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tiL12214_appe_643-698.indd Page 693 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

__ 22.14. Use the values provided and the formula in example 22.4 to
reflected solar radiation

α =
insolation solve for insolation.
∴ reflected solar radiation = insolation (α)
Morning
2 W _
reflected solar radiation = 8.7 × 10 (0.4) 3 W _ __

L stick

m 2 I max = 1.0 × 10 I = I max √ L + L )
m 2 (
2
2

2 W _
= 3.5 × 10 L stick = 0.75 m stick shadow

m 2
L shadow = 0.75 m 3 W _ __
0.75 m

Step 2: Determine Energy = 1.0 × 10 2)
2
I = ? m ( (0.75 m) + (0.75 m)
√
2

The energy from the absorbed solar radiation, in joules, is the area

(A) of the layer of sand times the time (t) in seconds times the 3 __ W _ _
2 )
m
0.75
2)

= 1.0 × 10

difference between the insolation and the reflected solar radiation. ( 0.75 + 0.75 m ( √ m

2

2
√

2 W _
insolation = 8.7 × 10 3 __ W _

)
0.75

m 2 = 1.0 × 10

+

0.56
0.56

1 W _ ( √ m 2

reflected solar radiation = 3.5 × 10
m 2 3 _ W _
)
0.75

A = 2.0 m 2 = 1.0 × 10
( √ m 2

1.12
t = 1 h
absorbed energy = ? 3 0.75 W _
_

= 1.0 × 10 ( )
absorbed energy = At (insolation − reflected solar radiation) 1.1 m 2
W _
Convert hours to seconds: = 1.0 × 10 (0.68)
3

m 2
)(
60 min _
(

1 h _ 60 s ) 2 W _

1 h 1 min = 6.8 × 10
m 2
360 s
2 W _
2)
(
2 W _
2

absorbed energy = 2.0 m (360 s) 8.7 × 10 − 3.5 × 10 Noon 3 W _
√ L + L )
L stick

m 2 m I max = 1.0 × 10 I = I max __

m 2 (
2
2

(

2 W _
2)
2

= 2.0 m (360 s) 5.2 × 10 L stick = 0.75 m stick shadow
m
L shadow = 0.01 m 3 W _ __
0.75 m

J
_ I = ? = 1.0 × 10 √ 2)
2
2

m ( (0.75 m) + (0.01 m)
s _
2
2
= 2.0 (360)(5.2 × 10 ) m (s) ( )
m 2 3 __ W _ _
2 )
0.75
m
2)

5

= 3.7 × 10 J = 1.0 × 10
2
2
√

( 0.75 + 0.01 m ( √ m

Step 3: Determine the Mass of Sand
3 __ W _
0.75

= 1.0 × 10 )

+
A = 2.0 m 2 m _ ( √ m 2

0.0001

0.56

ρ = and V = zA ∴ m = ρ zA

z = 10.0 cm V 3 _ W _
)
0.75

g

_ m = ρ zA = 1.0 × 10
2

ρ = 1.6 ( √ m
0.5601

cm 3 Convert m to cm :
2
2
m = ? 3 0.75 W _
_

(
4
2 1 × 10 cm 2 = 1.0 × 10 ( )
_
2

2.0 m 2 ) 0.75 m
1 m
W _
3
4
2.0 × 10 cm 2 = 1.0 × 10 (1.0)
2
m
g
_ 4 2
3 W _
m = 1.6 (2.0 × 10 cm )(10.0 cm) = 1.0 × 10

cm 3 2
m
g
_
2
4
= 1.6(2.0 × 10 )(10.0) (cm )(cm) 22.15. Use the equation provided in the Humidity section of the text.
cm 3
g
_
5
= 3.2 × 10 g absolute humidity = 4.0

m 3
Step 4: Determine Temperature Change of Sand T = 0°C
The energy can then be used with the specific heat equations in relative humidity = ?

chapter 4 to determine the change in temperature of the absolute humidity
×

relative humidity = ____ 100%

asphalt. The relationship between heat, the mass of the sand, maximum absolute humidity at temperature

and the temperature change is equation 4.4 in chapter 4. g
_

4.0
5
m
Q
m = 3.2 × 10 g _ = _ 3 × 100%

Q = mc ΔT ∴ ΔT = _

g
mc

_ 5.0
cal

c = 0.2 4 m 3
(8.8 × 10 cal)
g⋅°C ΔT = __

5
energy = 3.7 × 10 J 5 _ = 80%
cal
(
(3.2 × 10 g) 0.2 g⋅°C)

Convert energy to calorie heat:
4
(8.8 × 10 ) _
__ (cal)

1 cal
(
)
5 _ =
5
Q = 3.7 × 10 J 4.184 J (3.2 × 10 )(0.2) (g) _
cal

( g⋅°C)

4
= 8.8 × 10 cal
ΔT = ? = 1.4°C
E-51 APPENDIX E Solutions for Group A Parallel Exercises 693