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Cp of ice is 25 J/oC and that of water is 30 J/oC, what is the enthalpy change of water from – 10oC to 35oC at standard conditions, if the later heat of fusion of water is 110 J?

Cp of ice is 25 J/oC and that of water is 30 J/oC, what is the enthalpy change of water from – 10oC to 35oC at standard conditions, if the later heat of fusion of water is 110 J? Correct Answer

∆H = -10∫025.dT + 110 + 0∫3530.dT = 25*10 + 110 + 35*30 = 1410 J.

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