Question

0.56 g of isopropyl alcohol (C₃H₈O) evaporates from a 39.25 g aluminum block initially at 25.0 °C. Calculate the final temperature of the aluminum block. Assume 100% heat transfer.

Isopropyl alcohol: at its boiling point ΔH_vap = 39.9 kJ/mol; at 25 °C ΔH_vap = 45.4 kJ/mol. Heat capacity of Al(s) = 0.903 J/(g °C).

Answer 1

Heat lost by aluminum block = Heat gained by isopropyl alcohol (C₃H₈O) evaporates

Heat gained by isopropyl alcohol (C₃H₈O) evaporates = (moles of isopropyl alcohol (C₃H₈O))*ΔH_vap

₌ 45.4 kJ/mol*(0.56/60) = 423.7J

Heat lost by aluminum block = mCp*del(T)

m = 39.25 g , Cp = 0.903 J/(g °C) ,

mCp*del(T) = 423.7

del(T) = 423.7/(0.903*39.25) = 11.95

25-T_f   =  11.95

T_f = 13.05°C