This preview shows pages 1–3. Sign up to view the full content.
lim (kl9356) – Homework 22 – Weathers – (17104)
1
This printout should have 18 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
A square hole 10
.
5 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
76 K.
Correct answer: 0
.
284886 cm
2
.
Explanation:
Given :
L
= 10
.
5 cm
and
Δ
T
= 76 K
.
Each side oF the hole expands linearly by
Δ
L
=
Lα
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
αA
Δ
T
= 2
b
1
.
7
×
10
−
5
(
◦
C)
−
1
B (
110
.
25 cm
2
)
(76 K)
= 0
.
284886 cm
2
002
(part 1 oF 2) 10.0 points
At 9
◦
C, an aluminum ring has an inner diam
eter oF 3 cm and a brass rod has a diameter oF
3
.
06 cm.
IF the temperature coe±cient oF expansion
For brass is
α
b
= 1
.
9
×
10
−
5
(
◦
C)
−
1
and the
temperature coe±cient 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: 842
.
333
◦
C.
Explanation:
Given :
T
0
= 9
◦
C
,
d
a
= 3 cm
,
d
b
= 3
.
06 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
= 9
◦
C +
3
.
06 cm

3 cm
(3 cm)(2
.
4
×
10
−
5
(
◦
C)
−
1
)
=
842
.
333
◦
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: 4338
◦
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
=
b
2
.
4
×
10
−
5
(
◦
C)
−
1
B
(3 cm)

b
1
.
9
×
10
−
5
(
◦
C)
−
1
B
(3
.
06 cm)
= 1
.
386
×
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
= 9
◦
C +
3
.
06 cm

3 cm
1
.
386
×
10
−
5
m
/
◦
C
=
4338
◦
C
.
004
10.0 points
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document lim (kl9356) – Homework 22 – Weathers – (17104)
2
2 mole of xenon gas at 37
◦
C occupies
0
.
294 m
3
.
What is the pressure exerted by the Xe
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 09/07/2010 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.
 Fall '08
 Weathers
 Magnetism, Work

Click to edit the document details