Question

The density of liquid oxygen at its boiling point is 1.14 \rm{kg/L} , and its heat of vaporization is 213 \rm{kJ/kg} .

How much energy in joules would be absorbed by 3.0 L of liquid oxygen as it vaporized?

Answer 1

m = rho x V
where rho is the density of liquid oxygen, V volume
m= 1.14 kg/L x 3 L = 3.42 kg

Energy = m times heat of vaporization
Energy = 3.42 kg x 213 kJ/kg
Energy = 724.46 kJ