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tiL12214_appe_643-698.indd Page 656 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

4.13. (a) Step 1: Cool the water from 18.0°C to 0°C. 4.15. W = J (Q H – Q L )
_
J
Q 1 = mc ΔT = 4,184 (55.0 kcal – 40.0 kcal)
kcal
(
gC°)
cal
= (400.0 g) 1.00 _ (18.0C°) = 4,184 (15.0 kcal)

J
_
_ kcal
cal

= (400.0)(1.00)(18.0) g × × C°
J·kcal
gC° _
= 4,184 ×15.0
kcal
g·cal·C°
_

= 7,200
gC° = 62,760 J
= 7,200 cal = 62.8 kJ
= 7.20 kcal
Step 2: Find the energy needed for the phase change of CHAPTER 5
water at 0°C to ice at 0°C.
1 _
5.1. (a) f = 3 Hz T =

Q 2 = mL f f
g )
( cal _ λ = 2 cm 1 _
= (400.0 g) 80.0 =

1 _
T = ? 3

g·cal
_ s
= 400.0 × 80.0 ×
g 1 _ s _
=

= 32,000 cal 3 1
= 0.3 s
= 32.0 kcal
(b) f = 3 Hz v = λf
Step 3: Cool the ice from 0°C to ice at –5.00°C.
λ = 2 cm 1 _

Q 3 = mc ΔT = (2 cm)(3 )
s
T = ?
cal
(
gC°)
1 _
= (400.0 g) 0.500 _ (5.00C°) = 2 × 3 cm ×

s
_
cal
_
cm

= 400.0 × 0.500 × 5.00 g × × C° = 6

gC° s
g·cal·C° 5.2. Step 1:
_

= 1,000
g·C°
)
0.600 m/s
(

(
= 1,000 cal t = 20.0°C v TP(m/s) = v 0 + _ T P )
°C
= 1.00 kcal f = 20,000 Hz m _ _
0.600 m/s
s (

(20.0°C)

= 331 + )
Q total = Q 1 + Q 2 + Q 3 v = ? °C
= 7.20 kcal + 32.0 kcal + 1.00 kcal λ = ? m _ m _

= 331 + 12.0

s
s
= 40.2 kcal
m _
_ = 343
Q

(b) Q = mL v ∴ m = s
L v
40,200 cal
_ Step 2:

=
cal _

40.0 v _
g v = λf ∴ λ =

f
g
40,200 cal _ _
_

= × m _

343
40.0 1 cal _
s
=

1 _
cal·g
_ 20,000

= 1,005 s
cal
343 _ _
_ m s
×
3
= 1.01 × 10 g = 20,000 s 1
4.14. W = J (Q H – Q L ) = 0.01715 m
J
_ = 0.02 m
= 4,184 (0.3000 kcal – 0.2000 kcal)
kcal
5.3. f 1 = 440 Hz f b = f 2 – f 1
_
J
= 4,184 (0.1000 kcal) f 2 = 446 Hz = (446 Hz) – (440 Hz)
kcal
f b = ? = 6 Hz
J·kcal
_
= 4,184 × 0.1000
kcal Note that the smaller frequency is subtracted from the larger
one to avoid negative beats.
= 418.4 J
656 APPENDIX E Solutions for Group A Parallel Exercises E-14