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Supercool liquid tin is adiabatically contained at 495K. Calculate the fraction of the tin which spontaneously freezes, and entropy change of the process. Hm(Sn)=7070J at

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Supercool liquid tin is adiabatically contained at 495K. Calculate the fraction of the tin which spontaneously freezes, and entropy change of the process. Hm(Sn)=7070J at Tm=505K Cp,Sn(I()=34.79.2103TJ/K Cp,Sn(s)=18.5+26103TJ/K

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