Question #: 9 How much energy is required to warm 90.1 g (5.00 mol) of H2O(s),...

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Question #: 9 How much energy is required to warm 90.1 g (5.00mol) of H2O(s), initially at â€“19.4 oC, to H2O( l) at 100.0 oC?melting point = 0.00 oC boiling point = 100.0 oC Î”Hfus = 6.02kJ/mol Î”Hvap = 40.7 kJ/mol Cs of H2O(s) = 2.09 J/g oC Cs of H2O(l)= 4.18 J/g oC Cs of H2O(g) = 2.01 J/g oC

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3.9 Ratings (493 Votes)

The energy required for the whole process has to be calculated as follows:

H2O(s)Â Â ------->Â Â Â Â H2O(s)Â Â -------->Â Â Â Â Â H2O(l)Â Â Â ------> H2O(l)Â Â ----------> H2O(g)

-19.4Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â OÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â OÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 100Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 100 all in ^oC

Â Â Â Â Â Â Â Â Â 2.09J/gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â deltaH fusÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 4.18J/gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â deltaH vapÂ Â Â Â Â Â Â Â Â Â Â energy required

Energy for step I = 2.09 x 90.1 = 188.309J

energy for step II = 6.02x1000 x 5 =30100J

energy for step III = 4.18 x 90.1 =376.618J

energy for step Iv = 40.7 x 1000 x 5 = 203500J

Thus the totyal energy required for all the processes = 234164.927 J =234.16 kJ

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