Updated:
1/14/04
Answers to Selected Problems
3.1 (a) 268.8 K and 4.625
×
10
5
Pa; (b) at 90
o
(horizontal) 282.6 K and 4.863
×
10
5
Pa; (c) at 180
o
(vertical downward) 296.4 K and 5.1
×
10
5
Pa.
3.2 (a)
δ
Q
/
δ
t
= 3.51
×
10
8
J/hr [contractor pays]
(b)
δ
Q
/
δ
t
= 3.23
×
10
8
J/hr [contractor collects fee]
3.5
(a)
L
A
= 0.4352 m,
L
B
= 0.0218 m,
T
A
=
T
B
= 311 K,
P
A
=
P
B
= 7 bar
(b)
Same as (a)
(c) There is no real solution given the constraints, we must specify more about the expansion
(or compression) of A or B.
For example, if A expands adiabatically,
P
A
=
P
B
= 6.68 bar,
T
A
= 263.6 K,
T
B
= 881.3 K,
L
A
= 0.3917 m, and
L
B
= 0.0655 m.
3.7
Many variations are possible.
If we assume gas in the cylinder is ideal and that it expands
adiabatically, then (a) total time
t
= 0.9 s, (b) height
h
(
max
) = 4.78 m, by reducing tube
length to 1.64 m.
3.12 (a) After 6 s,
T
= 349.5 K,
P
= 1.139
×
10
5
Pa in the bulge.
(b) After 3 s,
T
= 454.6 K,
P
= 6.26
×
10
3
Pa in the large tank.
4.1 For the minimum condition
T
A
(final) = 132 K and for the maximum condition
T
A
(final) =
617 K, these extrema require using the object with the lowest mass x heat capacity product.
4.2
W
= 9.807
y
in J/kg evaporated, for typical values of ambient dry bulb and wet bulb
temperatures of 300 K and 280 K respectively,
y
= 16.8 km!
4.3 (a)
n
±
B
= 0.166 mol/s,
n
±
C
= 0.833 mol/s
(b)
T
D
= 299.1K
(c)
W
=
!
26.4 J/s
(d)
∆
S
= 13.3 J/K
(e) Same as (d).
(f)
max
W
=
!
4000 J/s
4.4
W
=
−
1.11
×
10
4
J,
T
f
= 288.7 K,
P
f
= 0.54 bar
4.6 (a) 21.6 s
(b) 391.8 K and 2.2 bar at 10 s
(c)
∆
S
gas
= 4.93
×
10
3
J/K =
∆
S
universe
(d)
∆
S
(gas in tank) =
!
813.6 J/K
∆
S
(gas vented) = +2020 J/K
∆
S
(surroundings) = +1203 J/K
(e)
W
min
=
+2.83
×
10
5
J