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12
4.9 × 10 cal
____ 4 cal _

ΔT = Q = 2.5 × 10 g ( 80 )
g
g
g⋅°C)
(
2 _
( 5.5 × 10 (8.4 × 10 m ) 0.17 _ 6
cal
3)
3
9

m = 2.0 × 10 cal

12
4.9 × 10
___ __ The concrete is transferring energy to melt the snow, so the heat
cal

=

g
2
9
3) ( g⋅°C)
cal
( (m )
(5.5 × 10 )(8.4 × 10 )(0.17) _ 3 _ is negative to indicate it is released from the concrete.

m
Step 2: Temperature Change of Concrete
12
_ 3
4.9 × 10
°
Q
= C m = 1.8 × 10 kg _

7.9 × 10 11 cal Q = mcΔT ∴ ΔT =
mc

= 6.2°C c = 0.16 _ Convert kg to g:
g⋅°C
6
Q = −2.0 × 10 cal 3
)
(
1 × 10 g
3

23.5. This problem should be solved in two steps. Determine the heat ΔT = ? 1.8 × 10 kg _

absorbed by the air to increase its temperature. Then use this 1 kg

6
1.8 × 10 g
value to determine the mass of condensing water. 6
−2.0 × 10 cal

ΔT = ___

(
g⋅°C)
Step 1: Heat to Change Temperature of Air (1.8 × 10 g) 0.16 _
cal
6

g
2 _

ρ = 7.6 × 10
m 3 __ _
6
−2.0 × 10
cal

4
V = 8.9 × 10 km 3 =
6
( g⋅°C)
cal
(1.8 × 10 )(0.16) _

ΔT = 5.4°C (g)
cal
_

c = 0.17 −2.0 × 10 6
°
g⋅°C = _

C
2.9 × 10 5
Q = ?
= −6.9°C
m _
Q = mcΔT and ρ =

V
23.7. The temperature trends at stations 2 and 3 show warm air over

Rearrange the density formula to solve for mass:
colder air. Examination of Figure 23.11 shows this pattern occurs
m _

ρ = ∴ m = ρV with a warm front. Based on the temperatures at an altitude of
V
1,500 m, the front is situated between stations 2 and 3.
Substitute this expression into the specific heat formula:
23.8. The temperature trend at station 4 shows warm air while the

Q = (ρV)cΔT
trends at the other stations are colder with a sharp distinction
3
Convert km to m : 3
between the trend at station 4 and at the other stations.
3 )
(
6
3 1 × 10 m
4

8.9 × 10 km _ 3 Examination of Figure 23.10 shows this pattern occurs with a

1 km cold front. The front is between stations 3 and 4.

10
8.9 × 10 m 3 g 23.9. The trend over time was toward higher pressure, and the wind
cal
( 2 _ 3) 10 3 ( _
Q = 7.6 × 10 (8.9 × 10 km ) 0.17 g⋅°C) direction indicated the air was circulating in a clockwise direction

(5.4°C)
m
g
( g⋅°C)
3)
_
(
3 _
2
10

= (7.6 × 10 )(8.9 × 10 )(0.17)(5.4) (km ) cal (°C) 23.10. so the student’s observations recorded an anticyclone or a high.
m
Use the values provided and the formula in example 22.4 to
13
=6.2 × 10 cal solve for insolation.
Step 2: Mass of Condensing Water Low Latitude
3 W _
13

Q = 6.2 × 10 cal _ I max = 1.0 × 10 I = I max __

L stick

Q
Q = mL v ∴ m = m 2 √ L + L )

2
2

cal _ L v (

shadow
L v = 540 __ L stick = 0.75 m stick
g
13
6.2 × 10 cal
m = L shadow = 0.33 m
3 W _ __
0.75 m
cal _
m = ? 540 I = ? = 1.0 × 10 √ 2)

g
2

2
m ( (0.75 m) + (0.33 m)
13
6.2 × 10 _
_ cal
= 0.75 m

3 __ W _ _
2 )
2)
540 cal _ = 1.0 × 10

( 0.75 + 0.33 m ( √ m

2
g
2
√

11
= 1.1 × 10 g 3 __ W _
0.75
)

= 1.0 × 10

+
( √ m 2

0.56
0.11
23.6. This problem should be solved in two steps. Determine the heat

absorbed from the concrete due to melting of the snow. Use the = 1.0 × 10 0.75
3 _ W _
)

heat to calculate the temperature change of the concrete. Th e ( √ m 2

0.67
specific heat of concrete is provided in Table 4.2.

_
3 0.75 W _

= 1.0 × 10 ( )
Step 1: Heat from Melting 0.82 m 2
m = 25 kg Q = mL f = 1.0 × 10 (0.91)
W _
3

cal _ m 2

L f = 80 Convert kg to g:
g
2 W _
= 9.1 × 10

Q = ? 1 × 10 g m 2
3
(
25 kg _ )

1 kg
4
2.5 × 10 g
E-53 APPENDIX E Solutions for Group A Parallel Exercises 695