Latent heat and specific heat in insulated container

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SUMMARY

The discussion centers on a thermodynamics problem involving 1.0 kg of ice, 3.0 kg of water, and 5.0 kg of iron in a sealed insulated container. The specific heat capacities are given as cFe = 400 J/kg°C, cWater = 4200 J/kg°C, and cIce = 2000 J/kg°C, with the latent heat of ice being 3.3 × 105 J. The equilibrium temperature is calculated to be 20°C, but a discrepancy arises as the expected temperature for the iron is stated to be 353°C instead of the calculated 331°C. The solution involves applying the equations Q = mcΔT and Q = ml.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with specific heat capacity and latent heat concepts.
  • Proficiency in using the equations Q = mcΔT and Q = ml.
  • Basic knowledge of thermal equilibrium in closed systems.
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  • Review the concept of thermal equilibrium in insulated systems.
  • Study the calculations involving latent heat and specific heat in detail.
  • Learn about the implications of phase changes on temperature calculations.
  • Explore advanced thermodynamic problems involving multiple substances.
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Students studying thermodynamics, physics educators, and anyone interested in understanding heat transfer and thermal equilibrium in closed systems.

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Homework Statement


A 1.0kg of ice at 0◦C, 3.0kg of water at 0◦C, and 5.0kg of iron at temperature T are placed in a sealed and insulated container. cFe =400J/ kg◦C ,cWater=4200J/ kg◦C , cIce= 2000 J/ kg◦C

latent heat for ice is 3.3×105 J .
The equilibrium temperature is 20◦C.

Homework Equations


Q=mcΔT
Q=ml

The Attempt at a Solution


0=Qice→water+Qice+Qwater+Qiron
=ml(ice)+mcΔT(ice)+mcΔT(water)+mcΔT(iron)
=(1.0)(330000)+(1.0)(2000)(20-0)+(3.0)(4200)(20-0)+(5.0)(400)(20-Ti)
Ti=331°C

BUT the answer says it should be 353°
 
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Once you have melted the ice, it is water.
 
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