maldonado (om2892) – Homework
1
2010 – Stone – (13750)
1
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001
(part 1 of 2) 5.0 points
A copper rod and a steel rod are heated. At
0
◦
C the copper rod has a length of
L
C
, the
steel one has a length of
L
S
. When the rods
are being heated or cooled, a difference of
7
.
25 cm is maintained between their lengths.
Find the value of
L
C
.
Correct answer: 13
.
2917 cm.
Explanation:
Given :
d
= 7
.
25 cm
,
α
C
= 1
.
7
×
10
−
5
(
◦
C)
−
1
,
and
α
S
= 1
.
1
×
10
−
5
(
◦
C)
−
1
.
For
d
=
L
S

L
C
= 7
.
25 cm to be constant,
the rods must expand by equal amounts:
α
C
L
C
Δ
T
=
α
S
L
S
Δ
T
L
s
=
α
C
L
C
α
S
or
d
=
α
C
L
C
α
S

L
C
α
S
d
=
α
C
L
C

α
S
L
C
Hence
L
C
=
α
S
d
α
C

α
S
=
bracketleftbig
1
.
1
×
10
−
5
(
◦
C)
−
1
bracketrightbig
(7
.
25 cm)
1
.
7
×
10
−
5
(
◦
C)
−
1

1
.
1
×
10
−
5
(
◦
C)
−
1
=
13
.
2917 cm
.
002
(part 2 of 2) 5.0 points
Find the value of
L
S
.
Correct answer: 20
.
5417 cm.
Explanation:
L
s
=
L
c
+
d
= 13
.
2917 cm + 7
.
25 cm
=
20
.
5417 cm
.
003
(part 1 of 3) 3.0 points
Consider neon, a noble gas whose molecules
consist
of
single
atoms
of
atomic
mass
0
.
02 kg
/
mol.
Avogadro’s number is 6
.
02
×
10
23
mol
−
1
,
and Boltzmann’s constant is 1
.
38
×
10
−
23
J
/
K.
What is the average kinetic energy of a
neon atom when the gas is at a temperature
of 480 K?
Correct answer: 9
.
936
×
10
−
21
J.
Explanation:
Let :
T
= 480 K
and
k
= 1
.
38
×
10
−
23
J
/
K
.
The average kinetic energy of a gas molecule,
one neon atom in this case, is
1
2
m v
2
av
=
3
2
k T
=
3
2
(1
.
38
×
10
−
23
J
/
K) (480 K)
= 9
.
936
×
10
−
21
J
.
004
(part 2 of 3) 3.0 points
What is the rootmeansquare speed of a neon
atom under such conditions?
Correct answer: 773
.
4 m
/
s.
Explanation:
Let :
M
= 0
.
02 kg
/
mol
and
N
A
= 6
.
02
×
10
23
mol
−
1
.
The mass of one neon atom is
m
=
M
N
A
=
0
.
02 kg
/
mol
6
.
02
×
10
23
mol
−
1
= 3
.
32226
×
10
−
26
kg
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maldonado (om2892) – Homework
1
2010 – Stone – (13750)
2
so the rootmeansquare speed is
v
rms
=
radicalbig
v
2
av
=
radicalbigg
m v
2
av
m
=
radicalbigg
3
k T
m
=
radicalBigg
3 (1
.
38
×
10
−
23
J
/
K) (480 K)
3
.
32226
×
10
−
26
kg
= 773
.
4 m
/
s
.
005
(part 3 of 3) 4.0 points
The internal energy of a monoatomic ideal gas
such as neon is simply the total kinetic energy
of all its atoms.
What is the internal energy of 9 liters of
neon at a temperature of 480 K and pressure
of 0
.
89 atm?
Correct answer: 1216
.
18 J.
Explanation:
Let :
P
= 0
.
89 atm = 90157 Pa
and
V
= 9 L = 0
.
009 m
3
.
At temperature 480 K and pressure 0
.
89 atm,
the number of moles of 9 liters of neon is
n
=
P V
R T
So, the internal energy is
E
= (
n N
A
)
parenleftbigg
1
2
m v
2
av
parenrightbigg
=
P V N
A
R T
parenleftbigg
3
2
k T
parenrightbigg
=
3
2
P V
=
3
2
(90157 Pa) (0
.
009 m
3
)
= 1216
.
18 J
.
006
10.0 points
One liter of water at 49
◦
C is used to make iced
tea.
How much ice at 0
◦
C must be added to
lower the temperature of the tea to 19
◦
C?
The specific heat of water is 1 cal
/
g
·
◦
C and
latent heat of ice is 79
.
7 cal
/
g.
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 Summer '10
 Stone
 Energy, Work, Heat, Correct Answer, maldonado

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