CH 03 Matter & Energy
Due: 11:59pm on Tuesday, February 10, 2015
To understand how points are awarded, read the Grading Policy for this assignment.
Classifying Matter
Matter is anything that takes up space and has mass. Matter is classified in many different ways, including by its state (solid, liquid, or gas) and by its
composition (pure substances or mixtures).
Part A
The blue spheres below represent atoms. What state of matter is depicted in each bin?
Drag each item to the appropriate bin.
Hint 1. The properties of solids, liquids, and gases
The following chart describes some of the properties of solids, liquids, and gases.
State
Solid
Atomic/molecular motion
Atomic/molecular spacing
oscillation/vibration about fixed point close together
Shape
definite
Volume
definite
Liquid free to move relative to one another close together
indefinite definite
Gas
indefinite indefinite
free to move relative to one another far apart
ANSWER:
Correct
Part B
Classify each type of matter as an element, a compound, a heterogeneous mixture, or a homogeneous mixture.
Drag each item to the appropriate bin.
Hint 1. How to approach the problem
First, identify each type of matter as a pure substance or a mixture. A pure substance has a specific chemical composition, and all samples of the
substance have the same composition. A mixture, in contrast, is a combination of substances in which each substance retains its own identity. In
addition, mixtures do not have constant composition.
Then, further classify each pure substance as either an element or a compound, and classify each mixture as either a heterogeneous mixture or a
homogeneous mixture. It may help to use the periodic table to determine whether a substance is an element.
Hint 2. Define the terms
Complete each sentence with the appropriate term.
Match the words in the left-hand column with the appropriate blank in the sentences in the right-hand column.
ANSWER:
Hint 3. Classify each type of matter as a pure substance or a mixture
Classify each type of matter as a pure substance or a mixture.
Drag each item to the appropriate bin.
Hint 1. Explanation of the different examples of matter
The eleven examples of matter in this item are the following:
Air: Air is composed of about 78% nitrogen and 21% oxygen, with traces of argon, carbon dioxide, and other gases. The gases
are not chemically bonded to each other.
Salt water: Salt water is water that contains dissolved salt.
Orange juice with pulp: Orange juice with pulp contains liquid juice, which is mostly water, and some of the fruit solids (pulp).
Soil: Soil contains air, minerals, water, and organic matter.
Carbon monoxide: Carbon monoxide has the formula CO.
Hot coffee: Hot coffee is typically prepared by steeping ground coffee in hot water and then removing the spent coffee grounds.
Silver: Silver is a soft white metal that is a good thermal conductor.
Aluminum: Aluminum is a silvery-white, lightweight metal.
Salt: Table salt has the chemical formula NaCl.
Calcium: Calcium is a moderately hard, silvery alkaline earth metal.
Acetone: Acetone has the chemical formula (CH3 ) 2 CO .
ANSWER:
Correct
You can identify elements because they appear in the periodic table. Any pure substance that is not an element must be a compound.
Compounds can often be recognized by their name, which references the elements present.
Hint 4. Descriptions of heterogeneous and homogeneous mixtures
Both heterogeneous mixtures and homogeneous mixtures contain two or more atoms or compounds in variable proportions. A heterogeneous
mixture has a composition that is not uniform. That is, one part of the mixture has a different composition from another part. A chocolate-chip cookie
is one example of a heterogeneous mixture. Heterogeneous mixtures may also have components in different phases.
A homogeneous mixture has a composition that is the same throughout. Wine is a homogeneous mixture.
ANSWER:
Correct
A mixture can contain substances in different states. For example, a glass of ice water contains solid ice and liquid water.
Properties and Changes of Matter
Matter has both physical and chemical properties and can undergo physical or chemical changes. Physical properties are those that a substance displays
without changing its composition, whereas chemical properties are evident only during a chemical change (also called a chemical reaction). In contrast, when a
substance undergoes a physical change, it may change appearance, but not its composition.
Part A
Classify each of the changes as a physical change or a chemical change.
Drag each item to the appropriate bin.
Hint 1. Changes in matter
When matter undergoes a physical change, the appearance or texture of the matter may change, but its composition remains the same. For
example, water is made up of H2 O molecules whether it is in liquid form or solid form (ice). Thus, the freezing of water is a physical change.
When matter undergoes a chemical change, the composition of the matter changes. For example, when iron rusts, the iron atoms combine with
oxygen atoms to form a new substance, iron oxide.
Hint 2. Classify descriptions as applying to physical changes or chemical changes
Classify the following descriptions by whether they apply to physical changes or chemical changes.
Drag each item to the appropriate bin.
ANSWER:
ANSWER:
Correct
When tarnish forms on silver, the outer layer of silver atoms has reacted with sulfur to become a new substance, silver sulfide.
Part B
Classify each of the properties as a physical property or a chemical property.
Drag each item to the appropriate bin.
Hint 1. The properties of matter
Physical properties are properties of matter that can be observed without the matter changing composition. For example, the appearance of a
substance can be observed without changing the composition of the substance.
Chemical properties are properties of matter that can only be observed when the matter changes composition. For example, the reactivity of a
substance can only be observed when the substance is reacting (or changing its composition).
A good way to identify chemical properties is to look for properties that seem to describe a chemical reaction. Those that don't appear to describe a
chemical reaction are likely to be physical properties. It may help if you think about what you know about the matter before and after the observation.
If the matter is the same, then it is a physical property. If the matter is not the same, then it is a chemical property.
Hint 2. Classify general properties as physical properties or chemical properties
Classify these properties by whether you would have to perform a chemical reaction to test that property.
Drag each item to the appropriate bin.
ANSWER:
ANSWER:
Correct
Pause and Predict Video Quiz: Chemical and Physical Changes
First, launch the video below. You will be asked to use your knowledge of chemistry to predict the outcome of a demonstration. Then, close the video window
and answer the questions at right. You can watch the video again at any point to review.
Always wear protective clothing, appropriate gloves, and eye gear when in the chemistry lab.
Part A
You happen to be visiting Northern California and you are driving by Suisun Bay, a notorious graveyard for old ships. You notice that all of these ships
appear to be rusting away. Which of the following statements is true?
Hint 1. What is the difference between a chemical change and a physical change?
A chemical change is the result of a chemical reaction. A physical change just changes the state of the substance.
ANSWER:
The rusting of the metal is neither a chemical change nor a physical change.
The rusting of the metal is both a chemical change and a physical change.
The rusting of the metal is a chemical change.
The rusting of the metal is a physical change.
Correct
Part B
Black ice is a thin layer of water on a sidewalk or road that has frozen after the temperature has dropped below freezing. It is called black ice because the
ice is nearly invisible, especially when driving in a car at night. Which of the following statements is true?
Hint 1. What is the difference between a chemical change and a physical change?
A chemical change is the result of a chemical reaction. A physical change just changes the state of the substance.
ANSWER:
Black ice is both a chemical change and a physical change.
Black ice is an example of a physical change.
Black ice is an example of a chemical change.
Black ice is neither a chemical change nor a physical change.
Correct
± Law of Conservation of Mass
Matter is neither created nor destroyed in a chemical reaction. Thus, the mass of the products of a chemical reaction must be equal to the mass of the starting
materials. Formally, this concept is called the law of conservation of mass.
Part A
A sample of sodium reacts completely with 0.213kg of chlorine, forming 351 g of sodium chloride. What mass of sodium reacted?
Express your answer to three significant figures and include the appropriate units.
Hint 1. Apply the law of conservation of mass
In this chemical reaction, two elements (sodium and chlorine) combine to form a compound (sodium chloride):
sodium + chlorine→sodium chloride
The total mass of the starting materials must equal the mass of the product:
(
mass of sodium
mass of chlorine
mass of sodium chloride
)+(
)=(
)
that reacted
that reacted
that formed
How can this equation be rearranged to solve for the mass of sodium that reacted?
ANSWER:
(
mass of sodium
)=
that reacted
(
mass of chlorine
mass of sodium chloride
)+(
)
that reacted
that formed
(
mass of chlorine
mass of sodium chloride
)−(
)
that reacted
that formed
(
mass of sodium chloride
mass of chlorine
)−(
)
that formed
that reacted
Hint 2. Convert the mass to grams
Convert 0.213kg to grams.
Express your answer to four significant figures and include the appropriate units.
Hint 1. Determine the conversion factor between kilograms and grams
How many grams are in 1 kg?
Express your answer to three significant figures and include the appropriate units.
ANSWER:
1000 g
ANSWER:
213.0 g
ANSWER:
138 g
Correct
Part B
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 15.6g of carbon were burned in the presence of 58.4g of oxygen, 16.8g
of oxygen remained unreacted. What mass of carbon dioxide was produced?
Express your answer to three significant figures and include the appropriate units.
Hint 1. Interpret the data
In this chemical reaction, two elements (carbon and oxygen) combine to form a compound (carbon dioxide):
carbon + oxygen→carbon dioxide
The total mass of the starting materials must equal the mass of the product:
(
mass of oxygen
mass of carbon
mass of carbon dioxide
)+(
)=(
)
that reacted
that formed
that reacted
Which of these values is given in the problem statement?
ANSWER:
mass of carbon dioxide that formed
mass of carbon that reacted
mass of oxygen that reacted
Hint 2. Determine the mass of oxygen that reacted
Of the 58.4g of oxygen present, 16.8g did not react. How much oxygen reacted?
Express your answer to three significant figures and include the appropriate units.
ANSWER:
41.6 g
ANSWER:
57.2 g
All attempts used; correct answer displayed
Conservation of Energy
The law of conservation of energy states that energy can be neither created nor destroyed. Energy changes from one form to another. The total amount of
energy in the universe always remains constant.
There are two basic types of energy in the universe: kinetic energy, which is the energy related to the motion of a body, and potential energy, which is either the
energy that is stored within the body or the energy due to chemical composition.
The forms of energy that you observe include heat, light, sound, electrical, chemical, mechanical, and nuclear energy. These forms of energy can be classified
as either potential energy or kinetic energy. For example, sound is produced when a body vibrates, so sound energy is kinetic energy. However, chemical
energy is the energy stored within chemical substances, so it is a form of potential energy.
Part A
Batteries are portable devices that can produce electric current. The electric current produced by a battery is due to the chemical reaction that occurs inside
it. A classic example of a battery is the Daniel cell, named after the scientist who invented it. The Daniel cell makes use of zinc metal and its ions in
conjunction with copper metal and its ions. Electrons that are produced due to the chemical reaction are transferred from the zinc metal to the copper metal
through a conducting wire producing the electric current.
Complete the following statements that relate to the fact that energy is conserved in a Daniel cell.
Drag the appropriate labels to their respective targets.
Hint 1. The chemical reaction occurring in a Daniel cell
In a Daniel cell, an electric current is produced due to the chemical reaction that occurs inside the cell. Half of the cell is made up of a zinc rod
immersed in zinc sulfate solution, and the other half of the cell is made up of a copper rod immersed in copper sulfate solution. During the chemical
reaction, the zinc atoms on the surface of the zinc rod lose two electrons (2e − ) to become Zn2+ ions, and the copper accepts the two electrons
from the zinc, forming Cu2+ ions.
When the two rods are connected by a wire, the electrons flow from the zinc rod to the copper rod. If a bulb is connected in the middle of the wire,
the bulb glows, indicating that an electric current is being produced.
The reaction can be summarized as:
Zn→Zn2+ + 2e −
Cu2+ + 2e − →Cu
Hint 2. Define the terms chemical energy and electrical energy
Sort the following statements about chemical and electrical energy.
Drag the appropriate items to their respective bins.
ANSWER:
ANSWER:
Correct
In a Daniel cell, also known as a galvanic cell, the chemical energy that is present in zinc and copper is converted to electrical energy, which is used to
light a bulb. This shows that the energy is conserved. The energy is not created, nor is it destroyed. It is converted from one form to another. In a
Daniel cell or in any other portable cell, the chemical energy is converted to electrical energy.
Part B
Identify the energy conversions associated with the following appliances.
Drag the appropriate labels to their respective targets.
Hint 1. How to approach the problem
The law of conservation of energy states that energy is converted from one form to other. An appliance works by converting one form of energy into
another form of energy.
To identify what energy conversions take place in each of the appliances, first identify the source of energy, that is, the form of potential energy that
is present. Next, identify what work the appliance does, from which you will be able to find the converted energy.
Using the source of energy present and the converted energy, you can identify the energy conversion associated with the appliance.
For example, in a loudspeaker, the source of energy is electrical energy and the converted form of energy is sound energy. So, the energy
conversion taking place is electrical energy to sound energy.
Hint 2. Identify the sources of energy
Complete the following statements to identify the source of energy for each of the given appliances.
Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before
submitting your answer.
ANSWER:
Hint 3. Identify the energy associated with work done
Energy is related to the work done. The associated energy of an appliance can be identified based upon the work done by the appliance. For
example, mechanical energy relates to the motion of an object, whereas electrical energy relates to the flow of electrons.
Complete the following statements to identify the energy associated with the work done by each of the appliances.
Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before
submitting your answer.
ANSWER:
ANSWER:
Correct
In all of the above examples, energy is converted from one form to another, and the law of conservation of energy is obeyed.
Introduction to Specific Heat
Heat is a form of energy related to the motion, or kinetic energy, of the particles in a substance. When a substance absorbs heat energy, the temperature of the
substance increases. The temperature change (∆T ) is dependent upon how much of the substance (mass) is present and the specific heat (SH ) of the
substance. The formula relating heat, mass, specific heat, and temperature change is
heat = mass × SH × ∆T
The following table lists the specific heats of some common substances:
Substance
Specific heat
[ J/(g ⋅ ∘ C) ]
aluminum (Al )
0.897
copper (Cu)
0.385
cast iron
0.460
gold ( Au)
0.129
silver (Ag)
0.235
sodium chloride (NaCl)
0.864
stainless steel
0.500
water (H2 O)
4.184
Part A
Rank the following species according to the decreasing energy needed to raise the temperature of 10.0 g of the substance by 25.0 ∘ C.
Rank from most to least energy needed. To rank items as equivalent, overlap them.
Hint 1. Calculate the energy needed to raise the temperature of water
How much energy does it take to raise the temperature of 10.0 g of water by 25.0 ∘ C ?
Express the energy numerically in joules. To rank items as equivalent, overlap them.
Hint 1. The relationship between energy and specific heat
To heat 1 g of water by 1 ∘ C requires 4.184 J . But the mass in the question is 10 times greater, and the given temperature change is 25
times greater. So the energy required will be 250 times greater than the specific heat of water.
ANSWER:
energy = 1050
J
Hint 2. Calculate the energy needed to raise the temperature of copper
How much energy does it take to raise the temperature of 10.0 g of copper by 25.0 ∘ C ?
Express the energy numerically in joules.
Hint 1. The relationship between energy and specific heat
To heat 1 g of copper by 1 ∘ C requires 0.385 J . But the mass in the question is 10 times greater, and the given temperature change is 25
times greater. So the energy required will be 250 times greater than the specific heat of copper.
ANSWER:
energy = 96.3
J
Hint 3. Identify the relationship between specific heat and energy
Consider the specific heat of copper versus the specific heat of water, as well the amount of energy needed to raise the temperature of each
substance. Which of the following statements is true?
ANSWER:
As the specific heat increases, the amount of energy decreases.
As the specific heat increases, the amount of energy increases.
As the specific heat increases, the amount of energy remains constant.
ANSWER:
Correct
Part B
If you pour equal amounts of scalding hot water into different metallic cups of equal temperature, which cup will heat up the most? Rank the cups from
hottest to coldest.
Rank from warmest to coolest final temperature.
Hint 1. Identify the relation between temperature change and mass
How is mass related to the temperature change?
Hint 1. How to approach the problem
The heat equation
heat = mass × SH × ∆T
can be rearranged to solve for the temperature change, ∆T :
∆T =
What happens to ∆T as the mass increases?
heat
mass×SH
ANSWER:
increases.
As mass increases, temperature change
decreases.
stays constant.
Hint 2. Identify the relation between temperature change and specific heat
How is specific heat related to the temperature change?
Hint 1. How to approach the problem
The heat equation
can be rearranged to solve for ∆T :
heat = mass × SH × ∆T
∆T =
heat
∆T =
What happens to ∆T as the specific heat increases?
heat
mass×SH
ANSWER:
increases.
As specific heat increases, temperature change
decreases.
remains constant.
ANSWER:
Correct
Energy Conversions
Learning Goal:
To practice converting between different units of energy.
Duncan is a college student studying for a chemistry exam this evening. He also plans to go swimming at the school gym and have a dinner of ramen noodles.
While the water for the ramen is heating, it absorbs energy as it gets hot. Later, while swimming, Duncan uses the potential energy in the food he has
consumed.
The following conversion factors may be useful in this problem:
1 cal = 4.184 J
1 kJ = 1000 J
Part A
Duncan takes a break from studying and goes to the gym to swim laps. If swimming burns 6.25×105cal per hour, how many kilojoules does swimming burn
in that same amount of time?
In other words, convert 6.25×105cal to kilojoules.
Express your answer numerically in kilojoules.
Hint 1. Convert calories to joules
Convert 6.25×105cal to joules. Recall that 1
cal = 4.184 J .
Express your answer in joules to four significant figures.
ANSWER:
6.25×105cal = 2.615×106
J
ANSWER:
6.25×105cal = 2620
kJ
Correct
Part B
Duncan knows that it takes 36400cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has prepared ramen noodles so
many times he does not need to measure the water carefully. If he happens to heat 0.600pint of room-temperature water, how many kilojoules of heat
energy will have been absorbed by the water at the moment it begins to boil?
Express your answer numerically in kilojoules.
Hint 1. Calculate the calories absorbed by the water
It takes 36400cal to heat a pint of water from room temperature to boiling. How many calories would it take to heat just 0.600pint of water?
Express your answer in calories to four significant figures.
Hint 1. The conversion factor
The heat required to boil water is 36400cal per pint. This can be expressed as the following conversion factor:
36400 cal
1 pint
If we had, for example, 2.00 pints of water, we can see that it would take twice as much heat as a single pint would:
2.00 pints × ( 36400 cal ) = 72800 cal
1 pint
Similarly, if we had 0.500 pint of water, we can see that it would take half as much heat as a full pint would:
0.500 pints × ( 36400 cal ) = 18200 cal
1 pint
Follow this same procedure with the value that you were given in the main question: 0.600pint .
ANSWER:
calories absorbed = 21840
cal
Hint 2. Convert calories to joules
Convert 21840cal to joules. Recall that 1
cal = 4.184 J .
Express your answer in joules to four significant figures.
ANSWER:
21840cal = 9.138×104
J
ANSWER:
energy absorbed = 91.4
kJ
Correct
If heating food is literally adding calories to food, why doesn't heating increase its "caloric content"? When you heat a can of soup, you are increasing
the kinetic energy of the molecules in the soup, and thus they move around more rapidly. The calories reported on the can's label refer to the potential
energy in the chemical bonds that hold those molecules together. Your body cannot utilize the kinetic energy of colliding molecules in food, but it can
utilize the potential energy in many of the chemical bonds holding those molecules together.
± Temperature Conversion
Learning Goal:
To express temperatures using the Fahrenheit, Celsius, and Kelvin scales.
Temperature is commonly measured in units of degrees Fahrenheit (∘ F), degrees Celsius ( ∘ C), or kelvins (K ). The most familiar temperature scale in the
United States is the Fahrenheit scale, in which water freezes at 32 ∘ F and water boils at 212 ∘ F . The scale often used by scientists is the Celsius scale, in
which water freezes at 0 ∘ C and boils at 100 ∘ C . A third temperature scale, called the Kelvin scale, avoids negative temperatures by assigning 0 K to the
coldest temperature possible, absolute zero.
Use the following equations to convert between these temperature scales:
TC =
(T F −32)
1.8
T K = T C + 273.15
where T F is the temperature in degrees Fahrenheit, T C is the temperature in degrees Celsius, and T K is the temperature in kelvins.
Part A
Convert 21.0 ∘ C to degrees Fahrenheit.
Express your answer numerically in degrees Fahrenheit.
Hint 1. Identify how to solve for the Fahrenheit temperature
Starting with the formula
TC =
what algebraic steps should you take to isolate T F ?
(T F −32)
1.8
ANSWER:
Multiply both sides by 1.8, then add 32 to both sides.
Add 32 to both sides, then multiply both sides by 1.8.
Swap the positions of T F and T C in the formula.
ANSWER:
69.8
∘
F
Correct
Part B
Becky is going to take her dog for walk in the park, but before she leaves, she watches the weather report. According to the weather report, it is 74.0 ∘ F
outside. What is the outside temperature in kelvins?
Express your answer numerically in kelvins.
Hint 1. How to approach the problem
Although you are not given an equation that allows you to directly convert from degrees Fahrenheit to kelvins, you can use the two equations given
in the introduction to complete the conversion.
Start by converting the temperature to the Celsius scale, and then convert to the Kelvin scale.
Hint 2. Calculate the temperature in degrees Celsius
The first step is to convert to degrees Celsius. Using the formula
substitute 74.0 ∘ F for T F ,
TC =
TC =
then calculate T C .
(T F −32)
1.8
(74.0 ∘ F−32)
1.8
Express your answer numerically in degrees Celsius.
ANSWER:
∘C
23.3
ANSWER:
296.5
K
Correct
Part C
Arrange the following in order of decreasing temperature.
Rank the temperatures from the highest to the lowest. To rank items as equivalent, overlap them.
Hint 1. How to approach the problem
Notice that the temperatures do not have the same units: Some are given in kelvins, some are given in degrees Celsius, and some are given in
degrees Fahrenheit. Convert all the temperatures to a common unit (either kelvins, degrees Celsius, or degrees Fahrenheit) so that you can
compare the values within the same scale.
Hint 2. Convert the temperature in degrees Fahrenheit to degrees Celsius
Convert 133 ∘ F to degrees Celsius.
Express your answer numerically in degrees Celsius.
ANSWER:
∘C
56.1
Hint 3. Convert the temperature in kelvins to degrees Celsius
Convert 351 K to degrees Celsius.
Express your answer numerically in degrees Celsius.
ANSWER:
78
ANSWER:
∘
C
Correct
Scientists rarely work with temperatures in degrees Fahrenheit. Instead, they usually work with temperatures in degrees Celsius or kelvins.
Using the Heat Equation
The specific heat of a substance (which will be given the abbreviation SH in this item) is the amount of heat (q ), in joules or calories, needed to change the
temperature of 1 g of the substance by exactly 1∘ C . The equation and units for specific heat are provided below:
SH =
q
m × ∆T
=
J
g ⋅ ∘C
The heat equation is used to determine the quantity of heat (q ) absorbed or lost by a substance. It is derived from the specific heat relationship by multiplying
both sides by m and ∆T :
q = m × ∆T × SH
Part A
Some homes that heat with baseboards use copper tubing. As hot water runs through this copper tubing, it heats the tubing, which in turn heats aluminum
fins. It is actually the aluminum fins that heat the air rising through the fins.
How much energy would it take to heat up just the copper tubing in a 5.00 ft section of pipe, which weighs about 675.0g , from 14.40∘ C to 23.66∘ C ?
Copper has a specific heat of 0.3850 (J/g) ⋅ ∘ C .
Express your answer to four significant figures.
Hint 1. How to approach the problem
Problems involving the heat equation can be solved using the following steps:
1.
2.
3.
4.
State the given and needed quantities.
Calculate the temperature change (∆T ).
Write the heat equation.
Substitute the given values and solve, making sure like units cancel.
Hint 2. Determine the change in temperature
What was the change in temperature for the copper tubing if it was initially 14.40∘ C and ended up being 23.66∘ C ?
Express your answer to two decimal places figures.
ANSWER:
∘C
∆T = 9.26
ANSWER:
Heat = 2406
J
Correct
The amount of heat absorbed by the 5.00 ft copper pipe (with a mass of 675.0g ) to raise it from 14.40∘ C to 23.66 ∘ C (which is equivalent to heating
from 57.92∘ F to 74.59∘ F ) is 2406J . This is calculated by substituting the known values into the heat equation and solving accordingly.
q = 675.0g × 9.26 ∘ C × 0.385 (J/g) ⋅ ∘ C
q = 2406J
The value is reported to four significant digits to prevent rounding errors in future calculations.
Since heat is absorbed, the energy change is positive (endothermic). When heat is released, such as when the copper pipe cools down, the energy
change will be negative (exothermic).
Part B
The heat equation, as given in the introduction, can also be rearranged to calculate the mass or temperature change for a substance. You would follow the
same steps used to calculate the quantity of heat gained or lost, but when you solve the equation, the term for mass or temperature change will need to be
isolated on one side of the equation.
What mass, in grams, of aluminum fins could 2406J of energy heat from 14.40∘ C to 23.66∘ C ? Aluminum has a specific heat of 0.897 (J/g) ⋅ ∘ C.
Express your answer to three significant figures.
Hint 1. How to approach the problem
Problems involving the heat equation can be solved using the following steps:
1.
2.
3.
4.
State the given and needed quantities.
Calculate the temperature change (∆T ).
Write the heat equation.
Substitute in the given values and solve, making sure like units cancel.
In this specific problem, you will need to rearrange the heat equation as you solve to isolate mass on one side of the equation.
Hint 2. Determine the change in temperature
What was the change in temperature for the aluminum fins if they were initially 14.40∘ C and ended up being 23.66∘ C ?
Express your answer to two decimal places.
ANSWER:
∆T = 9.260
∘C
Hint 3. Identify the equation that is rearranged appropriately to solve for mass
Which of the following equations represents the heat equation, shown here, properly rearranged to solve for mass?
q = m × ∆T × SH
ANSWER:
m =
∆T
q×SH
m = q × ∆T × SH
m =
m =
q
∆T ×SH
∆T ×SH
q
ANSWER:
Mass = 2.90×102
g
Correct
The mass of the aluminum fins that could be heated by 2406J from 14.40∘ C to 23.66∘ C (which is equivalent to heating from 57.92∘ F to 74.59∘ F ) is
calculated by substituting the known values into the rearranged heat equation solving for mass:
m =
m =
q
∆T × SH
2406J
9.26∘ C × 0.897 (J/g)⋅∘ C
m = 2.90×102g
Algebraic signs and energy
Reactions that release energy are said to be exothermic, and reactions that consume energy are said to be endothermic. Energy values associated with
exothermic reactions are negative values (since energy is lost by the system), and energy values associated with endothermic reactions are positive values
(since energy is absorbed by the system).
Part C
Typically, water runs through the copper tubing of the baseboards and, therefore, fresh hot water is constantly being run through the piping. However,
consider a pipe where water was allowed to sit in the pipe. What would be the change in temperature of the water if 196.5g of water sat in the copper pipe
from part A, releasing 2406J of energy to the pipe? The specific heat of water is 4.184 (J/g) ⋅ ∘ C.
Express your answer to four significant figures.
Hint 1. How to approach the problem
Problems involving the heat equation can be solved using the following steps:
1.
2.
3.
4.
State the given and needed quantities.
Calculate the temperature change (∆T ).
Write the heat equation.
Substitute in the given values and solve, making sure like units cancel.
In this specific problem, you will need to rearrange the heat equation as you solve to isolate ∆T on one side of the equation.
Hint 2. Identify the equation that is rearranged appropriately to solve for change in temperature
Which of the following equations represents the heat equation, shown here, properly rearranged to solve for the change in temperature?
q = m × ∆T × SH
ANSWER:
∆T =
m × SH
q
∆T = m × q × SH
∆T =
∆T =
q
m × SH
q × SH
m
Correct
The heat equation can be rearranged into the following format to solve for the change in temperature:
q
m × ∆T × SH
=
m × SH
m × SH
∆T =
q
m × SH
ANSWER:
∆T = -2.927
∘
C
Answer Requested
The change in temperature of the 196.5g of water after it has released 2406 J into the copper pipe is calculated by substituting the known values into
the rearranged heat equation solving for ∆T :
∆T =
∆T =
q
m × SH
−2406J
196.5g × 4.184 (J/g) ⋅ ∘C
∆T = -2.927∘ C
The change in temperature of the water is negative because heat is lost (it is transferred to the copper pipe) and the temperature of the water has
cooled down. Water has a specific heat that is over 10 times that of copper. This allows copper tubing to be a great acceptor of heat for water
traveling through it.
± Specific Heat Calculations
Learning Goal:
To become familiar with the concept and calculations of specific heat.
Specific heat (which can be represented as SH , C s , sp. ht. , or a number of other possibilities) is defined as the amount of energy needed to raise the
temperature of 1 g of a substance by 1 ∘ C. For example, 0.0920 cal is enough energy to raise 1 g of copper from 21.0 ∘ C to 22.0 ∘ C . Therefore, the specific
heat of copper is 0.0920 cal/(g ⋅ ∘ C) .
Using SH for specific heat, the formula for calculating specific heat is
SH =
heat
mass×∆T
Part A
How much heat energy is required to raise the temperature of 0.366kg of copper from 23.0 ∘ C to 60.0 ∘ C ? The specific heat of copper is
0.0920 cal/(g ⋅ ∘ C) .
Express your answer with the appropriate units.
Hint 1. Identify the unknown
Visualize the equation as having empty compartments that can be filled with information:
Consider the values given in the question. Fill in each square with the appropriate number and units. Which square will remain empty in this case?
ANSWER:
SH
heat
mass
∆T
Correct
The specific heat and mass are given. The temperature change can be calculated from the given temperatures:
∆T = T final − T initial
So, the only unknown is the amount of heat.
Hint 2. Calculate the temperature change
Calculate ∆T using the formula
∆T = T final − T initial
Express your answer with the appropriate units.
ANSWER:
∆T = 37.0 ∘ C
Correct
Hint 3. Convert the mass to grams
Convert 0.366 to grams. Recall that 1 kg = 1000 g.
Express your answer with the appropriate units.
ANSWER:
0.366 kg = 366 g
Correct
It is critical to check that any units you are given are consistent with the units in the formula you use to solve the problem and also that the
answer you calculate is expressed in the units requested.
ANSWER:
heat = 1250 cal
Answer Requested
Part B
If 125 cal of heat is applied to a 60.0- g piece of copper at 20.0 ∘ C , what will the final temperature be? The specific heat of copper is 0.0920 cal/(g ⋅ ∘ C)
.
Express your answer with the appropriate units.
Hint 1. How to approach the problem
Applying heat will increase the temperature of the copper. Use the specific heat equation to solve for the temperature change, ∆T :
heat
mass×∆T
Once you know how much the temperature will increase, you can determine the final temperature by using the value of ∆T and the given initial
temperature:
SH =
∆T = T final − T initial
Hint 2. Calculate the change in temperature
We know that heat
cal
= 125 cal, mass = 60.0 g , and SH = 0.0920 g⋅∘ C .
Solve for ∆T using the following formula:
SH =
heat
SH =
heat
mass×∆T
Express your answer with the appropriate units.
Hint 1. Rearrange the equation to solve for ∆T
Which of the following is a correct expression for
∆T ?
ANSWER:
∆T = mass × SH × heat
∆T =
∆T =
mass×SH
heat
heat
mass×SH
Correct
Plug the given values into this equation, and calculate the value of ∆T .
ANSWER:
∆T = 22.6 ∘ C
ANSWER:
T final = 42.6 ∘ C
Correct
± Heat Capacity
The following table lists the specific heat capacities of select substances:
Substance
Specific heat capacity
[J/(g ⋅ ∘ C)]
silver
0.235
copper
0.385
iron
0.449
aluminum
0.903
ethanol
2.42
water
4.184
Part A
Water (2310g ) is heated until it just begins to boil. If the water absorbs 5.13×105J of heat in the process, what was the initial temperature of the water?
Express your answer with the appropriate units.
Hint 1. How to approach the problem
Use the equation that relates heat and temperature:
q = m × C × ∆T
where q is the heat, m is the mass, C is the specific heat capacity, and ∆T is the change in temperature. The specific heat capacity, C , of water is
4.184 J/(g ⋅ ∘ C) .
Hint 2. Calculate the temperature change
∆T
What is the value of the temperature change, ∆T ?
Express your answer with the appropriate units.
Hint 1. Solve the equation for the temperature change
Rearrange this equation to isolate the temperature change, ∆T :
q = m × C × ∆T
ANSWER:
∆T =
m×C
q
∆T = q − (m × C)
∆T = q × m × C
∆T =
q
m×C
ANSWER:
∆T = 53.1 ∘ C
Hint 3. Determine the meaning of ∆T
What does the variable ∆T indicate?
ANSWER:
T final − T initial
T initial − T final
T initial
T final
Hint 4. Identify the final temperature of the water
At what temperature (in degrees Celsius) does water boil?
Express your answer with the appropriate units.
ANSWER:
100 ∘ C
ANSWER:
46.9 ∘ C
Correct
Part B
An unknown substance has a mass of 12.3g . When the substance absorbs 1.107×102J of heat, the temperature of the substance is raised from 25.0 ∘ C
to 45.0 ∘ C . What is the most likely identity of the substance?
Hint 1. How to approach the problem
In this case, the substance can be identified by its specific heat capacity. The specific heat capacity can be determined using the following equation:
q = m × C × ∆T
where q is the heat, m is the mass, C is the specific heat capacity, and ∆T is the change in temperature. Once you have determined the specific
heat capacity of the substance, compare the value to the data given in the chart in the introduction.
Hint 2. Calculate the temperature change
What is the temperature change, ∆T ?
Express your answer with the appropriate units.
Hint 1. Determine the meaning of
What does the variable
∆T
∆T indicate?
ANSWER:
T initial
T final − T initial
T initial − T final
T final
ANSWER:
∆T = 20.0 ∘ C
Hint 3. Solve the equation for the specific heat capacity
Rearrange this equation to isolate the specific heat capacity, C :
q = m × C × ∆T
ANSWER:
C = q × m × ∆T
C=
C=
m×∆T
q
q
m×∆T
C = m × ∆T − q
Hint 4. Calculate the specific heat capacity
Given that the equation that relates heat and temperature can be rearranged to solve for specific heat capacity, C ,
C=
what is the specific heat capacity of the unknown substance?
Express your answer with the appropriate units.
ANSWER:
C = 0.450 g⋅J∘ C
ANSWER:
q
m×∆T
iron
copper
ethanol
water
silver
aluminum
Correct
Interactive Worked Example 3.10: Relating Heat Energy to Temperature Changes
First launch this video. During the video, you’ll be asked a conceptual question about the example. After watching the video, answer the related questions at the
right. You can watch the video again at any point.
Part A
You find a 1979 copper penny (pre-1982 pennies are pure copper) in the snow and pick it up. How much heat does the penny absorb as it warms from the
temperature of the snow, -5.0∘ C to the temperature of your body, 37∘ C. Assume the penny is pure copper and has a mass of 3.10 g. (The specific heat
capacity of copper is 0.385 J/g∘ C.)
Express your answer in Joules to one decimal place.
ANSWER:
J
heat absorbed by the penny = 50.1
Correct
Part B
The temperature of a lead fishing weight rises from 26∘ C to 38∘ C as it absorbs 11.3 J of heat. What is the mass of the fishing weight in grams? (The
specific heat capacity of lead is 0.128 J/g∘ C.)
Express your answer in grams to two significant figures.
ANSWER:
mass of the fishing weight = 7.40
g
Correct
Part C
mL
kJ
J/g∘ C
What is the temperature change in 355
mL of water upon absorption of 34 kJ of heat? (The specific heat capacity of copper is 4.184 J/g∘ C.)
Express your answer in degrees Celcius.
ANSWER:
the temperature change of the water = 23.0
∘
C
Correct
Part D
Calculate the amount of heat required to raise the temperature of a 65-g sample of water from 32∘ C to 65∘ C . (The specific heat capacity of copper is
4.184 J/g∘ C.)
Express your answer in Joules.
ANSWER:
amount of heat = 9000
J
Correct
Exercise 3.31 with eText link
Classify each of the following pure substances as an element or a compound.
Part A
Aluminum.
ANSWER:
element
compound
Correct
Part B
Sulfur.
ANSWER:
compound
element
Correct
Part C
Methane.
ANSWER:
element
compound
Correct
Part D
Acetone.
ANSWER:
element
compound
Correct
Exercise 3.33 with eText link
Classify each of the following mixtures as homogeneous or heterogeneous.
Part A
Coffee.
ANSWER:
homogeneous
heterogeneous
Correct
Part B
Chocolate sundae.
ANSWER:
homogeneous
heterogeneous
Correct
Part C
Apple juice.
ANSWER:
homogeneous
heterogeneous
Correct
Part D
Gasoline.
ANSWER:
homogeneous
heterogeneous
Correct
Exercise 3.35 with eText link
Part A
Classify following substance as a pure substance or a mixture: helium gas.
ANSWER:
a pure substance
a mixture
Correct
Part B
Classify following pure substance as an element or a compound: helium gas.
ANSWER:
an element
a compound
Correct
Part C
Classify following substance as a pure substance or a mixture: clean air.
ANSWER:
a pure substance
a mixture
Correct
Part D
Classify following mixture as homogeneous or heterogeneous: clean air.
ANSWER:
homogeneous
heterogeneous
Correct
Part E
Classify following substance as a pure substance or a mixture: rocky road ice cream.
ANSWER:
a pure substance
a mixture
Correct
Part F
Classify following mixture as homogeneous or heterogeneous: rocky road ice cream.
ANSWER:
homogeneous
heterogeneous
Correct
Part G
Classify following substance as a pure substance or a mixture: concrete.
ANSWER:
a mixture
a pure substance
Correct
Part H
Classify following mixture as homogeneous or heterogeneous: concrete.
ANSWER:
heterogeneous
homogeneous
Correct
Exercise 3.37 with eText link
Classify each of the following properties as physical or chemical.
Part A
The tendency of silver to tarnish.
ANSWER:
chemical
physical
Correct
Part B
The shine of chrome.
ANSWER:
chemical
physical
Correct
Part C
The color of gold.
ANSWER:
chemical
physical
Correct
Part D
The flammability of propane gas.
ANSWER:
chemical
physical
Correct
Exercise 3.39 with eText link
The following list contains several properties of ethylene (a ripening agent for bananas). Which are physical properties and which are chemical?
Part A
Colorless.
ANSWER:
physical
chemical
Correct
Part B
Odorless.
ANSWER:
physical
chemical
Correct
Part C
Flammable.
ANSWER:
chemical
physical
Correct
Part D
Gas at room temperature.
ANSWER:
physical
chemical
Correct
Part E
1 L has a mass of 1.260 g under standard conditions.
ANSWER:
physical
chemical
Correct
Part F
Mixes with acetone.
ANSWER:
chemical
physical
Correct
Part G
Polymerizes to form polyethylene.
ANSWER:
chemical
physical
Correct
Exercise 3.43 with eText link
A block of aluminum is (a) ground into aluminum powder and then (b) ignited. It then emits flames and smoke. Classify (a) and (b) as chemical or physical
changes.
Part A
(a)
ANSWER:
chemical change
physical change
Correct
Part B
(b)
ANSWER:
chemical change
physical change
Correct
Exercise 3.61 with eText link
A common type of handwarmer contains iron powder that reacts with oxygen to form an oxide of iron. As soon as the handwarmer is exposed to air, the reaction
begins and heat is emitted.
Part A
Is the reaction between the iron and oxygen exothermic or endothermic?
ANSWER:
exothermic
endothermic
Correct
Exercise 3.63 with eText link
Determine whether each of the following is exothermic or endothermic.
Part A
Gasoline burning in a car.
ANSWER:
exothermic
endothermic
Correct
Part B
Isopropyl alcohol evaporating from skin.
ANSWER:
exothermic
endothermic
Correct
Part C
Water condensing as dew during the night.
ANSWER:
exothermic
endothermic
Correct
Exercise 3.22
Part A
Which has greater energy in an exothermic reaction, the reactants or the products?
ANSWER:
products
reactants
Correct
Exercise 3.32
Classify each of the following pure substances as an element or a compound.
Part A
Carbon.
ANSWER:
compound
element
Correct
Part B
Baking soda (sodium bicarbonate).
ANSWER:
compound
element
Correct
Part C
Nickel.
ANSWER:
element
compound
Correct
Part D
Gold.
ANSWER:
compound
element
Correct
Exercise 3.42
Determine whether each of the following changes is physical or chemical.
Part A
Sugar dissolves in hot water.
ANSWER:
physical
chemical
Correct
Part B
Sugar burns in a pot.
ANSWER:
chemical
physical
Correct
Part C
A metal surface becomes dull because of continued abrasion.
ANSWER:
chemical
physical
Correct
Part D
A metal surface becomes dull on exposure to air.
ANSWER:
chemical
physical
Correct
Exercise 3.46
Part A
In the explosion of a hydrogen-filled balloon, 0.20g of hydrogen reacted with 1.6g of oxygen. How many grams of water vapor are formed? (Water vapor is
the only product.)
Express your answer using two significant figures.
ANSWER:
m = 1.8 g
Correct
Exercise 3.52
Perform each conversion.
Part A
45.7J to calories
ANSWER:
10.9
cal
Correct
Part B
359cal to joules
ANSWER:
1500
J
Correct
Part C
43.3kJ to calories
ANSWER:
1.03×104
cal
Correct
Part D
219cal to kilojoules
ANSWER:
0.916
kJ
Correct
Exercise 3.64
Determine whether each of the following is exothermic or endothermic.
Part A
Dry ice subliming (changing from a solid directly to a gas).
ANSWER:
exothermic
endothermic
Correct
Part B
The wax in a candle burning.
ANSWER:
exothermic
endothermic
Correct
Part C
A match burning.
ANSWER:
exothermic
endothermic
Correct
Exercise 3.68
The warmest temperature ever measured in the United States was 134 ∘ F on July 10, 1913, in Death Valley, California.
Part A
Convert that temperature to degrees Celsius.
ANSWER:
T = 56.7
∘
C
Correct
Part B
Convert that temperature to degrees Kelvin.
ANSWER:
T = 3.30×102 K
Correct
Exercise 3.70
Liquid helium boils at 4.2
Part A
K.
Convert this temperature to degrees Fahrenheit.
ANSWER:
∘
T = -452
F
Correct
Part B
Convert this temperature to degrees Celsius.
ANSWER:
∘
T = -269
C
Correct
Exercise 3.78
Part A
Calculate the amount of heat required to heat a 4.0kg gold bar from 30∘ C to 60∘ C . Specific heat capacity of gold is 0.128
Express your answer using two significant figures.
ANSWER:
Q = 1.5×104 J
Correct
Exercise 3.82
A 46kg sample of water absorbs 3.30×102kJ of heat.
Part A
If the water was initially at 25.1∘ C, what is its final temperature?
Express your answer using two significant figures.
ANSWER:
T f = 27
∘
C
Answer Requested
Chapter 3 Question 7 - Multiple Choice
Part A
Which among the following statements is false?
ANSWER:
J/g∘ C.
A solid has a definite shape and a definite volume.
A liquid has a definite volume; but it has no definite shape.
Both solids and liquids are incompressible while gases are compressible.
A gas has neither definite volume nor definite shape.
none of the above
Correct
Chapter 3 Question 8 - Algorithmic
Part A
Given the table of specific heat values below, what is the identity of a 10.0 g metal sample that increases by 14.0∘ C when 32.9
Element Specific Heat(J/g∘ C)
Au
0.128
Ag
0.235
Cu
0.385
Fe
0.449
Al
0.903
ANSWER:
Al
Ag
Fe
Cu
none of the above
Correct
Chapter 3 Question 17 - Multiple Choice
Part A
How would you classify salt water?
ANSWER:
pure substance-compound
mixture-heterogeneous
pure substance-element
mixture-homogeneous
none of the above
Correct
Chapter 3 Question 18 - Multiple Choice
J of energy is absorbed?
Part A
How would you classify raisin bran?
ANSWER:
pure substance-compound
mixture-heterogeneous
pure substance-element
mixture-homogeneous
none of the above
Correct
Chapter 3 Question 23 - Multiple Choice
Part A
Which of the following items is a physical property?
ANSWER:
the corrosive action of acid rain on granite
the odor of spearmint gum
the tarnishing of a copper statue
the combustion of gasoline
none of the above
Correct
Chapter 3 Question 25 - Multiple Choice
Part A
Which of the following items is a chemical property?
ANSWER:
the paint color on a new red Corvette
the odor of spearmint gum
the melting and boiling point of water
the tarnishing of a copper statue
none of the above
Correct
Chapter 3 Question 28 - Multiple Choice
Part A
Which of the following is NOT a technique that could be used to separate a mixture into its components?
ANSWER:
decanting
stirring
distillation
filtration
none of the above
Correct
Chapter 3 Question 33 - Multiple Choice
Part A
Which type of energy is associated with motion?
ANSWER:
kinetic
electrical
chemical
potential
none of the above
Correct
Chapter 3 Question 34 - Multiple Choice
Part A
Which type of energy is associated with position?
ANSWER:
electrical
kinetic
potential
chemical
none of the above
Correct
Chapter 3 Question 35 - Multiple Choice
Part A
What type of energy is associated with the burning of gasoline?
ANSWER:
kinetic
potential
electrical
chemical
none of the above
Correct
Chapter 3 Question 39 - Multiple Choice
Part A
How many kilojoules are there in 95.0 Calories?
ANSWER:
2.27 × 107
3.97 × 10-4
397
22.7
none of the above
Correct
Chapter 3 Question 41 - Multiple Choice
Part A
An energy diagram that shows the reactants having greater energy than the products illustrates an:
ANSWER:
impossible reaction.
isothermic reaction.
endothermic reaction.
exothermic reaction.
none of the above
Correct
Score Summary:
Your score on this assignment is 79.8%.
You received 126.81 out of a possible total of 159 points.