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15.5. Determine the conversion ratio (CR) based on actual units ice, c is the specific heat (see Table 4.2), and Δ T is the
rounded up to the nearest whole number coeffi cient. Th e temperature change. Determine the heat needed to raise the
distance from the Sun to Mars is used because only the temperature of the ice to its melting point, and compare to the
terrestrial planets are being modeled. This distance is energy received from the Sun. If there is excess energy from the

8
2.28 × 10 km, so the conversion ratio would be determined Sun, use the latent heat relationship Q = mL f to determine the
8
based on a rounded actual distance of 3 × 10 km: mass of the ice that melts.
_ First, determine the mass of the ice:
d model
d model = d actual (CR) ∴ CR =
d actual 3 2 4 3
V ice = (2.5 × 10 cm )(30 cm) = 7.5 × 10 cm
)
_
100 cm model
g

2.5 m model _ m _

__ ρ ice = 0.917 ρ = ∴ m = ρV
( 1 m model

CR = 9 cm 3 V
3 × 10 km actual
m ice = ? m ice = ρ ice V ice
_ ) _ 4 3
g
cm model
m model (

2.5(100)

_ __ = 0.917 (7.5 × 10 cm )
m model
3

= 9 cm
3 × 10 km actual = 6.9 × 10 g
4
−8 cm model
= 8 × 10 _ Determine energy required to heat the ice to its melting point:

km actual 4
m ice = 6.9 × 10 g Q = mcΔ T
15.6. Determine the conversion ratio (CR) based on the radii of cal _
(
cal
4
(273°C)

c ice = 0.500 _ = (6.9 × 10 g) 0.500 g⋅°C)

Earth and Jupiter: g⋅°C
cal
3
( g⋅°C)
r Earth = 6.38 × 10 km T i = −273 °C = (6.9 × 10 )(0.500)(273)(g) _ (°C)
4

4
r Jupiter = 7.15 × 10 km T f = 0°C 6
_ Q = ? = 9.4 × 10 cal
r Jupiter
r Jupiter = r Earth (CR) ∴ CR =
r Earth
Use the solar energy received from Table 15.1 and multiply by
4
__
7.15 × 10 km

CR = the area and the duration of time to determine the energy

3
6.38 × 10 km received:
4
_ _
7.15 × 10 km

= cal

3
6.38 × 10 km solar energy received = 0.86 _

2
= 11.2 cm ⋅s
t = 1 h
3
Apply the conversion ratio to determine the diameter of the A = 2.5 × 10 cm 2
model of Jupiter from the diameter of the model of Earth. Q = ?
r Earth Model = 2.5 cm r Jupiter Model = r Earth Model (CR) Q = (solar energy received)(At)
CR = 11.2 = 2.5 cm(11.2) Convert time to seconds:
1
r Jupiter Model = ? = 2.8 × 10 cm
3 s _
)
(

= 28 cm t = 1 h 3.6 × 10
h
3
15.7. Use the solar energy received from Table 15.1, and multiply by = 3.6 × 10 s
the area and the duration of time this energy is received: Q = 0.86 _ 3 2 3
cal

(2.5 × 10 cm )(3.6 × 10 s)
2
_ cm ⋅s
cal

solar energy received = 0.86 cal

3 _
3
2
2
cm ⋅s = 0.86(2.5 × 10 )(3.6 × 10 ) (cm )(s)

2
t = 1 h 6 cm ⋅s
A = 1 m 2 = 7.7 × 10 cal
Q = ? Compare the energy received to the energy required to heat the
Q = (solar energy received)(At) ice to its melting point:
6
Convert time to seconds: 7.7 × 10 cal < 9.4 × 10 cal
6
3 s _
)
(
t = 1 h 3.6 × 10 ∴The ice will not reach the melting point, therefore the latent

h
3
= 3.6 × 10 s heat relationship does not apply.
Convert area to square centimeters: 15.9. Determine the heat needed to raise the temperature of the
griddle to the cooking temperature, and compare to the energy
2
(
4 cm
2

A = 1 m 1 × 10 _ 2) received from the Sun.
m
4
= 1 × 10 cm 2 First, determine the mass of the griddle:
_ 4 2 3 V = (8.0 × 10 cm )(1 cm) = 8.0 × 10 cm 3
2
2
2
cal
Q = 0.86 (1 × 10 cm )(3.6 × 10 s)

2
cm ⋅s ρ iron = 7.87
g
_
3 _
4
2
= 0.86(1 × 10 )(3.6 × 10 ) cal (cm )(s) cm 3

2
cm ⋅s m = ?
7
= 3.1 × 10 cal
m _
ρ = ∴ m = ρV

15.8. This problem is an application of the specific heat concept V _
g
3
2

described in chapter 4. The formula for the relationship m = 7.87 (8.0 × 10 cm )
3
cm
between heat supplied and the temperature change is = 6.3 × 10 g
3
Q = mcΔ T, where Q is the heat needed, m is the mass of the
672 APPENDIX E Solutions for Group A Parallel Exercises E-30