Piece of iron put into container with ice

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SUMMARY

The discussion centers on the thermodynamic equilibrium of a system containing 1 kg of iron at 100°C and 1 kg of ice at 0°C. The heat transfer from the iron to the ice is analyzed using the heat balance equation, leading to the conclusion that the final equilibrium temperature will be 0°C, with a mixture of ice and liquid water present. The coefficient of melting of ice (c_L) is 330 kJ/kg, and the heat transfer coefficient of iron (c_I) is 450 J/(kg*K). The entropy change is calculated based on the melting of ice and the cooling of iron, confirming that the total entropy change is positive.

PREREQUISITES
  • Understanding of thermodynamics, specifically heat transfer principles.
  • Familiarity with entropy calculations and the second law of thermodynamics.
  • Knowledge of specific heat capacities, particularly for iron and ice.
  • Ability to apply the heat balance equation in thermodynamic systems.
NEXT STEPS
  • Study the principles of thermodynamic equilibrium in closed systems.
  • Learn about the calculation of entropy changes in phase transitions.
  • Explore the concept of heat transfer coefficients in different materials.
  • Investigate the implications of the second law of thermodynamics in real-world applications.
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Students studying thermodynamics, physics enthusiasts, and professionals in engineering fields who require a solid understanding of heat transfer and entropy in phase change scenarios.

Lukasz Madry
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Homework Statement


We put 1kg iron of temperature 100 Celsius into container with 1kg of ice, temperature 0 Celsius. What is state of the system after reaching equilibrium? Calculate change of entropy.

Coefficient of melting of ice (c_L) is 330 kJ/kg, coefficient of heat transfer of iron (c_I) is 450 J/(kg*K).

Homework Equations


I don't really know. Heat balance, for sure. Entropy equation - Q = ∫ T dS.
S = c ln(T_f/T_i)
But I'm sure I'm missing something crucial.

The Attempt at a Solution


I tried to compute heat balance:

C_I ( 100 - T_f) = c_L

but this does not work, obviously, since leads to T_f = 100 - c_L/c_I = 100 - 733 which is far below zero temperature.

Another attempt - assume that only part of ice was melted. Let k = c_I 100 / c_L be coefficient that describes how much ice was melted into water. In this case, entropy would be:

S_{iron} = c_I ln(273/373)
S_{ice} = k c_L/273

after substition we can find that it's sum is greater than zero, which is expected.

Is this close to being correct?
 
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So the first thing that I would try to figure is if all the ice melted or not. It seems you determined the answer to that. If you have a container which has ice and liquid water and a piece of iron at equilibrium, what temperature do you think the whole thing will be? If the iron and water are different temperatures, is it at equilibrium? Once you figure the final temperature, then you can figure how much heat transferred from the iron, then go from there.
 
Essentialy, energy will flow from iron to ice until iron reaches temperatures of ice OR ice melts down. In this case, iron reaches 0 Celsius way before ice melts down. My only gripe with this is liquid water that keeps appearing - at the beginning it has temperature of ice, but is in contact with iron. It should make heat flow from iron not only to ice, but to this water too. However, ultimately this heat should also reach ice, since ice has temperature of 0 until it totally melts down.

This means that whole thing will be at the temperature of 0 Celsius, and there will be ice, liquid water and iron in the system. Proportions of ice to liquid water are determined by k coefficient I defined in the first post.
Is this correct?
 
Thank you for Your help
 

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