Energy and the Environment– 3rd
ST
Edition
UV
SOLUTION
IA
_A
MANUAL
PP
Robert A. Ristinen, Jack J. Kraushaar & Jeffrey T. Brack
RO
Complete Solution Manual for Instructors and
VE
Students
D?
© Robert A. Ristinen, Jack J. Kraushaar & Jeffrey T. Brack
All rights reserved. Reproduction or distribution without permission is prohibited.
??
©MEDGEEK
, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
Chapter 1: Energy Fundamentals
Odd-numbered problems
ST
•Problem 1.3 : Cart on a horizontal surface
Given: F = 10lb and d = 10f t
UV
F ×d=W
10lb × 10f t = 100f t · lbs
j
100f t · lbs × 1.36 = 136 joules
f t · lb
IA
•Problem 1.7 : Tons of coal per person
_A
From Table 1.1: 98 QBtu/yr used in U.S.
U.S. population approximately 300 × 106 people.
1 ton coal
98 × 1015 Btu × 7
= 3.6 × 109 tons
2.7 × 10 Btu
PP
3.6 × 109 tons tons
Then: 6
= 12 , approximately.
300 × 10 people yr · person
•Problem 1.11 : Windmill heats water
RO
f t3 lb
40 gallons × 0.1337 × 62.4 3 = 334 lb
gal ft
Btu
334 lb × 1 × 50°F = 16, 700 Btu needed
lb · °F
1, 055 j
VE
16, 700 Btu × = 17.6 × 106 j needed
Btu
j
1400 W = 1400
sec
6
17.6 × 10 j
D?
= 12, 600 seconds = 210 minutes = 3.5 hours
1400j/s
??
1
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, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
Multiple choice problems
•Problem 1.1 : Product with exponentials
Answer: f)
ST
(5 × 6 × 7) × 10(5+6+7) = 210 × 1018 = 2.1 × 1020
UV
•Problem 1.3: Mass on string
Answer: c)
Initially: P.E. = mgh (see page 113)
m
= 5kg × 9.8 2 × 2m
s
IA
2
m
= 98kg 2
s
= 98Joules
_A
•Problem 1.5 : U.S. consumption vs. India
Answer: d)
PP
50
From Fig. 1.3: = 12.5
4
•Problem 1.7 : Average personal power
RO
Answer: b)
3000C = 3 × 106 cal
cal j 1 day 1 hour j
3 × 106 × 4.184 × × = 145 = 145 watts
day cal 24 hours 3600 sec sec
VE
•Problem 1.9 : Fossil fuel percentage of energy use
Answer: c)
D?
From Table 1.1: 18.5 + 27.3 + 36.1 = 81.9
??
2
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, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
•Problem 1.11 : Energy in a pound
Answer: b)
1 lb = 0.454 kg
ST
mc2 = 0.454 × (3 × 108 )2 = 4.1 × 1016
UV
•Problem 1.13 : Math
Answer: c)
4.8 × 3.6
× 109+5−10 = 6.2 × 104
2.8
IA
_A
PP
RO
VE
D?
??
3
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ST
Edition
UV
SOLUTION
IA
_A
MANUAL
PP
Robert A. Ristinen, Jack J. Kraushaar & Jeffrey T. Brack
RO
Complete Solution Manual for Instructors and
VE
Students
D?
© Robert A. Ristinen, Jack J. Kraushaar & Jeffrey T. Brack
All rights reserved. Reproduction or distribution without permission is prohibited.
??
©MEDGEEK
, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
Chapter 1: Energy Fundamentals
Odd-numbered problems
ST
•Problem 1.3 : Cart on a horizontal surface
Given: F = 10lb and d = 10f t
UV
F ×d=W
10lb × 10f t = 100f t · lbs
j
100f t · lbs × 1.36 = 136 joules
f t · lb
IA
•Problem 1.7 : Tons of coal per person
_A
From Table 1.1: 98 QBtu/yr used in U.S.
U.S. population approximately 300 × 106 people.
1 ton coal
98 × 1015 Btu × 7
= 3.6 × 109 tons
2.7 × 10 Btu
PP
3.6 × 109 tons tons
Then: 6
= 12 , approximately.
300 × 10 people yr · person
•Problem 1.11 : Windmill heats water
RO
f t3 lb
40 gallons × 0.1337 × 62.4 3 = 334 lb
gal ft
Btu
334 lb × 1 × 50°F = 16, 700 Btu needed
lb · °F
1, 055 j
VE
16, 700 Btu × = 17.6 × 106 j needed
Btu
j
1400 W = 1400
sec
6
17.6 × 10 j
D?
= 12, 600 seconds = 210 minutes = 3.5 hours
1400j/s
??
1
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, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
Multiple choice problems
•Problem 1.1 : Product with exponentials
Answer: f)
ST
(5 × 6 × 7) × 10(5+6+7) = 210 × 1018 = 2.1 × 1020
UV
•Problem 1.3: Mass on string
Answer: c)
Initially: P.E. = mgh (see page 113)
m
= 5kg × 9.8 2 × 2m
s
IA
2
m
= 98kg 2
s
= 98Joules
_A
•Problem 1.5 : U.S. consumption vs. India
Answer: d)
PP
50
From Fig. 1.3: = 12.5
4
•Problem 1.7 : Average personal power
RO
Answer: b)
3000C = 3 × 106 cal
cal j 1 day 1 hour j
3 × 106 × 4.184 × × = 145 = 145 watts
day cal 24 hours 3600 sec sec
VE
•Problem 1.9 : Fossil fuel percentage of energy use
Answer: c)
D?
From Table 1.1: 18.5 + 27.3 + 36.1 = 81.9
??
2
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, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
•Problem 1.11 : Energy in a pound
Answer: b)
1 lb = 0.454 kg
ST
mc2 = 0.454 × (3 × 108 )2 = 4.1 × 1016
UV
•Problem 1.13 : Math
Answer: c)
4.8 × 3.6
× 109+5−10 = 6.2 × 104
2.8
IA
_A
PP
RO
VE
D?
??
3
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