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Example 23.4, p. 584
CHAPTER 23 Compare the monthly temperature and total precipitation to the crite-
ria in Table 23.1.
Example 23.2, p. 568
The average monthly temperature shows this location has a short,
The heat released from the freezing water is transferred to the air par-
mild winter, moderate to hot summers, and moderate precipitation.
cel. The change in temperature can be determined by equating the heat
This city is in a humid subtropical climate.
from the freezing calculated from the latent heat of fusion equation
This problem has made use of data obtained from NOAA National
(chapter 4, equation 4.5) to the change in air temperature calculated Climatic Data Center (NCDC) Monthly Global Surface Data. http://gis.
from the specific heat equation (chapter 4, equation 4.4). This problem ncdc.noaa.gov/website/ims-cdo/gcosmon/viewer.htm
can be solved in two steps.
Step 1 Heat from Freezing into Ice CHAPTER 24
3
m = 1.65 × 10 kg Q = mL f
Example 24.2, p. 602
cal _
L f = 80 Convert kg to g: Multiply the groundwater discharge component of the budget by the

g 3 drainage area to obtain volume. Perform the necessary unit conver-
)
(
1 × 10 g
3

Q = ? 1.65 × 10 kg _ sions so the units are in cubic meters.
1 kg
6
1.65 × 10 g Q G = 68 mm
cal _
(
g )
6
Q = 1.65 × 10 g 80 A = 12 km 2

volume = ?
8
= 1.3 × 10 cal
volume = Q G A
Step 2 Temperature change of air
Convert mm to m:
g
3 _
ρ = 1.1 × 10 1 m

3 )
(

m 3 68 mm _
V = 1.0 km 3 1 × 10 mm
−2
6.8 × 10 m
_
cal
c = 0.17

2
2
g⋅°C Convert km to m :
8
Q = 1.3 × 10 cal 2
2)
(
2 __
12 km 1 m

ΔT = ? 1 × 10 km
−6
7
m _ 1.2 × 10 m 2
Q = mc ΔT and ρ =

2
−2
7
V volume = 6.8 × 10 m (1.2 × 10 m )
5
Rearrange the density formula to solve for mass: = 8.2 × 10 m 3
m _
ρ = ∴ m = ρV Example 24.4, p. 613

V
Determine the water depth 165 m offshore, using the slope of the bottom.
Substitute this expression into the specific heat formula:
cm
_ _
ΔY

slope = 2.1 slope = ∴ΔY = ΔX slope

Q = (ρV )c ΔT m ΔX
ΔX = 165 m _ m _
cm
Rearrange this expression to solve for ΔT: Convert m to :

m
ΔY = ?
Q
_ _ 1 m
cm _

ΔT = 2.1 m( 2 )

ρVc 1 × 10 cm
−2
3
3
Convert km to m : 2.1 × 1 0
−2
6
(
_
3 1 × 10 m 3 ΔY = 165 m (2.1 × 10 )

1.0 km 3 )
1 k m = 3.5 m
6
1.0 × 1 0 m 3
The breakers occur where the water depth is about 1.33 times the wave
8
____
1.3 × 1 0 cal
ΔT = height, so the wave height can be calculated from the water depth.
3 g
g⋅°C)
_
3)
(
( 1.1 × 1 0 (1.0 × 1 0 m ) 0.17 _
cal
3
6

m water depth = 3.5 m
8
___ __ wave height = ?
cal
1.3 × 1 0
=

g
(1.1 × 1 0 )(1.0 × 1 0 )(0.17) _ 3) ( g⋅°C) water depth = 1.33 (wave height)
(
3 _
3
6
cal

( m )
m water depth

_ 8 ∴wave height = _
1.3 × 1 0
°
= C 1.33

8
1.9 × 1 0
3.5 m
wave height = _

= 0.7 °C 1.33
= 2.6 m
642 APPENDIX D Solutions for Follow-Up Example Exercises D-10