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Chemical Engineering Thermodynamics

  • Thread starter 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
 

Answers and Replies

  • #2
Mapes
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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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