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Question

Calculate the change in enthalpy for the following process at 1 atm
H2O(l,50oC)H2O(g,150oC)
given that ΔHv at 100oC is 40.7kJmol1
CP(H2O,l)=75.0Jmol1K1
CP(H2O,g)=33.3Jmol1K1

A
64.8KJ
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B
52.5KJ
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C
48.6KJ
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D
46.1KJ
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Solution

The correct option is A 46.1KJ
H2O(l,50C)H2O(l,100C)
H1=Cp(10050)=75×50=3.75kJmol1
H2O(l,100C)H2O(g,100C)
H2=Hv=40.7kJmol1
H2O(g,100C)H2O(g,150C)
H3=Cp(150100)=33.3×50=1.665kJmol1
Let no. of moles=1
H=H1+H2+H3
=40.7+3.75+1.665=46.1kJ

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