78. (a) Let
∆
T
be the change in temperature and
κ
be the coefficient of linear expansion
for copper. Then
∆
L =
L
∆
T
and
∆
∆
L
L
T
==
×
°
=
×
−−
(.
/
.
.
17 10
17 10
55
K)(1.0 C)
This is equivalent to 0.0017%. Since a change in Celsius is equivalent to a change on the
Kelvin temperature scale, the value of
used in this calculation is not inconsistent with
the other units involved. Incorporating a factor of 2 for the twodimensional nature of
A
,
the fractional change in area is
∆
∆
A
A
T
×
°
=
×
2
2 17 10
34 10
/
.
K)(1.0 C)
which is 0.0034%. For small changes in the resistivity
ρ
, length
L
, and area
A
of a wire,
the change in the resistance is given by
∆∆
∆
∆
R
RR
L
L
R
A
A
=
∂
∂
+
∂
∂
+
∂
∂
.
Since
R =
L
/
A
,
∂
R
/
∂
= L
/
A = R
/
,
∂
R
/
∂
L
=
/
A = R
/
L
, and
∂
R
/
∂
A = –
L
/
A
2
–R
/
A
.
Furthermore,
∆
/
=
α
∆
T
, where
is the temperature coefficient of resistivity for copper
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Rebello,NobelS
 Absolute Zero, Kelvin, Thermodynamic temperature, Kelvin Temperature Scale, fractional change

Click to edit the document details