Question

An air parcel initially at 290 K contains 10g/kg of water vapor. Half of the water vapor condenses into liquid water. The pressure does not change.

A) How much latent heat is released per kilogram of air?

B) What is the final air temperature?

Answer 1

a)

Heat of condensation of water is 2260 j/g of water.

Now as half of water is condensed so amount of heat liberated is given by ( 10/2 = 5 g. )

H = 2260 (j/g)   5 (g) = 11300 (j) = 11.3   (kj)

b)

Let the final temperature of air is T

Then heat released by water is absorbed by air parcel.

so ms t = 11300 (j)

where m is mass which is 1 kg

s is specific heat of air at constant pressure is 1 Kj/ Kg.K

t is the change in temperature.

so putting into the formula we will get

1 (kg)   1 (Kj/ Kg.K) (T - 290) = 11.300 (Kj)

This will give T - 290 = 11.300

T = 301.300 (K)

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