_—————————
21.17 The diffusion coefﬁcient of glucose is 5.7 x 10—10 m2 5". Calculate the time required for a
glucose molecule to diffuse through (a) 10,000 A and (b) 0.10 m.
_—_———-—————-
The root-mean-square distance is related to time via
J55 = V2Dt
T
3
3.13 System = tank and its contents (open
system)
(a) Steady state mass balance
.
M1
T1
dM
= 0 = M1 + M 2 + M 3
dt
M 3 = M1 + M 2 = 10 kg min
b
.
M2
T2
g
.
M3 , T3
Steady state energy balance
dU
= 0 = M1H1 + M 2 H2 + M 3 H3
dt
H3 = H exit stream = H at
3
3.12 System = contents of storage tank (open system)
Mass balance: M 2 M1 = M
Energy balance:
c MU h c MU h = (M )H
2
1
since Q = W = 0 and steam entering is of constant
in
properties.
Initially system contains 0.02 m3 of liquid water and (40 0.02) = 39
3
3.11 System = gas contained in piston and cylinder (closed)
0
Energy balance: U t2 U t1 = Q + Ws PdV
(a) V = constant,
z
z
d
i
a
PdV = 0 , Q = U t2 U t1 = N U t2 U t1 = NCV T2 T1
f
From ideal gas law
N=
.
PV
114,367 Pa 0120 m3
.
=
= 5539 mol (see note f
3
3.10
System = resistor
Energy balance: dU dt = Ws + Q
where Ws = E I , and since we are interested only in steady state dU dt = 0 . Thus
Q = Ws = 1 amp 10 volts = 0.2 (T 25 C) J s
and 1 watt = 1 volt 1 amp = 1 J s .
10 watt 1 J s watt
T =
+ 25 C = 750 C
3
3.9
a) Use kinetic energy = mv2/2 to find velocity.
v 2 m2
kg
1 kg
= 1000 J = 1000 2 2 so v= 44.72 m/sec
2 sec2
m sec
b) Heat supplied = specific heat capacity temperature change, so
1 mol
J
1000g
2510
.
T = 1000 J so T=2.225 K.
55.85g
mol K
3
3.7 (a) Consider a change from a given state 1 to a given state 2 in a closed system. Since initial and final
states are fixed, U1 , U 2 , V1 , V2 , P , P2 , etc. are all fixed. The energy balance for the closed
1
system is
z
U 2 U1 = Q + Ws PdV = Q + W
3
3.6
System: Contents of Drum (open
system)
mass balance: M t2 M t1 = M
energy balance:
MU
t2
but
Ws = 0
and
MU
Q=0
z
t1
steam
z
= MHin + Q + Ws PdV
by problem statement,
PdV = PV
is
negligible.
(Note
V (T = 25 C) = 1003 103 m3 kg ,
.
V (T = 80 C) = 102
3
3.5
a
f a
f
From Illustration 3.2-3 we have that H T1, P = H T2 , P2 for a Joule-Thomson expansion. On the
1
Mollier diagram for steam, Fig. 3.3-1a, the upstream and downstream conditions are connected
by a horizontal line. Thus, graphically, we find th
3
3.2
The velocity change due to the 55 m fall is
m
cv h = 2 9.807 sec
2
2
55 m
F km 3600 sec I
H 1000 m hr K
2
v f = 118.24 km/hr. Now this velocity component is in the vertical direction. The initial velocity of
8 km/hr was obviously in the horizontal
3
3.1
(a) By an energy balance, the bicycle stops when final potential energy equals initial kinetic energy.
Therefore
1 2
v2
mvi = mgh f or h f = i
2
2g
F 20 km 1000 m 1 hr I
H hr
km 3600 sec K
=
2 9.807
2
m
sec2
or h=1.57 m.
(b) The energy balance now i
1
1.2
(a) Water is inappropriate as a thermometric fluid between 0C and 10C, since
the volume is not a unique function of temperature in this range, i.e., two
temperatures will correspond to the same specific volume,
V (T = 1 C) ~ V (T = 7 C); V (T = 2 C)