** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #1
(20 Points Total)
CHAPTER 1
QUESTIONS FOR THOUGHT
2. What two things does a measurement statement always contain?
What do the two things tell you? (see p 4 and 5) (2 Pt)


A measurement statement always contains a number and the name of the referent unit.
The number tells “how many,” and the unit explains “of what.”
4. Describe the metric standard units for length, mass and time. (see p 4 to 8) (2 Pt)

The standard unit for length in the metric system is the meter. Currently the meter is
defined by the distance that light will travel in a vacuum during a time period of
1/299,792,458 seconds.

The standard unit for mass in the metric system is the kilogram. A metal cylinder kept by
the International Bureau of Weights and Measures in France defines the kilogram. This
is the only standard unit that is still defined in terms of an object.

The standard unit of time is the second. Currently a second is defined by the
characteristic vibrations of a Cesium-133 atom.
5. What Does the density of a liquid change with the shape of a container? Explain. (See p 10)
(2 Pt)

If matter is distributed the same throughout a volume (i.e. homogeneously), then the
density will remain the same throughout the volume independent of the shape of the
volume. Density is an intrinsic property of the3 material and does not depend of the
shape, volume or mass of the material.
7. What is an equation? How are equations used in the physical sciences? (See p 12) (2 Pt)
An equation is a mathematical statement that describes a relationship where the quantities on
one side of the equal sign are identical to the quantities on the other side.

9.
Equations are used to describe quantities, define concepts, and show how quantities
change together.
What is a model and how are models used?
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #1
(20 Points Total)
I “appropriated” the following discussion on scientific models from the University of Texas (url
http://www.utexas.edu/courses/bio301d/Table.of.contents.html )
The model is the most basic element of the scientific method. Everything done in science is done with models. A model is
any simplification, substitute or stand-in for what you are actually studying or trying to predict. Models are used because
they are convenient substitutes, the way that a recipe is a convenient aid in cooking. This section of the book is dedicated to
explaining what models are and how they are used.
Models are very common. The ingredients list on a bottle of ketchup is a model of its contents, and margarine is a model of
butter. A box score from a baseball game is a model of the actual event. A trial over an automobile accident is a model of
the actual accident. A history exam is a model designed to test your knowledge of history.
A model is a substitute, but it is also similar to what it represents. Thus the ingredients list is a fairly accurate guide to the
contents of the ketchup bottle. Margarine looks and spreads like butter, and can substitute for it in many recipes. The box
score contains most of the critical information about the baseball game---such as the winner, the final score, and the
pitchers. Similarly, trials and history exams contain the essence of the events they model. In fact, models are more than just
common, they are ubiquitous. Nearly everything we encounter is a model. To drive home this point, we list in Table 4.1
several objects or ideas that are models.
Models inside science
Scientific models are fundamentally the same as models outside of science, which will be introduced below. Many people
think mistakenly that scientific models are always complicated, impenetrable mathematical equations. But in truth, many
scientific models are just as understandable as are models found outside of science.
The USDA food pyramid, which recommends the proportions of different kinds of foods in a healthy diet, is a model of the
thousands of scientific studies that have been undertaken on the relation among cancer, heart disease and diet. The figure
summarizes these studies in a picture that recommends healthy diets. Thus, this figure is a substitute for the many scientific
studies on diet, and it is also a substitute for an actual diet.
As a second example, when scientists use rats to determine whether a food additive causes cancer, the rats become a model
of humans. Rats are convenient because they are relatively easy to raise in the lab (at least compared to humans), and one
can perform experiments on them relatively quickly (in a matter of months rather than years). Moreover, most people find it
more ethical to experiment on rats rather than humans.
Big Models and Small Models.
Even the most rudimentary science course contains some of the grand, all-encompassing, models that scientists have
discovered. The periodic table of the elements is a model chemists use for predicting properties of the elements. Physicists
use Newton's law to predict how objects will interact, such as planets and spaceships. In geology, the continental drift
model predicts the past positions of continents. But these three models are atypical because they are immensely successful.
Most models used are nowhere near so powerful or widely useful. But scientists use these less-successful ones anyway.
Models are used at every turn in a scientific study. Samples are models. Ideas are models. Methods are models. Every
attempt at a scientific study involves countless models, many of them small and of interest only to a small group of other
scientists. The primary activity of the hundreds of thousands of U.S. scientists is to produce new models, resulting in tens of
thousands of scientific papers published per year.
10. Are all theories completely accepted or completely rejected? Explain.
I appropriated the following from Wikipedia’s Entry on Theory
According to the United States National Academy of Sciences, “Some scientific explanations are so well established that
no new evidence is likely to alter them. The explanation becomes a scientific theory. In everyday language a theory means
a hunch or speculation. Not so in science. In science, the word theory refers to a comprehensive explanation of an important
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #1
(20 Points Total)
feature of nature supported by facts gathered over time. Theories also allow scientists to make predictions about as yet
unobserved phenomena.”
According to the American Association for the Advancement of Science, “A scientific theory is a well-substantiated
explanation of some aspect of the natural world, based on a body of facts that have been repeatedly confirmed through
observation and experiment. Such fact-supported theories are not "guesses" but reliable accounts of the real world. The
theory of biological evolution is more than "just a theory." It is as factual an explanation of the universe as the atomic
theory of matter or the germ theory of disease. Our understanding of gravity is still a work in progress. But the phenomenon
of gravity, like evolution, is an accepted fact.”
The primary advantage enjoyed by this definition is that it firmly marks things termed theories as being well supported by
evidence. This would be a disadvantage in interpreting real discourse between scientists who often use the word theory to
describe untested but intricate hypotheses in addition to repeatedly confirmed models. However, in an educational or mass
media setting it is almost certain that everything of the form X theory is an extremely well supported and well tested theory.
This causes the theory/non-theory distinction to much more closely follow the distinctions useful for consumers of science
(e.g. should I believe something or not?)
The term theoretical is sometimes informally used in place of hypothetical to describe a result that is predicted, but has not
yet been adequately tested by observation or experiment. A hypothesis is the application of a theory or theories to new
conditions which has yet to be tested while a theory is a prediction based on previous observations or experiments of the
same or similar circumstances. It is not, however, uncommon for a theory to produce predictions that are later confirmed or
proven incorrect by experiment. By inference, a prediction proved incorrect by experiment demonstrates the hypothesis is
invalid. This either means the theory is incorrect, or the experimental conjecture was wrong and the theory did not predict
the hypothesis.
So now…let’s answer the question. Are all theories always completely accepted or completely
rejected? The answer is clearly “No”. Good scientific theories are (should be) accepted by scientists
within the range of validity of the theory. For example, Newton’s theory of gravity (to be studied a bit
next week) is very good at describing the effect of gravity on objects in everyday life, even to the
extreme of navigating spacecraft to the surfaces of other planets! However, Newton’s theory of
gravity is not valid in regions of intensely strong gravitational fields around astronomical objects such
as neutron stars and black holes. Here, another theory of gravity (Einstein’s General Theory of
Relativity) must be used. In this case, scientists hold fast to Newton’s theory unless they are working
on problems in these extreme areas. So the Newton’s theory is not completely accepted in that it must
be rejected, in favor of Einstein’s theory, in situations of extreme gravity.
GROUP B SOLUTIONS
1. What is your mass in kilograms? In grams?
Givens:
Your weight (use my weight for this solution) = 165 pounds
Find:
mass
Simply use the conversion factor from the front of the text to convert from pounds to grams: 1 lb =
453.6 g
 453.6 g 
165 lb  165 lb  
  72,576 g  72.576 kg
 1 lb 
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #1
(20 Points Total)
My mass is 72, 576 grams or, equivalently, 72.576 kilograms
2. What is the mass density of iron if 5.0 cm3 has a mass of 39.5 g? (3 Pt)
Givens:
Volume (V) = 5.0 cm3
mass (m) = 39.5 g
Find:
density (ρ)
Since mass density is given by the relationship  = m/V, then
 

m
V
39.5
5.0
 7.9
39.5 g
5.0 cm 3

g
cm3
g
cm3
The mass density of iron is 7.9 g/cm3.
3. What is the mass of a 10.0 cm3 cube of copper?
Givens:
The mass density (ρ) of copper is given as 8.96 g/cm3.
Solving ρ = m/V for mass (M) gives
M
V
M
 V  V
V
 V  M

g
10.0 cm 3  89.6 g
3
cm
3
The mass of the 10.0 cm copper cube is 89.6 grams.
 M    V  8.96
4. If ice has a mass density of 0.92 g/cm3, what is the volume of 5,000 g of ice?
Givens:
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #1
(20 Points Total)
mass (m) = 5,000 g
mass density (ρ) = 0.92 g/cm3
Find:
Volume (V)
Solving the relationship  = m/V for volume gives V = m/, and
5,000 g
V 
g
0.92
cm3
5,000 g cm3


0.92
1
g
 5, 434.7826 cm 3
 5, 400 cm3
The volume of 5,000 g of ice is 5,400 cm3.
5. If you have 51.5 g in a 50.0 cm3 volume of one of the substances listed in Table 1.3, which one
is it?
Givens:
Volume (V) = 50.0 cm3
mass (m) = 51.5 g
Find:
density (ρ)
Since density is given by the relationship  = m/V, then

m
51.5 g
51.5 g
g


 1.03
3
3
V 50.0 cm
50.0 cm
cm3
The density of the substance is 1.03 g/cm3 which matches the density of seawater in Table
1.3
6. What is the mass of gasoline ( - 0.680 g/cm3) in a 94.6 L gasoline tank? (3 Pt)
Givens:
The mass density (ρ) of iron is given as 7.87 g/cm3.
Converting 94.6 L to cm3
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #1
(20 Points Total)
1000 cm 3
94.6 kg *
= 94,600 cm3 (See Figure 1.8 page 8)
1L
Solving ρ = m/V for mass (M) gives
M
V
M
 V  V
V
 V  M

g
 94,600 cm 3  64,328 g  64.3 kg
cm 3
The mass of gasoline in a 94.6 L gasoline tank is 64.3 kg.
 M   V  0.680
7. What is the volume of a 2.00 kg pile of iron cans that are melted, then cooled into a solid
cube?
Givens:
mass (m) = 2.00 kg = 2,000 g
mass density (ρ) = 7.87 g/cm3
Find:
Volume (V)
Solving the relationship  = m/V for volume gives V = m/, and
m
2000 g
V 
 254 cm 3
 7.87 g
cm 3
The volume of 2,000 g of iron is 254 cm3.
8. A cubic tank holds 1,000 kg of water. What are the dimensions of the tank in meters?
Explain your reasoning? (3 Pt)
Givens:
The mass of water in the tank is 1,000 kg = 1x106 g. (1 kg = 1,000 g = 1x 103 g)
The density of water is 1.0 g/cm3. (See Table 1.4, page 10)
The volume of a cube is V = l3, where l is the length of a side.
Solving ρ = m/V for volume (V) gives
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #1
(20 Points Total)
M
V
M
 V  V
V
 V M




1  106 g
 1  106 cm3
g
 1.00
cm3
The volume of the tank is 1x 106 cm3. To find the length of a side simple take the cube root of the
volume.
l3V
V 
M

l  3 1  10 6 cm 3  1  10 2 cm  1 m
The tank, therefore, has sides of length 1 meter.
9. A hot dog bun (volume 240 cm3) with a density of 0.15 g/cm3 is crushed in a picnic cooler into a
volume of 195 cm3. What is the new density of the bun?
This is a great example of a complex problem that can be easy by recognizing that it is really two
separate simple problems. First find the mass of the hot dog bun from its volume 240 cm3 and density
0.15 g/cm3, then find the new density using the lower “crushed” volume of 195 cm3.
M
M
Using  
it is easy to show that the mass of the hot dog is 36.0 g. Using  
again with the
V
V
new volume, it is easy to calculate the new density of 0.185 g/cm3. However, I’d like you to see how
the problem is even more interestingly solved with fewer calculations, by just exploring the
relationships algebraically.
Start with the expression for the mass of the hot dog bun M   Original  VOriginal , where the “Original”
subscript indicates the original volume and density of the bun. Now write the expression for the new
M
density of the crushed bun  Crushed 
. Since the mass of the bun is a constant in the problem,
VCrushed
you can substitute the expression for the mass in terms of the original properties into the density

V
V

relation for the crushed bun: Crushed  Original Original  Original   Original  . What we have now is an
VCrushed
 VCrushed 
expression for the new crushed density in terms of the original density and the ratio of the volumes –
no mass calculation is needed! Thus the new crushed density of the hot dog bun is given by
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #1
(20 Points Total)
 VOriginal 
g  240 cm3 
g
g
  0.15 3  
  0.15 3 1.23  0.185 3
3 
cm  195 cm 
cm
cm
 VCrushed 
Crushed  Original  
Thus the new density of the crushed hot dog bun is 0.185 g/cm3
10. According to Table 1.3, what volume of iron would be needed to balance a 1.00 cm3
This is another great example of a complex problem that can be easy by recognizing that it is really
two separate simple problems. First find the mass of 1cm3 of iron and then find the volume of copper
that would have that mass. Let’s do it algebraically to reduce the number of calculations needed.
Start with the expression for the mass of the iron M  Iron VIron . Now write the expression for the
M
volume of copper VCopper 
. Since the mass of the iron equals the mass of the copper, you can
Copper
substitute the expression for the mass in terms of the iron properties into the volume relation for the
 

 V
copper: VCopper  Iron Iron  VIron   Iron  . What we have now is an expression for the volume of


 Copper
 Copper 
copper in terms of the original volume of iron and the ratio of the densities – no mass calculation is
needed! Thus the volume of the copper is given by
 V
VCopper  Iron Iron
Copper
g

 7.87
  Iron 
cm3
  1 cm3  
 VIron  


 8.96 g
 Copper 

cm3



7.87 
3
  1 cm3  
  0.878 cm =

 8.96 


Thus the volume of copper needed to balance 1 cm3 of iron is 0.0.878 cm3.