roofner (bar784) – Homework 22 – Weathers – (17104)
1
This
printout
should
have
18
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A square hole 9
.
2 cm along each side is cut in
a sheet of copper.
Calculate the change in the area of this hole
if the temperature of the sheet is increased by
44 K.
Correct answer: 0
.
126621 cm
2
.
Explanation:
Given :
L
= 9
.
2 cm
and
Δ
T
= 44 K
.
Each side of the hole expands linearly by
Δ
L
=
L α
Δ
T
.
The area, therefore expands
by Δ
A
= (
L
+ Δ
L
)
2

A
. Since Δ
L
is much
smaller than
L
we should keep only the linear
term in Δ
L
, so
Δ
A
= 2
L
Δ
L
= 2
α A
Δ
T
= 2
bracketleftbig
1
.
7
×
10
−
5
(
◦
C)
−
1
bracketrightbig (
84
.
64 cm
2
)
(44 K)
= 0
.
126621 cm
2
002
(part 1 of 2) 10.0 points
At 17
◦
C, an aluminum ring has an inner di
ameter of 6 cm and a brass rod has a diameter
of 6
.
01 cm.
If the temperature coefficient of expansion
for brass is
α
b
= 1
.
9
×
10
−
5
(
◦
C)
−
1
and the
temperature coefficient of expansion for alu
minum is
α
a
= 2
.
4
×
10
−
5
(
◦
C)
−
1
, to what
temperature must the ring be heated so that
it will just slip over the rod?
Correct answer: 86
.
4444
◦
C.
Explanation:
Given :
T
0
= 17
◦
C
,
d
a
= 6 cm
,
d
b
= 6
.
01 cm
,
α
b
= 1
.
9
×
10
−
5
(
◦
C)
−
1
,
and
α
a
= 2
.
4
×
10
−
5
(
◦
C)
−
1
.
The new length will be
d
b
=
d
a
(1 +
α
Δ
T
)
d
b
=
d
a
+
d
a
α
Δ
T
T
1
=
T
0
+ Δ
T
=
T
0
+
d
b

d
a
d
a
α
a
= 17
◦
C +
6
.
01 cm

6 cm
(6 cm)(2
.
4
×
10
−
5
(
◦
C)
−
1
)
=
86
.
4444
◦
C
.
003
(part 2 of 2) 10.0 points
To what temperature must both be heated so
that the ring just slips over the rod?
Correct answer: 352
.
458
◦
C.
Explanation:
We need
L
Al
=
L
brass
for some Δ
T
. Using
the same law,
L
0
Al
(1 +
α
a
Δ
T
) =
L
0
brass
(1 +
α
b
Δ
T
)
d
a
+
d
a
α
a
Δ
T
=
d
b
+
d
b
α
b
Δ
T
Δ
T
=
T
0
+
d
b

d
a
α
a
d
a

α
b
d
b
α
a
d
a

α
b
d
b
=
bracketleftbig
2
.
4
×
10
−
5
(
◦
C)
−
1
bracketrightbig
(6 cm)

bracketleftbig
1
.
9
×
10
−
5
(
◦
C)
−
1
bracketrightbig
(6
.
01 cm)
= 2
.
981
×
10
−
5
m
/
◦
C
so
Δ
T
=
T
2

T
0
T
2
=
T
0
+ Δ
T
=
T
0
+
d
b

d
a
α
a
d
a

α
b
d
b
= 17
◦
C +
6
.
01 cm

6 cm
2
.
981
×
10
−
5
m
/
◦
C
=
352
.
458
◦
C
.
004
10.0 points
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
roofner (bar784) – Homework 22 – Weathers – (17104)
2
3
mole
of
xenon
gas
at
53
◦
C
occupies
0
.
154 m
3
.
What is the pressure exerted by the Xe
atoms on the walls of a container?
Correct answer: 52799
.
3 Pa.
Explanation:
Given :
n
= 3 mole
,
T
= 53
◦
C = 326 K
,
V
= 0
.
154 m
3
,
and
R
= 8
.
314 J
/
K
·
mol
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Weathers
 Thermodynamics, mechanics, Energy, Work, Correct Answer

Click to edit the document details