Chemical Engineering Thermodynamics

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SUMMARY

In the discussion on Chemical Engineering Thermodynamics, the behavior of supercooled liquid tin (Sn) at 495K is analyzed. The melting temperature of tin is 505K, and the enthalpy of fusion is 7070 J/mol. The heat capacities for both liquid and solid tin are provided, which are essential for calculating the fraction of liquid tin that will spontaneously freeze. The discussion concludes that the supercooled liquid tin will indeed freeze, releasing heat that raises the system's temperature to the freezing point.

PREREQUISITES
  • Understanding of phase transitions in thermodynamics
  • Familiarity with heat capacity calculations
  • Knowledge of enthalpy of fusion concepts
  • Ability to perform adiabatic process analysis
NEXT STEPS
  • Study the Clausius-Clapeyron equation for phase changes
  • Learn about adiabatic processes in thermodynamic systems
  • Explore the calculation of heat transfer during phase transitions
  • Investigate the properties of supercooled liquids and their stability
USEFUL FOR

Chemical engineers, thermodynamics students, and researchers in materials science will benefit from this discussion, particularly those focusing on phase behavior and heat transfer in materials.

sean/mac
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One mole of supercooled liquid tin (Sn) is adiabatically contained at 495K. Given:

the melting temperature of Sn, Tm, Sn = 505 [K]
the enthalpy of fusion of Sn, hfus, Sn = 7070 [J/mol]
the heat capacity of liquid Sn, Cp, Sn(l) = 34.7 – 9.2 x10-3 T [J / (mol K)]
the heat capacity of solid Sn, Cp, Sn(s) = 18.5 + 2.6 x 10-2 T [J / (mol K)]

(a) Briefly explain what will happen to the supercooled liquid tin.

(b) Calculate the fraction of liquid tin which spontaneously freezes. Provide a clearly labeled
hypothetical path used in your calculation. State any assumption(s) you make.

I know (a), I'm just unsure how to draw a hypothetical path and unsure where to even start the calculation, any help is appreciated
 
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I'm thinking that some of the supercooled liquid tin will freeze. (If it doesn't, then the problem is pretty uninteresting.) And I'm thinking that the freezing process will give off some heat, yes? Enough heat to raise the temperature of this adiabatic system to the freezing point? Does this help?
 

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