Laboratory Chapter 18 (Building a Planet) - Carlo Garcia
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Building a Planet
PRE-LAB QUESTIONS
1.
Which of the following are chemical layers? (Select all that apply.)
a.
Lithosphere
b.
Core
c.
Inner Core
d.
Mantle
e.
Crust
2.
Given what you know about how the Earth differentiated, what would you expect the
interiors of other planets in the solar system to look like?
Given what I know about how the Earth is differentiated, I would expect the other
planets of the solar systems interiors to differ based on their sizes, compositions,
and distances from the Sun. We assume metallic cores for rocky planets like
Mercury, Venus, and Mars, while gas giants like Jupiter and Saturn have gas and
liquid layers, and ice giants like Uranus and Neptune have rocky cores
surrounded by icy mantles.
©2016, eScience Labs
Building a Planet
EXERCISE 1 DATA SHEET
Table 1. Unknown Rock 1 Density Measurements
Trial 1
Trial 2
Trial 3
Mass (g)
37.9
31.6
32.6
Volume (V
0
, mL)
100
100
100
Volume (V
1
, mL)
112
112
111
Rock Volume (mL)
12
12
11
Table 2. Unknown Rock 2 Density Measurements
Trial 1
Trial 2
Trial 3
Mass (g)
11.1
8.8
3.3
Volume (V
0
, mL)
100
100
100
Volume (V
1
, mL)
102
101
101
Rock Volume (mL)
2
1
1
Table 3. Density of Substances
Material
Density
(ρ)
Continental Crust
2.7 g/cm
3
Oceanic Crust
2.9 g/cm
3
Mantle
3.3 g/cm
3
Water
1.0 g/cm
3
©2016, eScience Labs
Building a Planet
POST-LAB QUESTIONS
Use Tables 1, 2, and 3 to answer the following questions:
1.
Find the density for each trial of Unknown Rock 1 using the equation: ρ = M/V. Show
your work, and write the answer for each trial here.
Trial1: Density (ρ) = 3 .16g/cm
Trial2: Density (ρ) = 2 .63g/cm
Trial3: Density (ρ) = 2 .96g/cm
2.
Find the average for the three trials of Unknown Rock 1. Show your work, and write the
answer here.
AVG = (3.16+2.63+2.96) / 3 = 2.92
3.
Find the density for each trial of Unknown Rock 2 using the equation: ρ = M/V. Show
your work, and write the answer for each trial here.
Trial 1 = 11.1/2 = 5.55,
Trial 2 = 8.8/1 = 8.8
Trial 3 = 3.3/1 = 3.3
4.
Find the average for the three trials of Unknown Rock 2. Show your work, and write the
answer here.
AVG = (5.5+8.8+3.3) / 3 = 5.86
5.
Compare the average density you found for Unknown Rock 1 and Unknown Rock 2 to
Table 3 on the Data Sheet. What materials do they resemble?
Unknow Rock 1 = Continental Crust, Unknown Rock 2 is closest to Mantle.
6.
Why is it necessary to take the average of multiple trials to find density? Use your work
to justify your response.
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Building a Planet
Taking the average of multiple trials to find density is necessary because it helps
to reduce the impact of random errors and uncertainties in the measurements.
The only degree of inherent error due in the trials is due to the measurement
limitations, being instrumental.
7.
Apply what you have just learned from the exercise to answer and understand the
following questions: Imagine you have a rectangular wooden block with dimensions of
10 cm x 3 cm x 8 cm (L x W x H).
a.
What is the volume of your wooden block? Show your work.
Volume (V) = Length (L) × Width (W) × Height (H)
Length (L) = 10 cm
Width (W) = 3 cm
Height (H) = 8 cm Volume
(V) = 10 cm × 3 cm × 8 cm Volume
(V) = 240 cm^3
b.
What is the density of this wooden block if it has a mass of 168 g? Show your
work.
Density (ρ) = Mass (M) / Volume (V)
Given the mass (M) = 168 g
We already found the volume in part (a) to be 240 cm^3, so let's proceed with the
calculation: Density (ρ) = 168 g / 240 cm^3
Density (ρ) ≈ 0.7g/cm^3
The Density of the wooden block is approx. 0.7
g/cm^3.
©2016, eScience Labs
Building a Planet
8.
Your wooden block is shown here in a pool of water. Calculate how much of the block
lies below the water line using the equation:
r = (ρ_block / ρ_water) × h
r = (0.7 g/cm^3 / 1.0 g/cm^3) × 8 cm
r = 0.7 × 8 cm
r = 5.6 cm
9.
where
r
is the height below the water and
h
is the total height of the block. The density of
water is given in Table 3.
5.6 cm of the wooden block lies below the waterline in the pool.
10. A section of continental crust with a height of 25 km (h) floats in the mantle. Some of it
lies within the mantle (r), and the rest floats above the mantle. How much of the
continental crust lies within the mantle? Use the concepts you learned from the previous
question and the densities in Table 3 to solve the problem.
ρ_crust = density of the continental crust (given as 2.7 g/cm³ in Table 3)
ρ_mantle = density of the mantle (given as 3.3 g/cm³ in Table 3)
r / h = ρ_crust / ρ_mantle
©2016, eScience Labs
Building a Planet
r / 25,000 m = 2.7 g/cm³ / 3.3 g/cm³
r = (2.7 g/cm³ / 3.3 g/cm³) × 25,000 m
r ≈ 20,202.02 m
approximately 20,454.5.4 meters (or 20.5 km) of the continental crust lies within the
mantle. The rest of the continental crust = 4.5 km
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