CHEMISTRY 11
Chapter 14
Energy Trapped
in Hydrocarbons
Solutions for Practice Problems
Student Textbook page 584
1. Problem
The following equation shows the combustion of 3-ethyl-2,5-dimethylheptane:
C11H24 + 17O2 → 11CO2 + 12H2O
(a) Does this equation show complete or incomplete combustion ?
(b) Draw the structural formula for 3-ethyl-2,5-dimethylheptane.
What Is Required?
(a) You have to decide if the combustion is complete or incomplete.
(b) You have to draw the structural formula for the alkane.
What Is Given?
The balanced chemical equation is given.
Plan Your Strategy
(a) A complete combustion will produce only carbon dioxide and water. An incom-
plete combustion will produce other by-products.
(b) Follow the steps below.
Step 1 The root chain is heptane, so it will have 6 carbons.
Step 2 There will be a ethane group on C3 and two methane groups on C2 and
C5. Add the methane groups first at these two positions (you can arbitrarily choosing either end as C1).
Step 3 Add the ethane group last to C3.
Act on Your Strategy
(a) Since only CO2 and H2O are in the equation, it shows complete combustion.
(b)
CH3
CH3
CH3
CH
CH
CH2
CH
CH2
CH3
C2H5
3-ethyl-2,5-dimethylheptane
Check Your Solution
For (b), work backward. Use the naming rules for an alkane to name the structure
you have drawn. The answer should match.
2. Problem
(a) Write a balanced equation for the complete combustion of pentane, C5H12 .
(b) Write a balanced equation for the complete combustion of octane, C8H18 .
(c) Write two possible balanced equation for the incomplete combustion of ethane,
C2H6 .
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What Is Required?
You need to write balanced equations for the alkanes listed.
What Is Given?
The given compounds are the reactants. You know that O2 is a reactant too.
Equations (a) and (b) are complete combustion reactions. Equation (c) is an incomplete combustion.
Plan Your Strategy
For pentane and octane, the complete combustion will give rise to the products CO2
and H2O only.
Step 1 Write the equation.
Step 2 Balance the C atoms first.
Step 3 Balance the H atoms next.
Step 4 Balance the O atoms last.
For ethane, the reaction is incomplete, so the possible products are carbon, carbon
monoxide, carbon dioxide, and water. Since water is the only hydrogen-containing
product, you should balance the H atoms first, the C atoms next, and the O atoms
last.
Act on Your Strategy
(a) C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(g)
(b) C8H18(g) + 25O2(g) → 8CO2(g) + 9H2O(g)
(c) (i) C2H6(g) + 2O2(g) → C(s) + CO(g) + 3H2O(g)
(ii) C2H6(g) + 3O2(g) → CO(g) + CO2(g) + 3H2O(g)
Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equation.
3. Problem
(a) The flame of a butane lighter is usually yellow, indicating incomplete combustion
of the gas. Write a balanced chemical equation for the incomplete combustion of
butane in a butane lighter. Use the condensed structural formula for butane.
(b) If you supplied enough oxygen, the butane would burn with a blue flame. Write a
balanced chemical equation for the complete combustion of butane.
What Is Required?
(a) You need to write the balanced equation of the incomplete combustion of butane.
(b) You need to write a balanced equation for the complete combustion of butane.
What Is Given?
You have to check the formula for butane, which should be C4H10 . You know butane
and O2 are the reactants in the equation.
Plan Your Strategy
(a) For the incomplete reaction, the possible products are carbon, carbon monoxide,
carbon dioxide, and water. Since water is the only hydrogen-containing product,
you should balance the H atoms first, the C atoms next, and the O atoms last.
(b) For the complete reaction, only CO2 and H2O are the products. Balance the
equation first by C, then by H, and finally by O.
Act on Your Strategy
(a) CH3 −CH2 −CH2 −CH3 + 4 O2 → 2 C + CO + CO2 + 5 H2O
(b) CH3 −CH2 −CH2 −CH3 +
13
2
O2 → 4 CO2 + 5 H2O
Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equation.
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4. Problem
The paraffin wax in a candle burns with a yellow flame. If it had sufficient oxygen it
would burn with a blue flame, it would burn rapidly and release a lot of energy. It
might even be dangerous! Write the balanced chemical equation for the complete
combustion of candle wax, C25H52(s) .
What Is Required?
You need to write a balanced equation for the complete combustion of candle wax.
What Is Given?
the formula for candle wax is C25H52 . You know C25H52 and O2 are the reactants in
the equation
Plan Your Strategy
The complete combustion will give rise to the products CO2 and H2O only. Balance
first by C, then by H, and finally by O.
Act on Your Strategy
C25H52(s) + 38O2(g) → 25CO2(g) + 26H2O(g)
Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equation.
5. Problem
4-Propyldecane burns to give solid carbon, water vapour, carbon monoxide, and carbon dioxide.
(a) Draw the structural formula for 4-propyldecane.
(b) Write two different balanced equations for the reaction described in this problem.
(c) Name the type of combustion. Explain.
What Is Required?
(a) You need to draw the structural formula for 4-propyldecane.
(b) You need to write two possible balanced equations for this reaction.
(c) You have to explain whether the react was complete or not.
What Is Given?
The reactants are 4-propyldecane and O2 .
Plan Your Strategy
(a) From the name given, you know the root chain has 10 C atoms. A propyl group
is on C4. Since there is only one branch, you can arbitrarily choose either end of
the chain as C1 and place the propyl branch at C4.
(b) Count the number of C and H atoms in the structural diagram in (a). This gives
the formula for the molecule. Since the problem asks for 2 balanced equations,
you can guess it is an incomplete one, so the possible products are C, CO, CO2,
and H2O. Since water is the only molecule with H atoms in it, start by balancing
the H, then by C, and finally by O.
Act on Your Strategy
(a)
C3H7
CH3
CH2
CH2
CH
CH2
CH2
CH2
CH2
CH2
CH3
4-propyldecane
(b) C13H28(l) + 15O2(g) → 3C(s) + 4CO(g) + 6CO2(g) + 14H2O(g)
C13H28(l) + 12O2(g) → 5C(s) + 6CO(g) + 2CO2(g) + 14H2O(g)
(c) In this case, incomplete combustion has occurred as both solid carbon product
and carbon monoxide gas have been formed as well.
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Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equations.
Check the structure by working backward, using the naming rules of alkanes to name
the structure drawn.
Solutions for Practice Problems
Student Textbook page 591
6. Problem
The formation of propane from its elements is an exothermic reaction. The combustion of propane is also exothermic.
(a) Write the balanced thermochemical equation for the formation of propane.
(b) Write the balanced thermochemical equations for the combustion of propane.
(The balanced equation is on page 580.)
(c) Consider the combustion of propane. Compare the energies of bond breaking and
bond making to explain why the reaction is exothermic.
What Is Required?
(a) You need to give the thermochemical equation for the formation of propane.
(b) You need to give the thermochemical equation for the combustion of propane.
(c) You need to explain why the reactions are exothermic.
What Is Given?
The balanced equation for the complete combustion of propane is on Student
Textbook page 580:
C3H8 + 5O2(g) → 3CO2(g) + 4H2O(g)
You will assume the equation is a complete one.
Plan Your Strategy
(a) The reactants for the formation of hydrocarbons come in the form of solid carbon
and hydrogen gas. Write the equation for this and balance the C and H atoms on
either side of the equation. Since the reaction is exothermic, the energy input to
break the bonds of the carbon solid and hydrogen gas molecules is less than the
energy released from bonding the C and H atoms together. The energy term
therefore sits on the right hand side of the equation.
(b) The thermochemical equation is the equation with the indicator of where the
energy is. Since the reaction is exothermic, the energy used to break the propane
and O2 bonds is less than the energy released when the products are formed. The
energy term sits on the right hand side of the equation.
Act on Your Strategy
(a) The formation of propane: 3C(s) + 4H2(g) → C3H8 + energy
(b) The combustion of propane:
C3H8 + 5O2(g) → 3CO2(g) + 4H2O(g) + energy
(c) The reactant bonds are broken. This process absorbs enough energy to split the
reactants into separate atoms. The product bonds are made. This process releases
energy as the product molecules are formed. The combustion of propane is
exothermic. Therefore, the energy that is used when the reactant bonds are broken
must be less than the energy that is released when the product bonds are formed.
Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equations.
Both equations are exothermic, so both must have the energy term on the product
side of the equations.
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7. Problem
(a) Explain why the formation of ethene, C2H4(g) , from its elements is endothermic,
while the combustion is exothermic.
(b) Write the balanced thermochemical equations for the formation and combustion
of ethene.
What Is Required?
(a) You need to explain the differences in the thermodynamics of the formation and
combustion reactions of ethene.
(b) You need to write the thermochemcial equations for the formation and combustion of ethene.
What Is Given?
The formation of ethene is endothermic.
The combustion of ethene is exothermic.
The formula for ethene is C2H4 .
Assume the combustion is complete.
Plan Your Strategy
(a) The endothermic and exothermic nature of an equation is the net energy result of
the amount of energy input to break a bond and the energy released from forming
a bond.
(b) Assuming the combustion reaction is complete, then CO2 and H2O will be the
products. For formation of the ethene, carbon solid and hydrogen gas are the reactants. The energy term for an endothemic reaction appears on the left hand side
of the thermochemical equation. The energy term for an exothermic reaction
appears on the right hand side of the thermochemcial equation.
Act on Your Strategy
(a) In the case of the formation of ethene, more energy was required to break the
bonds of the C(s) and H2(g) than what was released from forming the C = C and
the C-H bonds of ethene. The formation was therefore endothermic. In the combustion reaction, there is more energy released in the formation of the CO2 and
H2O products than was used to break the bonds of the ethene and O2 . The reaction was therefore exothermic.
(b) Formation of ethene: 2C(s) + 2H2(g) + energy → C2H4g
Combustion of ethene: C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) + energy
Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equations.
In the endothermic equation, the energy term must be on the reactant side of the
equation. In the exothermic reaction, the energy term must be on the product side of
the equation.
Solutions for Practice Problems
Student Textbook page 597
8. Problem
100 g of ethanol at 25˚C is heated until it reaches 50˚C. How much heat does the
ethanol gain? HINT: Find the specific heat capacity of ethanol in Table 14.2.
What Is Required?
You need to find the amount of heat gained by ethanol in the reaction.
What Is Given?
m of ethanol = 100 g
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Ti = 25˚C
Tf = 50˚C
Specific heat capacity of ethanol = 2.46 J/g˚C at 25˚C
Plan Your Strategy
Use the heat formula to solve for the heat gained, Q.
Act on Your Strategy
Q = mc∆T = 100 g × 2.46 J/g·˚C × (50˚C − 25˚C) = 6150 J
Therefore, 6150 J or 6.15 kJ of heat is gained.
Check Your Solution
The ethanol gained heat so the heat value should be positive. When the units in the
heat equation cancel out, J remain.
9. Problem
In Part A of the ThoughtLab on page 594, the students added ice to 120.0 g of water
in beaker 2. Calculate the heat lost by the water. Use the information given for beaker
2, as well as specific heat capacities in Table 14.2.
What Is Required?
You need to find the amount of heat lost by water in the experiment in beaker 2.
What Is Given?
From the data in ThoughtLab, Factors in Heat transfer, Student Textbook page 594:
m of water in beaker 2 = 120.0 g.
Ti = 26.5˚C
Tf = 17.4˚C
Specific heat capacity of water (liquid) = 4.184 J/g˚C at 25˚C
Plan Your Strategy
Use the heat formula to solve for the heat gained, Q. Since you are only concerned
with the beaker 2, you do not have to worry about the ice conditions, only the liquid
water conditions.
Act on Your Strategy
Q = mc∆T
= 120 g × 4.184 J/g·˚C × (17.4˚C − 26.5˚C)
= −4568.9 J
= −4.75 kJ
Check Your Solution
The water lost heat so the heat value should be negative. When the units in the heat
equation cancel out, J remain.
10. Problem
A beaker contains 50 g of liquid at room temperature. The beaker is heated until the
liquid gains 10˚C. A second beaker contains 100 g of the same liquid at room temperature. This beaker is also heated until the liquid gains 10˚C. In which beaker does
the liquid gain the most thermal energy? Explain.
What Is Required?
You have to explain which beaker of liquid gained the most thermal energy from the
conditions given.
What Is Given?
m of beaker1 liquid = 50 g
Ti of beaker1 liquid = 25˚C
Tf of beaker1 liquid = 35˚C
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m of beaker2 liquid = 100 g
Ti of beaker2 liquid = 25˚C
Tf of beaker2 liquid = 35˚C
Specific heat capacity of liquid = c J/g˚C for both liquids at 25˚C
Plan Your Strategy
Use the heat formula for both liquids and compare the value.
Act on Your Strategy
The beaker containing 100g of liquid will gain twice as much thermal energy
compared to the to the 50 g sample. This can be shown through examination of the
formula Q = mc∆T. Since c and ∆T are constant for both liquids, Q is only affected
by the value of m in both beakers. So if m doubles, Q will double too. Since ∆T is a
positive value (35˚C − 25˚C), the net result is a gain of thermal energy for both
liquids.
Check Your Solution
The liquids gained heat so the heat value should be positive.
value is doubled in beaker 2. Your explanation is reasonable.
100 g
50 g
= 2, so the Q
11. Problem
As the diagram on the next page illustrates, the sign of the heat value tells you
whether a substance has lost or gained heat energy. Consider the following descriptions. Write each heat value, and give it the appropriate sign to indicate whether heat
was lost or gained.
(a) In Part A of the ThoughtLab on page 594, the ice gained the heat that was lost by
the water. When ice was added to 60.0 g of water, it gained 4.22 kJ of energy.
When ice was added to 120.0 g of water, it gained 45.6 kJ of energy.
∆T
Tfinal − Tinitial
heat lost
heat gained
(b) When 2.0 L of water was heated over a campfire, the water gained 487 kJ of
energy.
(c) A student baked a cherry pie and put it outside on a cold winter day. There was a
change of 290 kJ of heat energy in the pie.
What Is Required?
You have to give the sign and the heat value for each of the situations listed.
What Is Given?
(a) When m of ice added to water = 60.0 g, 4.22 kJ is gained
When m of ice added to water = 120.0 g, 4.22, 4.6 kJ is gained
(b) When 2.0 L water is heated, 487 kJ is gained
(c) When a baked cherry pie is cooled, 290 kJ heat change is observed
Plan Your Strategy
Each example already gives the value of Q, the heat value. If energy is lost from the
substance, Q has a negative sign. If energy is gained by the substance, Q has a positive
sign.
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CHEMISTRY 11
Act on Your Strategy
(a) When ice was added to 60.0 g of water Q = 4.22 kJ , since energy was gained by
the water.
When ice was added to 120.0 g of water, Q = 4.6 kJ, since energy was gained by
the water
(b) Q = 487 kJ, since energy was gained by the water.
(c) Q = −290 kJ. Since the pie should cool down in the cold winter day, we expect
that thermal energy is lost from the pie.
Check Your Solution
For Q values, a gain of thermal energy should have a positive sign, while a loss of
thermal energy should have a negative sign.
Solutions for Practice Problems
Student Textbook page 599
12. Problem
Solve the equation Q = mc∆T for the following quantities.
(a) m
(b) c
(c) ∆T
What Is Required?
You have to use the heat formula and solve for the quantities listed.
What Is Given?
The quantity to be solved in the heat formula is given.
Plan Your Strategy
Simply rearrange the heat formula to solve for the quantity indicated.
Act on Your Strategy
(a) m =
(b) c =
(c) ∆T
Q
c∆T
Q
m∆T
Q
= mc
Check Your Solutions
Put in the units for each quantity into the formula. For example, in part (a), Q has
the unit of J, c has units of J/g·˚C, and DT has the unit of ˚C. When these units can, only g should remain.
cel out according to the formula given, i.e., J J
g ·˚C × ˚C
13. Problem
You know that ∆T = Tf − Ti . Combine this equation with the heat equation,
Q = mc∆T, to solve for the following quantities.
(a) Ti (in terms of Q, m, c, and Tf)
(b) Tf (in terms of Q, m, c, and Ti )
What Is Required?
You have to use the heat formula and solve for Ti and Tf.
What Is Given?
The T quantity to be solved is given.
Plan Your Strategy
Simply rearrange the heat formula to solve for the quantity indicated.
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CHEMISTRY 11
Act on Your Strategy
Q = mc(Tf − Ti)
Q
= Tf − Ti
(a)
mc
Q
Ti = Tf − mc
(b) From the result in part (a), we can rearrange the T values to get Tf =
Q
mc
+ Ti
Check Your Solutions
Put the units for each quantity into each formula. Q has the unit of J, c has units of
J/g·˚C, both Ti and Tf have the unit of ˚C, and m has the unit of g. When these
units cancel out according to the formula given, only ˚C should remain in each case.
14. Problem
How much heat is required to raise the temperature of 789 g of liquid ammonia,
from 25˚C to 82.7˚C?
What Is Required?
You have to calculate the heat energy needed to raise the temperature of the
ammonia.
What Is Given?
m of ammonia = 789 g
Ti = 25˚C
Tf = 82.7˚C
Specific heat capacity of liquid ammonia = 4.70 J/g˚C (from Table 14.2, Student
Textbook page 595)
Plan Your Strategy
Enter the values into the heat formula and solve for Q.
Act on Your Strategy
Q = mc∆T = (789 g) × (4.70 J/g·˚C × (82.7˚C − 25.0˚C)
Q = 213 968 J = 214 kJ
Check Your Solution
When the units cancel out, J should remain. The value of Q is positive since heat
energy is gained by the liquid ammonia.
15. Problem
A solid substance has a mass of 250.00 g. It is cooled by 25.00˚C and loses 4937.50
kJ of heat. What is its specific heat capacity? Look at Table 14.2 to identify the
substance.
What Is Required?
You need to find the specific heat capacity of the unknown substance and then identity the substance with known data.
What Is Given?
Q = −4937.50 J
∆T = −25˚C
m = 250.00 g
Plan Your Strategy
Since heat is lost, the heat value must be negative. Since c is always a positive value,
then ∆T must be negative too. Input the values into the heat formula and solve for c.
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CHEMISTRY 11
Act on Your Strategy
c =
=
Q
m∆T
−4937.50 J
(250.00 g)(−25.00˚C)
= 0.79 J/g˚C
From Table 14.2, the solid substance with the same specific heat capacity value is
granite.
Check Your Solution
When the units cancel out, J/g·˚C should remain and its value should be positive.
Granite, like the unknown substance, is a solid. The result is reasonable.
16. Problem
A piece of metal with a mass of 14.9 g is heated to 98.0˚C. When the metal is placed
in 75.0 g of water at 20.0˚C, the temperature of the water rises by 28.5˚C. What is
the specific heat capacity of the metal?
What Is Required?
You need to find the specific heat capacity of the metal.
What Is Given?
m of the metal = 14.9 g
Ti of metal = 98.0˚C
m of the water = 75.0 g
Ti of water = 20.0˚C
∆T of water = 28.5˚C
Specific heat capacity of water (liquid) = 4.184 J/g·˚C at 25˚C
Plan Your Strategy
The thermal energy (Q value) gained by the water is the same amount lost by the
metal. Therefore, we can equate the two heat formulas for the metal and water to
solve for c.
When the system is in final equilibrium, the temperature of the metal is the same as
the temperature of the water. We know the temperature of the water raised by
28.5˚C, so its final temperature must have been 20˚C + 28.5˚C = 48.5˚C. This
would be the final temperature of the metal as well.
Hence, ∆Tmetal = Tf − Ti = 48.5˚C − 98˚C = −49.5˚C .
For this calculation, we will ignore the negative signs for the Q value of the metal
(heat lost) and ∆Tmetal , since these would cancel out in an equation to solve for
cmetal .
Act on Your Strategy
Q lost by metal = Q gained by water
(mmetal)(cmetal)(∆Tmetal) = (mwater)(cwater)(∆Twater)
(14.9 g)(cmetal)(49.5˚C) = (75.0 g)(4.184 J/g·˚C)(28.5˚C)
cmetal = 12.12 J/g·˚C
Check Your Solution
When the units cancel out, J/g·˚C should remain and its value should be positive.
17. Problem
A piece of gold (c = 0.129 J/g·˚C) with mass of 45.5 g and a temperature of 80.5˚C
is dropped into 192 g of water at 15.0˚C. What is the final temperature of the system
(Hint: Use the equation Qwater = −Qgold .)
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What Is Required?
You need to find the final temperature of the system.
What Is Given?
m of gold = 45.5 g
Ti of gold = 80.5˚C
m of water = 192 g
Ti of water = 15.0˚C
∆T of water = 28.5˚C
Specific heat capacity of water (liquid) = 4.184 J/g·˚C at 25˚C
Specific heat capacity of gold = 0.129 J/g·˚C at 25˚C
Plan Your Strategy
In the final temperature of the system, Tf of the gold = Ti of the water. The heat lost
by the gold is the heat gained by the water, so the Q values are the same. Equate the
two heat formulas for the gold and the water in terms of Q, and rearrange the equation to solve for Tf.
Act on Your Strategy
mc∆Tgold = mc∆Twater
(45.5 g)(0.129 J/g·˚C)(Tf − 80.5˚C) = (192 g)(4.184 J/g·˚C)(Tf − 15.0˚C)
5.87 J/˚C Tf − 472.5 J = 803.8 J/˚C Tf − 12049.92 J
−797.93 J/˚C Tf = −11577.42 J
Tf = 14.5˚C
Check Your Solution
When the units cancel out, ˚C remains. The value is positive since the system gained
heat, and the temperature would have increased.
Solutions for Practice Problems
Student Textbook page 607
18. Problem
A reaction lowers the temperature of 500.0 g of water in a calorimeter by 1.10˚C.
How much heat is absorbed by the reaction?
What Is Required?
You need to calculate the amount of heat absorbed by the reaction.
What Is Given?
m of water = 500.0 g
∆Twater = −1.10˚C (a negative value since its temperature is lowered)
You know the specific heat capacity of water = 4.184 J/g·˚C
Plan Your Strategy
The heat lost by the water is the heat gained by the reaction, i.e., Qreaction = −Qwater .
So by calculating the Q value for water, we have the Q value for the reaction.
Act on Your Strategy
Qwater = mc∆T = (500.0 g)(4.184 J/g·˚C)(−1.10˚C) = −2301 J
The water lost 2301 J of energy. Therefore, 2301 J of energy was absorbed by the
reaction.
Check Your Solution
When the units cancel out in the heat formula, J remain. The sign for Q is negative
for the heat lost by water, and is positive for the heat gained by the reaction.
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19. Problem
Aluminum reacts with iron(III) oxide to yield aluminum oxide and iron. The temperature of 1.00 kg of water in a calorimeter increases by 3.00˚C during the reaction.
Calculate the heat that is released in the reaction.
What Is Required?
You have to calculate the amount of heat released from the reaction.
What Is Given?
m of water = −1.00 kg = 1000 g
∆Twater = 3.00˚C(a positive value since its temperature is increased)
You know the specific heat capacity of water = 4.184 J/g·˚C
Plan Your Strategy
The heat gained by the water is the heat lost by the reaction, i.e., −Qreaction = Qwater .
So by calculating the Q value for water, we have the Q value for the reaction.
Act on Your Strategy
Qwater = mc∆T = (1000 g)(4.184 J/g·˚C)(3.00˚C) = 1255.2 J = 12.55 KJ
Therefore, the heat that is released by the reaction is 12.55 kJ
Check Your Solution
When the units cancel out in the heat formula, J remain. Note the following difference in terminology: Since we are talking about the amount of heat released by the
reaction, we do not have to associate the negative sign with it. It is only when we
refer to the heat being lost by the substance that we give it the negative sign.
20. Problem
5.0 g of an unknown solid was dissolved in 100 g water in a polystyrene calorimeter.
The initial temperature of the water was 21.7˚C, and the final temperature of the
solution was 29.6˚C.
(a) Calculate the heat change caused by the solid dissolving.
(b) What is the heat of solution per gram of solid dissolved?
What Is Required?
(a) You have to calculate the heat change of the system.
(b) You have to calculate the heat of solution per gram of the solid.
What Is Given?
m of solid = 5.0 g
m of water = 100 g
Ti of water = 21.7˚C
Tf of solution or system = 29.6˚C
You know the specific heat capacity of water = 4.184 J/g·˚C at 25˚C.
Plan Your Strategy
(a) The heat change of the system is the heat gained by the water after the solid
dissolved in it. Substitute the values for water into the heat formula and solve
for Qwater .
(b) The heat of solution is the heat lost upon the solid dissolving in the water, so it
will be a negative value of the Qwater . Divide this by the given mass of the solid to
obtain the heat of solution per gram of solid.
Act on Your Strategy
(a) Qwater = mc∆T = (100.0 g)(4.184 J/g·˚C)(29.6˚C − 21.7˚C) = 3305.4 J
(b) Heat of solution per gram
Qwater
mass of solid
=
−3305.4 J
5.0 g
= −661.1 J/g
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Check Your Solution
(a) When the units cancel out in the heat formula, J remain. The value of Qwater is
positive because the solution increased in temperature.
(b) The units for heat of solution per gram solid should be J/g. The result is negative
because the process of dissolution results in a loss of energy to the solution.
21. Problem
A 92.0 g sample of a substance, with a temperature of 55.0˚C, is placed in a polystyrene calorimeter. The calorimeter contains 1.00 kg of water at 20.0˚C. The final
temperature of the system is 25.2˚C.
(a) How much heat did the substance lose? How much heat did the water gain?
(b) What is the specific heat capacity of the substance?
What Is Required?
(a) You have to calculate the heat energy lost and gained by the participating sub-
stances in the system.
(b) You have the calculate the specific heat capacity of the unknown substance.
What Is Given?
m of substance = 92.0 g
Ti of substance = 55.0˚C
m of water = 1.00 kg = 1000 g
Ti of water = 20.0˚C
Ti of the system = 25.2˚C
You know the specific heat capacity of water = 4.184 J/g·˚C.
Plan Your Strategy
(a) The amount of heat lost by the substance would be the amount of heat gained by
the water. Since we have all the values needed for the heat formula for water, we
can calculate Qwater to determine the heat gained.
(b) Knowing the Qwater value, we can use this value as Qsubstance , and solve for the
csubstance using the heat formula for the substance.
Act on Your Strategy
(a) Qwater = mc∆T = (1000 g)(4.184 J/g·˚C)(25.2˚C − 20.0˚C) = 21756.8 J
Qwater = Qsubstance = −21756.8 J or −21.8 J
The water gained 21.8 kJ of energy and the substance lost 21.8kJ of energy.
Q
(b) c = substance
m∆T
=
−21756.8 J
(92.0g)(25.2˚C − 55.0˚C)
= 8.01 J/g·˚C
Check Your Solution
(a) When the units cancel out in the heat formula, J remain. The value of Qwater is
positive because the water increased in temperature.
(b) When the units rearrange in the heat formula, J/g·˚C results. The result is positive
because the negative values of the heat loss (−21756.8 J) and change in temperature (−29.8˚C) cancel out in the formula.
Solutions for Practice Problems
Student Textbook page 611
22. Problem
Use the heat equation for a calibrated calorimeter, Qcal = C ∆T . Recall that
∆T = Tf − Ti . Solve for the following quantities.
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(a)
(b)
(c)
(d)
C
∆T
Tf (in terms of C, ∆T, and Ti)
Ti (in terms of C, ∆T, and Tf)
What Is Required?
You have to solve for the quantities listed, using the heat equation for the calibrated
calorimeter.
What Is Given?
The quantity to be solved for is given.
Plan Your Strategy
Simply rearrange the heat equation for the calibrated calorimeter to solve for the
quantity required.
Act on Your Strategy
Q
(a) C = cal
∆T
(b) ∆T =
(c) Tf =
(d) Ti =
Qcal
C
Qcal
C
Qcal
C
+ Ti
+ Tf
Check Your Solution
Put in the units for each quantity into the formulas. For example, in part (a), Qcal has
the unit of kJ and ∆T has the unit of ˚C. When these units rearrange according to
the formula given in (a), you have kJ˚C, which are indeed the units for the calibration of the calorimeter.
23. Problem
A lab technician places a 5.00 g food sample into a bomb calorimeter that is calibrated at 9.23 kJ˚C. The initial temperature of the calorimeter system is 21.0˚C. After
burning the food, the final temperature of the system is 32.0˚C. What is the heat of
combustion of the food in kJ/g?
What Is Required?
You need to find the heat of combustion per gram of the food sample.
What Is Given?
m of food sample = 5.00 g
calibration (C) of the calorimeter = 9.23 kJ˚C
Ti = 21.0˚C
Tf = 32.0˚C
Plan Your Strategy
Step 1 Calculate the value of Qcal for the sample using the heat equation for a calibrated calorimeter. The energy released by the burned sample (Qsample) was the
heat transferred to the calorimeter, so the food sample suffered an energy loss
during combustion. Its Q value should therefore be the negative of Qcal.
Step 2 Divide the Qsample value by the mass of the food sample to obtain its heat
combustion per gram.
Act on Your Strategy
Step 1 Qcal = C ∆T = (9.23 kJ˚C)(32.0˚C − 21.0˚C) = 101.5 kJ
Therefore, Qsample = −Qcal = −101.5 kJ
kJ
= −101.5 kJ/g
Step 2 Heat of combustion per gram −101.5
5.00 g
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Check Your Solution
When the units cancel out in the heat equation for a calibrated calorimeter, kJ
remain.
When the units rearrange in the heat of combustion per gram equation, it becomes
kJ/g. The answer is negative in step 2 since the food loses thermal energy to the
calorimeter during the combustion process.
24. Problem
A scientist places a block of ice in an uncalibrated bomb calorimeter. The ice melts,
gains 10.5 kJ (10.5 × 103 J) of heat and undergoes a temperature change of 25.0˚C.
The calorimeter undergoes a temperature change of 1.2˚C.
(a) What mass of ice was added to the calorimeter? (Use the heat capacity of liquid
water.)
(b) What is the calibration of the bomb calorimeter in kJ/˚C?
What Is Required?
(a) You need to find the initial mass of the ice.
(b) You need to find the C value of the calorimeter.
What Is Given?
Q of ice = 10.5 kJ = 10.5 × 103J
Specific heat capacity of water (liquid) = 4.184 J/g·˚C at 25˚C
∆T of water = 25.0˚C
∆T of calorimeter = 1.2˚C
Plan Your Strategy
(a) Apply the heat formula for water, inputting the values for water, which are all
given. Rearrange the formula to solve for m.
(b) Qwater = −Qcal since the calorimeter lost energy to the ice, which gained the energy for melting. ∆T also becomes a negative value because the change in temperature was a lowering of the temperature in the calorimeter. Applying −Qcal to the
heat equation for a calibrated calorimeter, you can rearrange the equation to solve
for C.
Act on Your Strategy
(a) mwater =
=
Q
c∆T
10.5 × 103 J
(4.184 J/g˚C))(25.0˚C)
= 100.4g
(b) Qwater = −Qcal = −10.5 kJ˚C
C = Qcal / ∆T = −10.5 kJ / −1.2˚C = 8.75 kJ/˚C
Check Your Solution
(a) When the units cancel out in the heat formula, g remain.
(b) When the units rearrange in the heat equation for a calibrated calorimeter, the
result is kJ/˚C, which are the units for calibration. The value for the calibration
should rightfully be positive, since the negative signs of Q and ∆T cancel each
other out.
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Solutions for Practice Problems
Student Textbook page 622
25. Problem
Your town is considering dumping its plastic waste in a nearby lake. Identify possible
risks and benefits for this plan. Explain where you could find more information to
help your town make a decision.
What Is Required?
You need to perform a risk-benefit analysis of plastic waste dumping. This includes
identifying, researching, and weighing the risks and benefits of dumping the plastic
waste in the nearby lake. You also have to research more ways to increase the information you need to help the town come to a decision about their plan.
What Is Given?
The plastic waste is to be dumped in a nearby lake. The plan is only under consideration, so your research and information could help to either persuade or dissuade the
decision.
Plan Your Strategy
Step 1 Identify possible risks and benefits.
Step 2 Research the risk and benefits. Use the Internet, reference books such as encyclopedias, and other sources to help you. Note to choose reliable sources only
— be especially wary of this when using the Internet as your information
source.
Step 3 Weigh the risks and benefits.
Step 4 Examine other sources of information that could support your decisions.
Act on Your Strategy
Possible Risks
Possible Benefits
Organic chemicals could possibly leach out
of the plastic and contaminate the lake and
groundwater.
Plastic waste would have a storage site
where it could be easily recovered once
recycling technology for plastics becomes
readily available.
Once the possible risks and benefits have been identified, more information on this
topic could be researched at the local library and on the Internet. A visit to the local
municipal office can provide further information on zoning regulations and environmental laws and regulations for the area.
Check Your Solution
It is always a good idea to share and discuss your results with others. The discussions
may reveal or lead to insights into things you may have missed yourself.
26. Problem
Earth’s reservoir of fossil fuels, including natural gas, will not last forever. As well,
burning fossil fuels releases carbon dioxide (a greenhouse gas) and other pollutants
that can cause acid rain. On the other hand, alternate energy sources, such as solar
panels, are more expensive. They can also be less reliable than using fossil fuels.
Perform a risk-benefit analysis to decide if you should heat your home using natural
gas or solar panels. You will need to do more research to make an informed decision
(See the Internet Link on this page.)
What Is Required?
You need to perform a risk-benefit analysis of the uses of natural gas and solar panels
for home heating. This includes identifying, researching, and weighing the risks and
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CHEMISTRY 11
benefits of using either type of energy source. You also have to research more ways to
increase the information you need to come to a decision about which type of energy
source to choose.
Plan Your Strategy
Step 1 Identify possible risks and benefits.
Step 2 Research the risk and benefits. Use the Internet, reference books such as encyclopedias, and other sources to help you. Note to choose reliable sources only
— be especially wary of this when using the Internet as your information
source.
Step 3 Weigh the risks and benefits.
Step 4 Examine other sources of information that could support your decisions.
Act on Your Strategy
You can use BLM 14-10: Risk-Benefit Analysis Worksheet to work through a list of
risks and benefits associated with home heating. Internet sites containing information
on heating with natural gas and solar panels can be found at the following web sites.
Solar panels:
• http://www.flasolar.com/
• http://www.ases.org/
• http://solstice.crest.org/renewables/wlord/index.html
Natural gas heating:
• http://www.naturalgas.org/
• http://www.cga.ca/energy/ehhome.htm
• http://www.strategies-tactics.com/nyngemail.htm
Check Your Solution
It is always a good idea to share and discuss your results with others. The discussions
may reveal or lead to insights into things you may have missed yourself.
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