Asked by ntombiningi
Calculate the enthalpy change, ÄH DeltaH, for the process in which 12.7g g of water is converted from liquid at 7.3 ∘ C \, ^\circ C to vapor at 25.0 ∘ C \, ^\circ C .
For water, ÄH vap \Delta H_{\rm vap} = 44.0kJ/mol kJ/mol at 25.0 ∘ C \, ^\circ C and s s = 4.18J/(g⋅ ∘ C) J/(g\cdot{ ^\circ C}) for H 2 O(l)
For water, ÄH vap \Delta H_{\rm vap} = 44.0kJ/mol kJ/mol at 25.0 ∘ C \, ^\circ C and s s = 4.18J/(g⋅ ∘ C) J/(g\cdot{ ^\circ C}) for H 2 O(l)
Answers
Answered by
DrBob222
dH for H2O moving from 7.3C to 25C is
dH = mass H2O x specific heat H2O x (Tfinal-Tinitial)
dH for vaporization at 25C is
dH = mass H2O x Hvap.
Add the two together for the total.
dH = mass H2O x specific heat H2O x (Tfinal-Tinitial)
dH for vaporization at 25C is
dH = mass H2O x Hvap.
Add the two together for the total.
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