PHYS 2101
HW#12
WileyPlus Problem Solutions
1
HW #12 Solutions – or at least most of them…
(USING BOOK VALUES 9
th
Edition in problems; problem # indicated by [])
1) [187] We assume scale X is a linear scale in the sense that if its reading is
x
then it is related to a reading
y
on the Kelvin scale by a linear relationship
y = mx + b
. We determine the constants
m
and
b
by solving the
simultaneous equations:
373.15
=
m
(
!
53.5)
+
b
&
273.15
=
m
(
!
170)
+
b
which yield the solutions
m
= 100/(170 – 53.5) = 0.858 and
b
= 419. With these values, we find
x
for
y
= 340:
2) [1813] Since a volume is the product of three lengths, the change in volume due to a temperature change
!
T
is given by
!
V
= 3
!
V
!
T
, where
V
is the original volume and
is the coefficient of linear expansion. See
Eq. 1811. Since
V
= (4
"
/3)
R
3
, where
R
is the original radius of the sphere, then
The value for the coefficient of linear expansion is found in Table 182. The change in volume can be
expressed as
, where
is the coefficient of volume expansion. For aluminum, we have
.
3) [1817] If
V
c
is the original volume of the cup,
a
is the coefficient of linear expansion of aluminum, and
!
T
is the temperature increase, then the change in the volume of the cup is
!
V
c
= 3
a
V
c
!
T
. See Eq. 1811. If
"
is
the coefficient of volume expansion for glycerin, then the change in the volume of glycerin is
!
V
g
=
V
c
!
T
.
Note that the original volume of glycerin is the same as the original volume of the cup. The volume of glycerin
that spills is
!
V
g
" !
V
c
=
#
"
3
$
a
( )
V
c
!
T
=
5.1
%
10
"
4
/ C
°
( )
"
3 23
%
10
"
6
/ C
°
( )
'
(
)
100cm
3
( )
6.0
°
C
( ) =
0.26cm
3
.
Note: Glycerin spills over because
, which gives
. Note that since liquids in general
have greater coefficients of thermal expansion than solids, heating a cup filled with liquid generally will cause
the liquid to spill out.
4) [1823] The mass
m
= 0.100 kg of water, with specific heat
c
= 4190 J/kg·K, is raised from an initial
temperature
T
i
= 23°C to its boiling point
T
f
= 100°C. The heat input is given by
Q
=
cm
(
T
f
– T
i
). This must be
the power output of the heater
P
multiplied by the time
t
;
Q = Pt
. Thus,
5) [1827] The melting point of silver is 1235 K, so the temperature of the silver must first be raised from
15.0° C (= 288 K) to 1235 K. This requires heat
Now the silver at its melting point must be melted. If
L
F
is the heat of fusion for silver, this requires
The total heat required is ( 2.91
#
10
4
J + 1.36
#
10
4
J ) = 4.27
#
m
i
. Then
,
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 Spring '07
 GROUPTEST
 Physics, Thermodynamics, WileyPlus Problem Solutions

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