Calculating final temperature in insulated container

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SUMMARY

The discussion centers on calculating the final temperature in an insulated container containing 0.60 kg of ice at 0 degrees Celsius, 2.0 kg of water at 0 degrees Celsius, and 3 kg of iron at 325 degrees Celsius. The specific heat capacities are provided: iron at 400 J/kg°C, water at 4200 J/kg°C, and ice at 2000 J/kg°C, with the latent heat of ice being 3.3 x 10^5 J/kg. The initial calculation for heat transfer, Q=mcΔT, was incorrectly applied, as the heat absorbed by the ice was not included in the total energy balance. The correct approach requires incorporating the phase change of ice and ensuring all components are accounted for in the energy equation.

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  • Understanding of thermodynamics, specifically heat transfer and specific heat capacity.
  • Familiarity with the concept of latent heat and phase changes of substances.
  • Proficiency in algebra for solving equations involving multiple variables.
  • Knowledge of the first law of thermodynamics as it applies to closed systems.
NEXT STEPS
  • Review the principles of heat transfer and specific heat calculations in thermodynamics.
  • Study the concept of latent heat and its role in phase changes, particularly for ice and water.
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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 celsius, 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 celsius, 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
 

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