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tiL12214_ch23_565-596.indd Page 568 9/23/10 11:07 AM user-f465
tiL12214_ch23_565-596.indd Page 568 9/23/10 11:07 AM user-f465 /Volume/201/MHDQ233/tat78194_disk1of1/0073378194/tat78194_pagefile

Step 1 Heat from condensation
3
m = 1.65 × 10 kg Q = mL v
cal _ Convert kg to g:
L v = 540 1.65 × 10 kg _ )
g
3
Increasing altitude Parcel Q = 1.65 × 10 g 540
(
1 × 10 g
3

Q = ?
1 kg
6
of air
1.65 × 10 g
cal _
g )
(
6
Temperature
8
a rising parcel
of surrounding air Temperature of = 8.9 × 10 cal
of air
Step 2 Temperature change of air
g
3 _
ρ = 1.1 × 10

m 3
Increasing temperature V = 1.0 km 3
_
cal

FIGURE 23.4 In a state of atmospheric instability, a parcel c = 0.17
g⋅C°
of air will always be warmer, and therefore less dense, than the 8
surrounding air at any altitude. The parcel will, therefore, continue Q = 8.9 × 10 cal
in the direction pushed when the upward force is removed. ΔT = ?
m _
Q = mcΔT and ρ =

V
whether or not clouds form. When the dew point temperature Rearrange the density formula to solve for mass:
is reached, the water vapor in the air tends to condense, but
m _
now it requires the help of tiny microscopic particles called ρ = ∴ m = ρV

V
condensation nuclei. Without condensation nuclei, water vapor
condenses to tiny droplets that are soon torn apart by collisions Substitute this expression into the specific heat formula:
with other water vapor molecules. Thus, without condensation
Q = (ρV )cΔT
nuclei, further cooling can result in the air parcel becoming
supersaturated, containing more than its normal saturation Rearrange this expression to solve for ΔT:
amount of water vapor. Condensation nuclei promote the con- _
Q

densation of water vapor into tiny droplets that are not torn ΔT =
ρVc
apart by molecular collisions. Soon the stable droplets grow on
3
3
Convert km to m :
a condensation nucleus to a diameter of about one-tenth the
6
diameter of a human hair, but a wide range of sizes may be 3 1 × 10 m 3
(
_

1.0 km 3 )
present. This accumulation of tiny droplets can easily remain 1 k m
suspended in the air from the slightest air movement. An av- 1.0 × 1 0 m 3
6
erage cloud may contain hundreds of such droplets per cubic ____
8
8.9 × 1 0 cal
3
centimeter, with an average density of 1 g/m . So, a big cloud ΔT = ( 3) 6 _
3 g
_
(
cal
3

3
that has a volume of 1 km would contain about a million liters 1.1 × 1 0 (1.0 × 1 0 m ) 0.17 g⋅°C)
m
(about 264,000 gal) of water. ___ __
8
8.9 × 1 0
cal
=
6
g
3
(1.1 × 1 0 )(1.0 × 1 0 )(0.17) _ 3) ( g⋅°C)
3 _
(
cal
( m )

EXAMPLE 23.1 8 m
8.9 × 1 0
°
3
3

A parcel of air with a volume of 1.0 km that contains 1.65 × 10 kg of = _ 8 C
1.9 × 1 0
water vapor rises to an altitude where all the water in the parcel con-
denses. What is the change in temperature of the parcel of air due to = 4.7°C
condensation? (Assume the density of air at the condensation altitude
3
3
is 1.1 × 10 g/m . The specific heat of air is provided in Table 4.2, and
the latent heat of vaporization of water is provided in Table 4.4.) EXAMPLE 23.2
What is the change in temperature of this parcel of air if the water in
Example 23.1 freezes into ice crystals? (The latent heat of fusion of wa-
SOLUTION
ter is provided in Table 4.4.) (Answer: 0.7°C.)
The heat released from the condensing water is transferred to the air
parcel. The change in temperature can be determined by equating the
heat from the condensation, calculated from the latent heat of vapor-
ization equation (chapter 4, equation 4.6), to the change in air tempera- A single water droplet in a cloud is so tiny that it would be dif-
ture, calculated from the specific heat equation (chapter 4, equation 4.4). ficult to see with the unaided eye. When water droplets accumulate
This problem can be solved in two steps. in huge numbers in clouds, what you see depends on the size of the
568 CHAPTER 23 Weather and Climate 23-4