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Find AH, q, w, and AE for the freezing of water at -47.80 °C. The specific heat of ice is 2.087 and its heat of fusion is -33
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Answer #1

q = 2.087 J/g.oC * {0 - (-47.8)} oC + (333.6 J/g) = 433.36 J/g

Now, \DeltaH = 433.36 J/g

Now, w = -P.\DeltaV = -R.\DeltaT = -8.314 J/mol.K * 47.8 K = -8.314 J/18 g * 47.8 = -22.08 J/g

Finally, \DeltaE = \DeltaH + w = 433.36 J/g + (-22.08 J/g) = 411.28 J/g

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