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Problem- What is the heat change in J associated with 45 g of liquid propane at -168 oC changing to gas at 12 oC?
c(propane) (g) = 2.40 J/(g.K)
c(propane) (liq) = 1.65 J/(g.K)
c(propane) (s) = 1.20 J/ (g.K)
Hfus(propane) = 3.526 kJ/mol
Hvap(propane) = 15.7 kJ/mol
The freezing point of propane is -182 oC.
The boiling point of propane is -42 oC.
Using textbook reference solve the reactions given in above problem
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