The density of liquid oxygen at its boiling point is 1.14 , and its heat of vaporization is 213 .
How much energy in joules would be absorbed by 3.0 of liquid oxygen as it vaporized?
Express your answer numerically in joules.
Units are neccessary...
3.0 what?
213 what?
1.14 what?
whats the answer
To answer this question, we need to use the formula for heat of vaporization:
q = m * ΔHvap
where q is the energy absorbed (in joules), m is the mass of the substance being vaporized (in grams), and ΔHvap is the heat of vaporization (in J/g).
First, let's convert the given mass of liquid oxygen (3.0 g) to kilograms:
m = 3.0 g = 0.003 kg
Now, we can substitute the values into the formula:
q = (0.003 kg) * (213 J/g)
To calculate the energy absorbed, we multiply the mass of the substance by the heat of vaporization:
q = 0.003 kg * 213 J/g = 0.639 J
Therefore, 3.0 g of liquid oxygen would absorb approximately 0.639 joules of energy as it vaporizes.