Question
Asked Dec 11, 2019
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The specific heat of ice is 2.087 Jg·°C and its heat of fusion is −333.6 J/g.

Find ΔH, q, w, and ΔE for the freezing of water at−19.10 °C.

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Expert Answer

Step 1

According to first law of thermodynamics,

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AE = q + W Where, AEischangein energy qis heat W is work done W = -Px AV AV = Changein volume P = Pressure

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Step 2

Calculation of change in enthalpy:

The following processes take place.

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1н.0(1-19.10°С) > н,о(10°C) 9-тxs(water)x {0'С -(-19.10°c)} 2Н,0(1,0°С) — н,о(iоe, 0' C) -> 93- тхдН. ЗН.о (iсе, 0° с) >н,о (ice, -19.10°C) 9--пхs (ice)x{(-19,10'с) -0'с} q -m x ΔΗ fus

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Step 3

The negative sign indicates that heat is released and positive sign indicates that he...

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AH. (-19.10°C) =m xs(water)x{0°C-(-19.10°C)} - mxs(ice) x{(-19.10°C)– o°C}–m× AH, AH (-19.10° C) = (1g x 4.184J/g°Cx19.10°C)+(lgx 2.087J/g°Cx19.10°C) fus fus fus +{lg x(-333.6J /g)} AH9 (-19.10°C) = 79.91J +39.86.J– 333.6J=-213.83J. fus

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