Final answer:
The mass of water in the air parcel is 7.787 × 1012 grams, calculated by equating the heat released by condensation to the heat absorbed by the air parcel due to the condensation of water vapor.
Step-by-step explanation:
To calculate the mass of water condensed in an air parcel, we will use the energy released by the condensation process and equate it to the heat needed to raise the temperature of the air parcel. The heat released by the condensation of water vapor is the mass of water times the latent heat of vaporization (540 cal/g), which then equals to the heat gained by the air which is the mass times the specific heat of air (0.17 cal/g°C) times the change in temperature (4.6°C).
First, we must find the mass of the air using the density and the volume:
mair = density x volume
= (8.1 × 102 g/m3) x (6.7 × 104 km3 x 109 m3/km3)
= 5.427 × 1014 g
Now, equate the heat released by water to the heat absorbed by the air:
Qcondensation = Qheating air
(mwater x 540 cal/g) = (mair x 0.17 cal/g°C x 4.6°C)
Substituting the mass of air (5.427 × 1014 g) and solving for mwater:
mwater = (mair x 0.17 cal/g°C x 4.6°C) / 540 cal/g
= (5.427 × 1014 g x 0.17 cal/g°C x 4.6°C) / 540 cal/g
= 7.787 × 1012 g
Therefore, the mass of water in the air parcel is 7.787 × 1012 grams