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tiL12214_appe_643-698.indd Page 673 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

12
Determine the energy to heat the griddle to cooking 15.11. r P = 1.400 × 10 m
temperature: t P = 29.46 yr
_
3
m = 6.3 × 10 g Q = mc ΔT G = 6.67 × 10 −11 N⋅m 2

kg 2
_ _
(
cal
cal
3

c iron = 0.11 = (6.3 × 10 g) 0.11 g⋅°C) m S = ?
(70°C)
g⋅°C
2 3
_
4π r
T i = 20°C 3 _ m S =
( g⋅°C)
cal

= (6.3 × 10 )(0.11)(70)(g) (°C) Gt 2

T f = 90°C
4
Q = ? = 4.9 × 10 cal Convert time to seconds.
7 s _
yr)
(

t P = 29.46 yr 3.154 × 10
Use the solar energy received from Table 15.1 to determine the
8
time required to heat the pan: t P = 9.292 × 10 s
2
12
3
___
4π (1.400 × 10 m)
_ m S =
cal

solar energy received = 2.0 −11 N⋅m 2

2 )
8
2
cm ⋅s ( 6.67 × 10 _ (9.292 × 10 s) 2

2
A = 8.0 × 10 cm 2 kg
4
2
12 3
Q = 4.9 × 10 cal ___ __
3
4π (1.400 × 10 )
(m)

t = ? = (6.67 × 10 kg⋅m 2
−11
8 2
)
)(9.292 × 10 )
_
⋅

m

2
Q = (solar energy received)(At) ( s _ s) 2
(

kg 2
Rearrange to solve for t. ___
2
36
4π (2.744 × 10 )

kg

Q
__ = −11
17
t = (6.67 × 10 )(8.634 × 10 )

(solar energy received)(A) (1.083 × 10 )
38
4

kg
___ = __
4.9 × 10 cal
7
t = cal (5.759 × 10 )
(8.0 × 10 cm )
2.0 _ 2 2 30
2
cm ⋅s = 1.88 × 10 kg
4
__ _
cal
4.9 × 10
= 15.12. Remember that when you apply Kepler’s third law, the time is

2
cal
2.0(8.0 × 10 ) _ 2 compared to the time of Earth’s orbit and that astronomical units
(cm )

2
cm ⋅s
must be used for distances between planets.
4.9 × 10 4
_
=

s

d(planet)
1.6 × 10 3 t(Earth) = 1 yr _ 2 _ 3
t(planet)

2
= 3.1 × 10 s d(Earth) = 1 AU t(Earth) 2 = d(Earth) 3
= 31 s d(Jupiter) = 5.2 AU 3
(
2 d(planet)
2

t(Jupiter) = ? t(planet) = t(Earth) _ 3)
11
15.10. r P = 2.28 × 10 m d(Earth)
t P = 1.88 yr √ 2 d(planet) 3)

(
3
_

_
G = 6.67 × 10 −11 N⋅m 2 t(planet) = t(Earth)

d(Earth)
kg 2
m S = ? 3
2 (5.2 AU)
√
_

= (1 yr) 3

2 3
4π r
_ (1 AU)
m S =
Gt 2 3
3
√
2 (AU)
2 5.2 _

= 1 (yr) _
Convert time to seconds. 1 3 (AU) 3
2 2
= √ 1.4 × 10 yr
7 s _
yr)
(
t P = 1.88 yr 3.154 × 10

1
= 1.2 × 10 yr
7
t P = 5.93 × 10 s = 12 yr
2
11
3
4π (2.28 × 10 m)
___
m S = 15.13. Remember when you apply Kepler’s third law, the time is
2
( 6.67 × 10 −11 N⋅m 2 ) 7 2 compared to the time of Earth’s orbit and that astronomical
_

(5.93 × 10 s)

kg
units must be used for distances between planets.
3
11 3
2
4π (2.28 × 10 )
(m)
___ __
= t(Earth) = 1 yr d(planet) 3 _ 2

t(planet)
_
7 2
−11
(6.67 × 10 )(5.93 × 10 ) kg⋅m m 2 ) d(Earth) = 1 AU d(Earth) 3 = t(Earth) 2

_
⋅

2
(

( s _ s) 2 2

2]
[
3 t(planet)
kg 2 t(Mars) = 1.88 yr d(planet) = d(Earth) _
3
d(Mars) = ? t(Earth)
34
2
4π (1.19 × 10 )
___
= 3
kg

2
t(planet)
15
(6.67 × 10 −11 )(3.52 × 10 ) √ 3 _ ]

d(planet) = d(Earth) [ 2
35
(4.70 × 10 ) t(Earth)
_
=

kg
5
(2.35 × 10 ) 3 2
(1.88 yr)
√
3 _

= 2.00 × 10 kg = (1 AU) (1 yr) 2
30
3
(yr)
√ 3 1.88 2 2 _ 2
_
= 1 1 2 (A U) (yr) 2

3 3
= √ 3.53 AU
= 1.52 AU
E-31 APPENDIX E Solutions for Group A Parallel Exercises 673