What amount of enthalpy is required to transform 100.0 g of liquid ethanol at 23°C to ethanol vapor at 95 °C?

Question

The specific heat of liquid ethanol, C2H5OH(l), is 2.46 J/g·°C and that of ethanol vapor is 1.88 J/g·°C; the heat of vaporization is 39.3 kJ/mol. The boiling point of ethanol is 78 °C.

Answer 1

Step 1: raise temperature of 100 g ethanol from 23º to 78º

q = mC∆T = (100 g)(2.46 J/gº)(55º) = 13,530 J = 13.5 kJ

Step 2: convert 100 g of liquid ethanol to vapor @ 78º. Phase change with no change in temperature.

q = m∆Hvap = (100 g)(39.3 kJ/mol) = 100 g x 1 mol/46 g x 39.3 kJ/mol = 85.4 kJ

Step 3: raise temperature of100 g gaseous ethanol @ 78º to 95º

q = mC∆T = (100 g)(1.88 J/gº)( 17º) = 3196 J = 3.20 kJ

Sum of all enthalpy = 13.5 kJ + 85.4 kJ + 3.20 kJ = 102 kJ