final temperature in an insulated container with ice, water, and iron at different temperatures. The user attempted to apply the heat transfer equation Q=mcΔT but neglected to account for the phase change of ice and its interaction with the other materials. It was pointed out that the calculations must include the heat absorbed by the ice as it melts and the subsequent temperature changes. The user is encouraged to detail their calculations more thoroughly to identify errors. Accurate consideration of all components is crucial for determining the final temperature correctly.
Gaith
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Homework Statement


0.60kg of ice at O degrees C, 2.0 kg of water at 0 degrees 0 degrees, 3kg of iron at 325 degrees C are place in a sealed and insulated container. The specific heat for iron is c=400J/kg degree C, for water is c=4200J/kgdegree, for water is c=4200J/kgdegree celcius, for ice is c=2000 and the latent heat for ice is 3.3x10^5J/kg.

What is the final temperature inside the insulated container

Homework Equations

[/B]
Q=mcΔT

The Attempt at a Solution


Q=0.6x2000(Tf)+2(4200)(Tf)+3x400(Tf-325) = 0
Q=1200Tf + 8400Tf + 1200Tf - 390000=0
I solved for Tf and got the wrong answer. Help me please.[/B]
 
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Gaith said:

Homework Statement


0.60kg of ice at O degrees C, 2.0 kg of water at 0 degrees 0 degrees, 3kg of iron at 325 degrees C are place in a sealed and insulated container. The specific heat for iron is c=400J/kg degree C, for water is c=4200J/kgdegree, for water is c=4200J/kgdegree celcius, for ice is c=2000 and the latent heat for ice is 3.3x10^5J/kg.

What is the final temperature inside the insulated container

Homework Equations

[/B]
Q=mcΔT

The Attempt at a Solution


Q=0.6x2000(Tf)+2(4200)(Tf)+3x400(Tf-325) = 0
Q=1200Tf + 8400Tf + 1200Tf - 390000=0
I solved for Tf and got the wrong answer. Help me please.[/B]
It would help you and us if you would detail your calculations more carefully.

That said, what happens to the ice in the container after the hot iron is put in? It looks like you have taken care of the temp. change of the iron and the liquid water there initially, but the ice appears to be missing from your calculations. :L
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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