PHYSICS 140A : ASSIGNMENT #1 SOLUTIONS
Problem 1.
Solution:
(a) Qw = m w C w "Tw = 4.2 ! 10 3 J , where the specific heat of water is :
C w = 4.2 " 10 3 J / kg !o C
(b) When we neglect the heat loss, the heat received by water should be the exact the same as the
heat lost by the metal. Therefore:
Qm = Qw = 4.2 ! 10 3 J
(c) Capacity of the metal is:
Cm =
Qm
= 55.3 J / K
T0 ! T1
(d) Specific heat is:
cm =
Cm
= 0.55 J / g ! K
m
Problem 2.
Solution:
When you toss pasta into boiling water, there will be heat transferring from boiling water to
pasta. In this process, the total thermal energy should be conserved.
m p C p !T p + m w C w !Tw = 0 , where p=pasta, w=water
Necessary constants are:
m w = !V = 1.0kg / L * 1.5 L = 1.5kg
C w = 4.2 " 10 3 J / kg !o C
! TFinal = 92.5 o C
So if the stove provides no heat to the system before thermal equilibrium, the final
o
temperature would be 92.5 C .
Problem 3.
Solution:
#
(a) % p +
$
!
$
,$ #p ' 2a+ 2 /
a" 2 &
a+ 2 '
(V
)
"
b)
=
"
RT
"
*
(V
*
+
b)
+
p
+
&
)
(
&
)=0
.
1
V2 '
V3 0
V2 (
-% #V (T
%
!
$ #p '
+RT
2a+ 2
"& ) =*
+
% #V (T
(V * +b) 2
V3
uRT
2a
' (p $
) !V %
! 2
" =
2
u
& (V #T (u ! b)
!
#
(b) % p +
$
$ #p '
a" 2 &
(V ) "b) = "RT " & ) (V * +b) = +R
2 (
% #T (V
V '
$ #p '
R
"& ) =
% #T (V (u * b)
!
!
#
(c) % p +
$
&
a" 2 &
2a$ 2 & %V )
a$ 2 )& %V )
(V
)
"
b)
=
"
RT
"
#
(V
#
$
b)
+
p
+
(
+
(
(
+( + = $R
V2 '
V 3 ' %T * p
V 2 *' %T * p
'
!
!
2a$ 2 & %V )
$RT & %V )
" # 3 ( + (V # $b) +
( + = $R
V ' %T * p
V # $b ' %T * p
!
"
!
where u =
1 $ #V '
R
& ) =
V % #T ( p 2ab * 2a + RTu
u2
u u*b
V
is the molar volume.
v
!
Problem 4.
! Solution:
# "E &
( +
$ "T ' p
(a) C p = %
! cp =
!
# "V & )*
p% ( = 0 eT /T0 + )R
$ "T ' p T0
" 0 T / T0
e
+R
T0
# "V &
( can be obtained by the similar method employed in Problem 3.
$ "T ' p
The term p%
TB
(b) Q =
!
# "c
TA
TB
p
dT =
% "$0
# '& T
TA
0
(
eT /T0 + "R*dT = E(TB ,V," ) + E(TA ,V," ) + "R(TB + TA )
)
= 3.58 !10 4 J
!
2
$ "p '
, ! is density defined by M /V = M 0 /u , M 0 is molar mass.
)
% "# ( dQ= 0
(c) c s = &
!
!
c s2 =
"p "u
u 2 % "p (
=$
' *
"u "#
M 0 & "u ) dQ= 0
(1)
Therefore, in order to compute c s , we must know the relationship between p and v.
!
dE = dQ ! pdV = ! pdV , also we know:
E = !" 0 e T / T0
" dT = #
pT0 #T /T0
e
du
$0
(2)
From equation of state,
(u " b)dp + pdu = RdT , then substituating (2) into this expression, we get:
!
dp = "
!
RT $ RT0 "T /T0 '
e
&1+
)du
(u " b) 2 %
#0
(
(3)
Then from (1) we can get the complete solution:
c s2 =
!
2
RT # u & # RT0 "T /T0 &
e
%
( %1+
(
M0 $ u " b ' $
)0
'
Problem 5.
!
Solution:
(a) Travel time is not a state function because it depends on the path taken and not on the
endpoints alone.
(b) Elevation is a state function.
No matter what path you take between A and B, the net
elevation change will be the same.
© Aggravation most certainly is not a state function. You can become more aggravated if your
path takes you along a highly congested stretch of road.
Problem 6.
Solution:
For a differential as: dW =
! A dx
i
i
i
A simple rule to identify it “exact” or “inexact” is:
!Ai !A j
=
!x j
!xi
(1)
dW = 2 x 2 ydx + x 3 dy
Q
!Ay
!Ax
= 2 x 2 " 3x 2 =
!y
!x
! Inexact differential.
It is easy to find if we employ a factor “
1
”, the new differential becomes:
x
dU = 2 xydx + x 2 dy = d ( x 2 y )
Now it is an exact differental. L = ln x
(2)
dW =
y2z
x2 z
xy
dx
+
dy +
dz
2
2
x+ y
( x + y)
( x + y)
According to the rule above, it is an exact differential.
" xyz %
'
# x + y&
In fact, it can be written as dW = d$
(3)
x
dW = e xy dx + (2 x + e xy )dy
y
!
Obviously, it is an inexact differential.