CHAPTER X HW (CHEM 110) SOLUTIONS
ENERGY
Kinetic and Potential Energy
1.)
Answer the questions about the following scenarios.

a. What kind of energy does ball A have initially when at rest at the top of the hill?
Potential energy due to its position at the top of the hill
b. Describe the changes in energy as ball A moves down the hill.
As Ball A rolls down the hill, its potential energy is converted to kinetic energy (movement), heat
(through friction), and possibly a small amount of sound.
What kind of energy does ball A have when it reaches the bottom of the hill and stops moving after
hitting ball B?
Potential energy due to its new position (not as much)
c. Where did the energy gained by ball B come from, which allowed it to move up the hill?
B gained potential energy (moved up the hill) and was transferred it from Ball A when they hit.
2.)
3.)
Explain how each substance has both potential and kinetic energy.
Substance
Potential energy
Kinetic energy
A rock falling off the roof
The rock held above the ground
could fall and squish something
(do “work”) or cause the ground
to heat up if it falls.
The rock is moving, so has kinetic
energy related to its speed during
the fall.
A stick of dynamite at 25 oC
The dynamite has the potential to
explode and release heat, sound
and light.
Although the dynamite may not
be moving to the eye, the
molecules are moving because the
temp is above absolute zero. This
is kinetic energy.
Explain how the molecular or atomic motion changes when boiling water is poured into a room
temperature metal container and causes the container to become warm. Discuss how both the water and
metal change.
The atoms in the metal are “jiggling” slowly (low temperature) in the beginning, and the molecules in the
hot water are moving past each other quickly (high temperature). When the hot water contacts the metal,
energy is transferred, and the atoms in the metal jiggle faster (increase their kinetic energy or
temperature), and the water molecules slow down (lower their temperature).
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Endothermic and Exothermic Reactions
4.)
Define the following terms:
a. Energy:
The ability to do work (push or pull something) or to create heat.
b. Law of Conservation of Energy:
5.)
Energy cannot be created or destroyed, only changed in form.
c. Exothermic reaction:
A chemical reaction that releases energy in the form of heat.
d. Endothermic reaction:
A chemical reaction that absorbs energy in the form of heat.
Are each of these processes endothermic or exothermic? Answer these from the process’ perspective (ex.
the wood in the first problem, not you on the outside feeling the burning).
a. Wood burning on a bonfire.
Exothermic (heat released)
b. NH4NO3 dissolving in water causes the water temperature to decrease. Endothermic (heat absorbed)
c. Sweat evaporates off my skin and causes me to feel cooler.
Endothermic
d. A few milliliters of concentrated sulfuric acid added to a small amount of water causes a lot of steam
to be produced.
Exothermic
Specific Heat Capacity
6.)
Are each of these statements true or false?
a. The products of an exothermic reaction have lower potential energy than the reactants. True
b. The sign of “q” is positive when a substance (ex. reactant) gains heat (energy).
True
c. A calorie is a measure of fat density.
False (energy)
d. The specific heat capacity of water is 4.184 J/gK
True:
o
Since Celsius and Kelvin have the same spacing, a delta T in C is the same as a delta T in K.
e. It takes the same amount of energy to warm 1 g of 25 oC water to 26 oC as it does to warm 1 g of 72
o
C water to 73 oC.
True
7.)
It takes 654 J of energy to warm a 5.51 g sample of water. How much energy would be required to warm
55.1 g of water by the same amount?
It would take ten times more energy for ten times more water (q = m c ΔT; q and m are directly related),
so it take 6540 J.
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8.)
250 J is applied to a 10 g sample of aluminum (cAl = 0.900 J/g°C), and the same amount of energy is
applied to 10 g of iron (cFe = 0.473 J/g°C). Without doing any calculations, which metal will have a
greater increase in temperature? Explain your answer.
Al has the higher specific heat capacity so it requires more energy to increase its temperature. Therefore
Fe will have the larger increase in temperature.
9.)
What is one reason that water is the most common coolant used in industrial processes and nuclear
reactors?
Water has an incredibly high specific heat capacity, which means it can absorb a lot of energy before it
raises it temperature. This makes it ideal for cooling something: it’s like an energy sponge.
(Note another good reason is that it is nontoxic, noncorrosive, nonreactive, and just a generally good
substance for the environment.)
Energy Calculations
10.) The specific heat capacity of mercury is 0.140 J/g ºC. What is this value in cal/g ºC?
!.!"# ! ! °!
x
! !"#
!.!"# !
= 0.0335
!"#
! °!
11.) What quantity of heat energy must have been applied to a 42.7 g block of aluminum (cAl = 0.89 J/g°C) if
the temperature of the block increased by 15.2 oC?
q = m c ΔT
= (42.7 g)(0.89
!
! °!
= 5.8 x 102 J
)(15.2 °C)
12.) How much thermal energy (in joules) is absorbed when 203.0 g of mercury (cHg = 0.140 J/g°C) warms
from room temperature (25.0 ºC) to 65.0 °C?
q = m c ΔT
= (0.140
!
! °!
)(203.0 g)( 65.0 °C – 25.0 °C)
= (0.140
!
! °!
)(203.0 g)(40.0 °C)
= 1.14 x 103 J
13.) If 100. J of heat energy is applied to a 25 g sample of mercury, by how many degrees will the temperature
of the sample of mercury increase? (The specific heat capacity of mercury is 0.140 J/g°C)
q = m c ΔT
ΔT =
!
! !
=
!"". !
!" ! (!.!" !
)
! °!
= 29 °C (this is how many degrees the T will rise)
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14.) It takes 1.25 kJ of energy to heat a sample of pure silver from 12.0 oC to 15.2 oC. Calculate the mass of
the sample of silver. (cAg = 0.236 J/g°C)
q = 1.25 kJ x
!""" !
! !"
q = m c ΔT
= 1.25 x 103 J (or 1250 J)
m=
!
!"#
=
!.!" ! !"! !
!
(!.!"# )(!".!!!".! °!)
! °!
=
!.!" ! !"! !
(!.!"# !
)(!.! °!)
! °!
= 1.7 x 103 g (or 1700 g)
15.) If 69.5 kJ of heat is applied to a 1012 g block of metal, the temperature of the metal increases by 11.4 oC.
Calculate the specific heat capacity of the metal.
q = 69.5 kJ x
!""" !
! !"
q = m c ΔT
= 6.95 x 104 J (or 69,500 J)
c=
!
=
! !"
!.!" ! !"! !
= 6.02
(!"!# !)(!!.! °!)
!
! °!
16.) 3939 J is added to a 125.0 g sample of iron (cFe = 0.450 J/g°C) initially at 33.0°C. What is the final
temperature of the iron?
q = m c ΔT
Tf =
q = m c (Tf – Ti)
!"!" !
(!.!"# !
)
! °!
!"#.! !
+ 33.0 oC
Tf =
!
! !
+ !! =
= 70.0 °C + 33.0 °C
= 103.0 °C
Note: it’s fine to do this problem in steps. Meaning to solve for ΔT first (70.0 oC), and then to add that to
the initial temperature to get the final T.
17.) Given that the density of water is 0.997 g/mL, what volume of water (cH2O = 4.184 J/g°C) can be heated
from 25.0 ºC to 100.0 ºC with 4.300 kJ?
q = m c ΔT
m=
!
!"#
13.7 g x
=
q = 4.300 kJ x
!.!"" ! !"! !
!
(!.!"# )(!"".! °!! !".! °!)
! °!
! !"
!.!!" !
!""" !
! !"
=
= 4.300 x 103 J (or 4300. J)
!.!"" ! !"! !
(!.!"# !
)(!".! °!)
! °!
= 13.7 g
= 13.7 mL
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