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CHAPTER TWO
2.1 (a)
wk 7 d h s 1000 ms 1 wk 1 d h 1 s =. × ms
(b)
1 ft / s 0.0006214 mi s 3.2808 ft 1 h = mi / h ⇒ mi / h
(c)
1000 g
m 1 d h kg 10 cm d kg 24 h 60 min 1 m
85 10 cm g
4 8 4 4
4 4 ⋅
=. × / min⋅
2.2 (a) 760 mi 3600
1 m 1 h h 0.0006214 mi s = m / s
(b) 921 kg 35.3145 ft
2.20462 lb 1 m 57 5 m 1 kg
m lb / ft 3 3 3 m
=.^3
(c) 5 37^10 1000 J^1 1
× kJ 1 min .34 × 10 hp=. ⇒ min 60 s 1 kJ J / s
hp hp
2.3 Assume that a golf ball occupies the space equivalent to a 2 in × 2 in × 2 incube. For a classroom with dimensions 40 ft × 40 ft × 15 ft: n balls ft 3 (12) 3 in 1 ball 6 ft in = 40 ×^40 ×^15 = × 10 ≈5 million balls 2
3 3 3 3. The estimate could vary by an order of magnitude or more, depending on the assumptions made.
2.4 4 3 24 3600 s 1 0 0006214
light yr 365 d h 1.86 10 mi 3.2808 ft 1 step 1 yr 1 d h 1 s mi 2 ft
× 5 = 7 × 10 16 steps
2.5 Distance from the earth to the moon = 238857 miles
238857 mi 1 4 10 11
1 m report 0.0006214 mi 0.001 m = × reports
km 1000 m mi L 1 L 1 km 1 m gal mi / gal
Calculate the total cost to travel miles.
Total Cost
gal (mi) gal 28 mi
Total Cost gal (mi) gal 44.7 mi
Equate the two costs 4.3 10 miles
American
European
5
⇒ = ×
x x x x x x
6 3 3 5
5320 imp. gal 14 h 365 d 10 cm 0.965 g 1 kg 1 tonne plane h 1 d 1 yr 220.83 imp. gal 1 cm 1000 g 1000 kg
1.188 10 tonne kerosene plane yr
⋅
= × ⋅ 9 5
4.02 10 tonne crude oil 1 tonne kerosene plane yr yr 7 tonne crude oil 1.188 10 tonne kerosene 4834 planes 5000 planes
× ⋅
×
2.8 (a) 25 0.^ lb^ 32.1714 ft / s^1 lb 25 0. 32.1714 lb ft / s
m lb 2 f m^2 ⋅ = f
(b)
2 2
25 N 1 1 kg m/s 2.5493 kg 2.5 kg 9.8066 m/s 1 N
(c) 10 ton 1 lb 1000 g 980.66 cm / s 1 dyne 9 10 9 5 10 ton 2.20462 lb 1 g cm / s
m dynes 2 × -4 m ⋅ 2 =^ ×
2.9^50 15 2 85 3^ 32 174
× × 4 5 10 6
m 35.3145 ft lb ft 1 lb = × 1 m 1 ft s 32.174 lb ft s
lb
(^3 3) m f 3 3 2 m 2 f
500 lb 5 10 1 2
m (^225) 3
m
1 kg 1 m 3 2.20462 lb 11.5 kg
≈ × FHG IKJFHG IKJ ≈ m
2.11 (a) m m V V h r H r h H
f f c c f c
c
f
displaced fluid cylinder cm cm g / cm^33 30 cm
g / cm
= = −^ =
2 2
( 30 14 1. )(. 100 ) (^) 0 53.
(b) ρ f ρ c^ H
h
= = (^30 cm)(.^ 0 53^ g / cm )^ = 171. (30 cm - 20.7 cm)
g / cm
(^33)
H
h
ρf
ρc
V R H^ V R H^ r h^ R H
r h r R H h
V R H^ h^ Rh H
R (^) H h H
V V R^ H h H
R H
H
H h H
H
H h (^) h H
s f
f
f f s s f s
f s s s
⇒ = − F
HG^
I
KJ^
F
HG^
I
KJ
F
HG^
I
KJ^
− FHG IKJ
2 2 2
2 2 2 3 2 2 3 2
2
3 2
3 3 3 3
ρ^ ρf s R
r
h H
2.15 (a) F = ma ⇒ FHG IKJ = ⋅
fern = (1 bung)(32.174 ft / s bung ft / s fern 5.3623 bung ft / s
2 2
2
(b) On the moon: 3 bung 32.174 ft 1 fern 6 s 5.3623 bung ft / s fern
On the earth: = 18 fern
W 2 2
W
2.16 (a) ≈ =
=
(b)
(^45)
4 6
− −
− −
×
≈ ≈ ×
× = ×
(c) ≈ + =
(d) ≈ × − × ≈ × ≈ × × − × = ×
3 3 3 4
.^4 2
(^1 5 42 ) 6 3
exact 3812.5^3810 3.81 10
R
R
× − × ×
≈ ≈ × ≈ ×
×
= ⇒ ⇒ ×
(Any digit in range 2-6 is acceptable)
2.18 (a) A: C
C
C
o
o
o
R
X
s
= +^ +^ +^ +^ =
= −^ +^ −^ +^ −^ +^ −^ +^ −
2 2 2 2 2
B: C
C
C
o
o
o
R
X
s
= +^ +^ +^ +^ =
= −^ +^ −^ +^ −^ +^ −^ +^ −
2 2 2 2 2
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.
2.19 (a)
X
X
s
X
X s X s
i = i^ = = i
= =
∑ ∑ 1
(^122) 1
12
C
C
min= max= (b) Joanne is more likely to be the statistician, because she wants to make the control limits stricter. (c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor temperature (failure of reactor control system), problems with the color measurement system, operator carelessness
2.20 (a), (b)
(c) Beginning with Run 11, the process has been near or well over the upper quality assurance limit. An overhaul would have been reasonable after Run 12.
2.21 (a)
4 2 2 2 2
2.36 10 kg m 2.20462 lb 3.2808 ft 1 h ' h kg m 3600 s
Q
× − ⋅
(b)
(^4) ( 4 3) 6 2 approximate (^3) 6 2 2 exact
' (2^10 )(2)(9) 12 10 1.2 10 lb ft / s 3 10 ' =1.56 10 lb ft / s 0.00000156 lb ft / s
Q
Q
− (^) − − −
−
≈ × ≈ × ≈ × ⋅
×
× ⋅ = ⋅
(a) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 X 134 131 129 133 135 131 134 130 131 136 129 130 133 130 133 Mean(X) 131. Stdev(X) 2. Min 127. Max 136.
(b) Run X Min Mean Max 1 128 127.5 131.9 136. 2 131 127.5 131.9 136. 3 133 127.5 131.9 136. 4 130 127.5 131.9 136. 5 133 127.5 131.9 136. 6 129 127.5 131.9 136. 7 133 127.5 131.9 136. 8 135 127.5 131.9 136. 9 137 127.5 131.9 136. 10 133 127.5 131.9 136. 11 136 127.5 131.9 136. 12 138 127.5 131.9 136. 13 135 127.5 131.9 136. 14 139 127.5 131.9 136.
126
128
130
132
134
136
138
140
0 5 10 15
2.26 (a) 70 5. lb (^) m/ ft 3 ; 8.27 × 10 -7^ in 2 / lbf
(b)
7 2 6 2 3 f m (^2 5 ) f 3 3 m (^3 3 6 3) m
8.27 10 in 9 10 N 14.696 lb / in (70.5 lb / ft )exp lb m 1.01325 10 N/m 70.57 lb 35.3145 ft 1 m 1000 g 1.13 g ft m 10 cm 2.20462 lb
ρ
⎡ × − × ⎤
⎢⎣ × ⎥⎦
= = /cm^3
(c) ρ^ lb ρ ρ
ft
g lb cm cm g 1 ft
m 3
m^3 3 3
F
HG^
I
KJ^
P^ lb P P in
N .2248 lb m m N 39.37 in
f 2
f^2 2 2 2
F
HG^
I
KJ^
= ' 0 1 =. × − '
(^24)
⇒ 62 43. ρ′ = 70 5. exp d8 27. × 10 −^7 id 145. × 10 − 4 P ' i ⇒ ρ′ = 113. exp d 120. × 10 −^10 P 'i
P ' = 9 00. × 10 6 N / m 2 ⇒ ρ' = 113. exp[(. 120 × 10 − 10 )( .9 00 × 10 6 )] = 113. g / cm^3
2.27 (a) V
V
cm V in 28,317 cm in
3 3 3
d i 3
= '^ d^ i =. '
16 39 ; t b gs = 3600 t ′b ghr
⇒ 16 39. V ' = exp b 3600 t ′ ⇒g V ' = 0 06102. expb 3600 t ′g
(b) The t in the exponent has a coefficient of s-^.
2.28 (a) 3 00. mol / L, 2.00 min -
(b) t C C
exp[(-2.00)(0)] = 3.00 mol / L t = 1 exp[(-2.00)(1)] = 0.406 mol / L For t=0.6 min: C C
int
mol / L
exact exp[(-2.00)(0.6)] = 0.9 mol / L For C=0.10 mol/L: t
t
int
exact
min
= - 1
ln C
= -^1
ln 0.
= 1.70 min
(c)
0
1
2
3
0 1 2 t (min)
C (mol/L)
(t=0.6, C=1.4) (t=1.12, C=0.10)
C (^) exact vs. t
2.29 (a) p *
..
185 166 2 20 42 mm Hg
(b) c MAIN PROGRAM FOR PROBLEM 2. IMPLICIT REAL*4(A–H, 0–Z) DIMENSION TD(6), PD(6) DO 1 I = 1, 6 READ (5, *) TD(I), PD(I) 1 CONTINUE WRITE (5, 902) 902 FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X,
- ‘ (C) (MM HG)’/) DO 2 I = 0, 115, 5 T = 100 + I CALL VAP (T, P, TD, PD) WRITE (6, 903) T, P 903 FORMAT (10X, F5.1, 10X, F5.1) 2 CONTINUE END SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6) I = 1 1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2 I = I + 1 IF (I.EQ.6) STOP GO TO 1 2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I)) RETURN END DATA OUTPUT 98.5 1.0 TEMPERATURE VAPOR PRESSURE 131.8 5.0 (C) (MM HG)
# 100.0^ 1.
215.5 100.0 105.0 1.
215.0 98.
2.30 (b) ln ln (ln ln ) / ( ) (ln ln ) / ( ). ln ln ln. ( )..
y a bx y ae b y y x x a y bx a y e
bx
x
2 1 2 1
(c) ln ln ln (ln ln ) / (ln ln ) (ln ln ) / (ln ln ) ln ln ln ln ( ) ln( ) /
y a b x y ax b y y x x a y b x a y x
= + ⇒ = b = − − = − − = − = − = − − ⇒ = ⇒ =
(d) /^ / 2 1 2 1 3 /
ln( ) ln ( / ) ( / ) [can't get ( )] [ln( ) ln( ) ]/[( / ) ( / ) ] (ln 807.0 ln 40.2) /(2.0 1.0) 3 ln ln( ) ( / ) ln 807.0 3ln(2.0) 2 2 [can't solve explicitly for
by x by x
y x
xy a b y x xy ae y a x e y f x b xy xy y x y x a xy b y x a xy e
y x ( )]
2.32 (a) A plot of y vs. R is a line through ( R = 5 , y = 0 011. ) and ( R = 80 , y = 0169. ).
0
0 20 40 60 80 100 R
y
y a R b a b
y R
= ×
= − × = ×
U
V
W|
⇒ = × + ×
− − −
− −
3 3 4
3 4
d ib g
(b) R = 43 ⇒ y = d 2 11. × 10 −^3 ib g 43 + 4 50. × 10 −^4 =0 092. kg H O kg 2
b 1200 kg h gb0 092. kg H O kg 2 g=110 kgH O h 2
2.33 (a) ln ln ln (ln ln ) / (ln ln ) (ln ln ) / (ln ln ). ln ln ln ln (. ) ln( )...
T a b T a b T T a T b a T
= + ⇒ = b = − − = − − = − = − = − − ⇒ = ⇒ = −
2 1 2 1 1 19
(b) T T T C T C T C
1 19 0.
φ. φ φ φ φ
b g
b g
b g
b g
o o o
L / s L / s L / s (c) The estimate for T =175°C is probably closest to the real value, because the value of temperature is in the range of the data originally taken to fit the line. The value of T =290°C is probably the least likely to be correct, because it is farthest away from the date range.
2.34 (a) Yes, because when ln[( C (^) A − C (^) Ae ) / ( C (^) A 0 − CAe )]is plotted vs. t in rectangular coordinates, the plot is a straight line.
-1.
-0.
0
0 50 100 150 200
ln ((C t (m in)
-CA
)/(CAe
-CA
))Ae
Slope = -0.0093 ⇒ k = 9.3 × 10 -3min−^1 (b) 3
0 0 (9.3 10 )(120) -
ln[( ) /( )] ( ) (0.1823 0.0495) 0.0495 9.300 10 g/L 9.300 10 g 30.5 gal 28.317 L = / = 10.7 g L 7.4805 gal
−
− − ×
= − + = ×
×
kt A Ae A Ae A A Ae Ae A
C C C C kt C C C e C C e
C m V m CV
2.35 (a) ft 3 and h -2, respectively
(b) ln( V ) vs. t^2 in rectangular coordinates, slope=2 and intercept= ln(. 353 × 10 −^2 ); or V (logarithmic axis) vs. t^2 in semilog coordinates, slope=2, intercept= 353. × 10 −^2 (c) V ( m 3 ) = 100. × 10 − 3 exp(. 15 × 10 −^7 t^2 )
2.36 PV k^ = C ⇒ P = C / V k ⇒ ln P = ln C − k ln V
ln P = -1.573(ln V ) + 12.
6
7
8
2.5 3 3.5 4 lnV
lnP
slope (dimensionless) Intercept = ln mm Hg cm 4.
k C C e
= ⇒ = = × ⋅
2.37 (a) G^ G G G (^) K C
G G
G G
K C G^ G
G G
L K m C L m L
L m L
L
0
(^1 0) ln 0 ln ln
ln (G 0 -G )/(G -G L )= 2 .4 8 3 5 ln C - 1 0 .0 4 5
0
1
2
3
3 .5 4 4 .5 5 5. ln C
ln(G
0 -G)/(G-G
L^ )
2.39 (a)
s n x y
s n x
s n
x s n
y
a
s s s s s
xy i i i
n
xx i i
n
x i i
n y i i
n
xy x y xx x
=
=
= =
1 2 1
2 2 2
1 1
2
[(. )(. ) (. )(. ) (. )(. )] /.
b g
2
2 2
b s s s s s s y x
xx y xy x
xx b g x
(b) a s s
xy y x xx
y = 1.0065x
y = 0.936x + 0.
0
1
2
3
4
0 1 2 3 4 x
y
2.40 (a) 1/C vs. t. Slope= b, intercept=a
(b) b = slope = 0.477 L / g h⋅ ; a =Intercept = 0.082 L / g
1/C = 0.4771t + 0.
0
1
2
3
0 1 2 3 4 5 6 t
1/C
0
1
2
1 2 3 4 5 t
C
C C-fitted (c) C a bt t C a b
/ ( ) / [.. ( )].
g / L h (d) t=0 and C=0.01 are out of the range of the experimental data. (e) The concentration of the hazardous substance could be enough to cause damage to the biotic resources in the river; the treatment requires an extremely large period of time; some of the hazardous substances might remain in the tank instead of being converted; the decomposition products might not be harmless.
2.41 (a) and (c)
1
10
0.1 1 10 100 x
y
(b) y = ax b ⇒ ln y = ln a + b ln x ; Slope = b, Intercept = ln a ln y = 0.1684ln x + 1.
0
1
2
-1 (^0 1) ln x 2 3 4 5
ln y
b a a
slope Intercept = ln
2.42 (a) ln(1- Cp /CA0 ) vs. t in rectangular coordinates. Slope=- k , intercept=
(b)
ln(1-Cp/Cao) = -0.0062t Lab 1
0 200 400 600 800
t
ln(1-Cp/Cao) Lab 2 ln(1-Cp/Cao) = -0.0111t-
0
0 100 200 300 400 500 600
t
ln(1-Cp/Cao)
k = 0 0062. s -1^ k = 0 0111. s -
Lab 3 ln(1-Cp/Cao) = -0.0063t-
0
0 200 400 600 800
t
ln(1-Cp/Cao) Lab 4
ln(1-Cp/Cao)= -0.0064t-
0
0 200 400 600 800
t
ln(1-Cp/Cao)
k = 0 0063. s -1^ k = 0 0064. s - (c) Disregarding the value of k that is very different from the other three, k is estimated with the average of the calculated k ’s. k = 0 0063. s - (d) Errors in measurement of concentration, poor temperature control, errors in time measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty reactor.
2.45 (a) E (cal/mol), D 0 (cm^2 /s)
(c) Intercept = ln D 0 (^) = -3.0151 ⇒ D 0 = 0.05 cm 2 / s. 1/T ln D
- (b) ln D vs. 1/T, Slope=- E/R, intercept=ln D
- ln D = -3666(1/T) - 3. Slope = − E / R = -3666 K ⇒ E = (3666 K)(1.987 cal / mol K) = 7284 cal / mol⋅
- -14.
- -13.
- -12.
- -11.
- -10.0 2.0E-032.1E-032.2E-032.3E-032.4E-032.5E-032.6E-032.7E-032.8E-032.9E-033.0E- - T D 1/T lnD (1/T)(lnD) (1/T)* (d) Spreadsheet - 347 1.34E-06 2.88E-03 -13.5 -0.03897 8.31E-
- 374.2 2.50E-06 2.67E-03 -12.9 -0.03447 7.14E-
- 396.2 4.55E-06 2.52E-03 -12.3 -0.03105 6.37E-
- 420.7 8.52E-06 2.38E-03 -11.7 -0.02775 5.65E-
- 447.7 1.41E-05 2.23E-03 -11.2 -0.02495 4.99E-
- 471.2 2.00E-05 2.12E-03 -10.8 -0.02296 4.50E- - Sx 2.47E- - Sy -12. - Syx -3.00E- - Sxx 6.16E- - -E/R - - ln D 0 -3. - D - E 0.
CHAPTER THREE
3.1 (a) m =
× ×
≈ × ≈ ×
m kg m
kg
3
3 b^ gb gb gd^ i
(b) m = ≈ × ×
8 10 ≈ ×
6 6 3
oz 1 qt cm 1 g 2 s oz 1056.68 qt cm
g / s
3
3 b gd i
(c) Weight of a boxer ≈220 lb (^) m W max ≥
×
220 stones lb 1 stone 14 lb
m m
(d) dictionary V = D L =
× × × × × × ×
× ×
≈ ×
π 2 2
2 3 7
.. ft 800 miles 5880 ft 7.4805 gal 1 barrel 1 mile 1 ft 42 gal
barrels
2 3
d i d i
(e) ( i ) V ≈ 6 ft^ ×^ 1 ft^ ×^ 0.5 ft^ 28,317 cm ≈ 3 × 3 × 10 4 ≈ 1 × 105 1 ft
cm
3 3
3
( ii ) V ≈ 150 28 317^ ≈ 150 ×^3 ×^10 ≈ × 60
lb 1 ft cm^45 62.4 lb 1 ft
m cm 3 3 m^3
(f) SG ≈ 105.
3.2 (a) (i)^995 1 0 028317 0 45359 1
kg lb m 62 12 m kg ft
m lb / ft 3 3 3 m
(ii)
kg / m lb / ft kg / m
lb / ft
3 m 3 3 m
(b) ρ = ρ H O 2 × SG = 62 43. lb m/ ft 3 × 5 7. = 360 lb m/ ft^3
3.3 (a)^50 10
L 0.70 10 kg 1 m 3 35 m L
kg
3 3 3
× =
(b)^1150 1 60
kg m 1000 L 1 min 27 0.7 1000 kg m s
L s
3 min × 3 =
(c)
gal ft lb min gal ft
lb / min
3 m 3 m
×
3.6 (a) V = (^) × =
kg H SO kg solution L kg H SO kg L
2 4 2 4 (b) V ideal^2 2 4 2 2 4
kg H SO L kg kg H SO kg H O L kg H SO kg L
= ×
% error = 470 −^445 × =. 445
3.7 Buoyant force b gup =Weight of block bdown g
E
Mass of oil displaced + Mass of water displaced = Mass of block
ρoil b0 542. g V + ρH O 2 b 1 − 0 542. g V =ρc V
From Table B.1: ρ c = 2 26. g / cm 3 , ρ w = 100. g / cm 3 ⇒ ρo il =3 325. g / cm^3
m oil = ρ oil× V = 3 325. g / cm 3 × 35 3. cm^3 =117 4. g
m oil + flask = 117 4. g + 124 8. g = 242 g
3.8 Buoyant force b gup = Weight of block bdown g
⇒ W displaced liquid = W block ⇒ ( ρ Vg ) disp. Liq =( ρ Vg )block
Expt. 1: ρ w 15 A g ρ B 2 A g ρ B ρ w^15
b.^ g =^ b g ⇒^ =^ ×.
ρ ρ
w B SG B
= = ⇒ =
1 0 75 0 75
.
..
g/cm (^33)
g / cm b g
Expt. 2: ρsoln b g A g = ρ B b 2 A g g ⇒ ρ soln= 2 ρ B = 15. g / cm^3 ⇒ b SG gsoln= 15.
W + WA B h^ s
h (^) b h^ ρ^1 Before object is jettisoned
1
1
Let ρ w = density of water. Note: ρ A > ρ w (object sinks)
Volume displaced: V d 1 = A hb si = Ab d h p 1 − hb 1 i (1)
Archimedes ⇒ ρ wV d 1 g = WA + WB
weight of displaced water
Subst. (1) for Vd 1 , solve for d h p 1 − hb 1 i
h h W^ W p b p gA A B w b
1 −^1 =^
Volume of pond water: V (^) w A hp p V (^) d V A h A h h
i = 1 − 1 ⇒ (^) w = (^) p p 1 − (^) b p 1 − b 1
b g
d i
for
subst. 2 b h w^ p^ p
A B w
p w p
A B p b w p
V A h W^ W p g h V A
W W
1 − 1 1 1 p gA
= − +^ ⇒ = + +
b g
2 solve for
subst. 3 for in
b g
b g b g
, h
h b w p
A B b w p b
p h
V
A
W W
1 p g^ A^ A
1 1
= + +^ L^1 −^1
N
M
M
O
Q
P
P
3.9 (cont’d)
W h (^) s B
h (^) b h^ ρ^2
After object is jettisoned
WA^2
2 Let V (^) A = volume of jettisoned object = W g
A
ρ A
Volume displaced by boat: V d 2 = Ab d h p 2 − hb 2 i (6)
Archimedes ⇒ =
E
ρ W Vd (^) 2 g WB
Subst. for Vd 2 , solve for d h p 2 − hb 2 i
h h W p b p gA B w b
2 −^2 =^ (7)
Volume of pond water: V A h V V V A h W p g
W
w p p d A w p p p g B w
A A
5 6 7 2
b g b g,^ &b g
2
solve for h (^) p p^2 w^ B^ A p w p A p
h V^ W^ W A p gA p gA
for in 7 ⇒ solve for^ =^ +^ +^ −
subst. 8 h h b
w p
B w p
A A p
B p b w b h
V
A
W
p gA
W
p gA
W
2 b g 2 2 p gA
b g
,
( a) Change in pond level
( )^8 ( )^3 ( )
2 1
A^1 1 A^ W^ A W A 0
p p p A W A W p
W W^ p^ p h h A g p p p p gA − =−^ ⎡^ − ⎤= − ⎯⎯⎯⎯^ ρ^ <ρ→ < ⎢ ⎥ ⎣ ⎦
⇒ the pond level falls
(b) Change in boat level
h h W A g p A p A p A
V
A
p p
A
p p A A p A p W p W b
A p
A W
p b
2 1
9 4 5
0 0
− = L^1 − 1 +^1 1 1
N
M
M
O
Q
P
P
F
HG^
I
KJ^
F
HG^
I
KJ
F
HG^
I
KJ
L
N
M
M
M
M
O
Q
P
P
P
P
−
b g b g b g
⇒ the boat rises
3.10 (a) ρ bulk^3
3
2.93 kg CaCO 0.70 L CaCO L CaCO L total
= = 2 05. kg / L
(b) Wbag = (^) bulkVg = ⋅
ρ 2 05.^ kg^ 50 L^ 9.807 m / s^1 N = 100. × 10 3
L 1 kg m / s
N
2 2 Neglected the weight of the bag itself and of the air in the filled bag.
(c) The limestone would fall short of filling three bags, because
- the powder would pack tighter than the original particles.
- you could never recover 100% of what you fed to the mill.
