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 2.0L of liquid oxygen as it vaporized?
q = \rm{mass O2 x heat vap} = ?
To calculate the energy absorbed by 2.0L of liquid oxygen as it vaporizes, we need to use the formula:
Energy = mass × heat of vaporization
First, let's calculate the mass of 2.0L of liquid oxygen:
Density = mass / volume
Rearranging the equation, we have:
mass = density × volume
mass = 1.14 kg/L × 2.0 L
mass = 2.28 kg
Now, let's calculate the energy absorbed:
Energy = mass × heat of vaporization
Energy = 2.28 kg × 213 kJ/kg
To convert kJ to J, we multiply by 1000:
Energy = 2.28 kg × 213,000 J/kg
Energy = 485,640 J
Therefore, 2.0L of liquid oxygen would absorb 485,640 J of energy as it vaporizes.
To find the energy absorbed by 2.0 L of liquid oxygen as it vaporizes, we can use the formula:
Energy absorbed = Mass of liquid oxygen x Heat of vaporization
To calculate the mass of liquid oxygen, we need to know the density. Since the density is given in kg/L, we can convert the volume (2.0 L) to mass:
Mass of liquid oxygen = Density x Volume of liquid oxygen
Substituting the given values:
Mass of liquid oxygen = 1.14 kg/L x 2.0 L
Mass of liquid oxygen = 2.28 kg
Now, we can calculate the energy absorbed:
Energy absorbed = Mass of liquid oxygen x Heat of vaporization
Energy absorbed = 2.28 kg x 213 kJ/kg
To convert kilojoules to joules, we need to multiply the result by 1000:
Energy absorbed = 2.28 kg x 213,000 J/kg
Energy absorbed = 486,840 J
Therefore, 2.0 L of liquid oxygen would absorb 486,840 joules of energy as it vaporizes.