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 4.0 L of liquid oxygen as it vaporized?

Calculate the number of kg in 4.0 liters of liquid, using the density. Then multipy that by the heat of vaporization.

density of a ring mass 38.6g and volume 2cm

To calculate the energy absorbed by 4.0 L of liquid oxygen as it vaporizes, you need to multiply the volume by the density to find the mass of the liquid oxygen. Then, you can multiply the mass by the heat of vaporization to calculate the total energy absorbed.

Given:
Density of liquid oxygen = 1.14 kg/L
Heat of vaporization = 213 kJ/kg

Step 1: Convert the density to kg/L:
Density = 1.14 kg/L

Step 2: Calculate the mass of the liquid oxygen:
Mass = Density × Volume
Mass = 1.14 kg/L × 4.0 L
Mass = 4.56 kg

Step 3: Convert the heat of vaporization to J/kg:
Heat of vaporization = 213 kJ/kg
1 kJ = 1000 J
Heat of vaporization = 213 kJ/kg × 1000 J/kJ
Heat of vaporization = 213,000 J/kg

Step 4: Calculate the energy absorbed:
Energy absorbed = Mass × Heat of vaporization
Energy absorbed = 4.56 kg × 213,000 J/kg
Energy absorbed = 972,480 J

Therefore, 4.0 L of liquid oxygen will absorb 972,480 J of energy as it vaporizes.

To find the energy absorbed by 4.0 L of liquid oxygen as it vaporizes, we need to use the equation:

Energy = Mass x Heat of Vaporization

First, we need to find the mass of 4.0 L of liquid oxygen. We can do this by multiplying the density by the volume:

Mass = Density x Volume

Mass = 1.14 kg/L x 4.0 L

Mass = 4.56 kg

Now, we can calculate the energy absorbed using the mass and the heat of vaporization:

Energy = Mass x Heat of Vaporization

Energy = 4.56 kg x 213 kJ/kg

To convert kJ to J, we multiply by 1000:

Energy = 4.56 kg x 213 kJ/kg x 1000 J/kJ

Energy = 973,080 J

Therefore, the energy absorbed by 4.0 L of liquid oxygen as it vaporizes is 973,080 Joules.