Week 1; Collected 9/13

Chapter 1

#3. False. There is usually no competition between public utilities;
their fees are set by state utility boards to yeild costs plus "reasonable"
profits.

#9. False. The costs of clean up, or maintaining clean air and water,
are not part of the corporate price-setting process, thus not part of competition.

#22. TANSTAAFL - objections to taxes, highway, airport, gasosline,
etc., fees.

T o C - new highwa,
promising easy access and quick travel to all, soon becomes choked up;
new, inexpensive sewage disp[osal for new homebuilders, new commerce and
industry, soon becomes inadequate.

Chapter 2

#13. (1.3 x 10^{14} ) x ( 1.4 kW/m^{2}) = 1.8 x 10^{14}
kW

#14. 80/10 = 2^{3}; 3 x 20 = 60 yr. (d) 2040

#17. (a)

#18. 2^{3} = 8, 2^{4} = 16, (c) 4

#20. mass of earth = 6 x 10^{24} kg; 2^{m} = (6 x 10^{24}
kg)/ (2 x 10^{-2} kg) = 3 x 10^{26}= 3 x 10^{2}
x 10^{24};

10^{12} ~ 2^{40};
10^{24} ~ 2^{80}; 3 x 10^{2} = 300 ~ 16 x
16 = 2^{8}; 2^{m} = 2^{8} x 2^{80}
= 2^{88}; thus m ~ 88.

#31. (860 trillion kWh)/(19.7 trillion kWh) =43.7 ~ 2^{5};
thus 5 doubling times = 50 years.

Week 2; Collected 9/22

Chapter 3

#17. (d); A and B do the same work, but A does it in half the time.

#20. (b); work in = 20 N x 50 mm =1000 N-mm; work out = 100N x 8mm
= 800N-mm;

efficiency = work out/ work in = 800N-mm/1000N-mm = 8/10 = 80%.

#23. Conservation of energy says that the __total__ amount of energy
in a __closed system __ neither increases or decreases (Equivalently,
the __change__ in the __total__ amount of energy in an __open system__
is equal to the __difference__, between the energy __coming in__
and the energy __going out__, t__hrough the boundaries__ of the system.)
Thus the battery may lose its energy to the attached electrical circuit;
the circuit may lose its energy to resistance heating; some of the heat
energy may turn into light energy in an attached light bulb; etc.

#27. (a) AV. speed = 125km / 1.5hr = 83.33 km/hr. (b) AV velocity =
83.33 km/hr due North

(c) Time of travel = 125
km / 100 km/hr = 1.25 hr; total time elapsed = 1.5 hr; stop time = 1.5
hr - 1.25 hr

= 0.25 hr = 15 minutes.

#30. (a) a = (110 km/hr) / (6.60 s) = 16.67 (km / hr-s ) x (
1 hr / 3600 s ) = 0.00463 km/ s^{2} = 4.63 m/ s^{2 }.

(b) v = at =
16.67 (km/hr-s) x 3.30 s = 55 km /hr.

(c) AV speed
= (1/2)( initial speed + final speed) = 0.5 x ( 0 + 110 km/hr) = 55 km/hr

(d) distance = AV
speed x time = 55 km/hr x 6.60 s x (1 hr / 3600s ) = 0.1008 km = 100.8
meters.

#31. (a) F=ma=22kg x 3.50 m/s^{2 }= 77 N.

(b) F = ma = (22 kg + 55
kg) x 3.50 m/s^{2 }= 269.5 N.

(c) The seat has to accelerate
both man and pack.

#33. (a) P.E.(Johnie) = 750N x 3.00m = 2250 joules; (b) P.E. ( Janie)
= 500N x 3.00m = 1500joules;

(c) 2250joules + 1500joules
= 3750joules.

#36. (a) weight of car = mg = 2000kg x 10m/s^{2 }= 20,000N.
Work to lift it 23.0m =20,000N x 23.0m =

46 x10^{4}joules. (b) Power = work/time = 46 x10^{4}joules
/ 30s = 15333.3 watts = 15.33 kW.

(c) gain in KE = loss in
PE = mg (23.0m - 10.0m) = 20,000N x 13.0m = 260,000joules = (1/2) m v^{2};

v^{2} = (2/2000kg)(260,000joules) = 260 (m/s)^{2}; thus
v = 16.1m/s.

(d) KE(top of hill 2)+PE(top
of hill 2) = PE(top of hill 1); so (1/2)v^{2 }+g(height of hill
2)= g(height of hill 2)

v^{2 }= 2x10m/s^{2}( 20m - 10m) = 2x100(m/s)^{2};
so v = 14.1m/s.

(e) KE(bottom of hill 2)
+ PE(bottom of hill 2) = PE(top of hill 1) = mg x 23.0m;

KE(bottom of hill 2) = mg x (23.0m - 5.0m); thus v^{2}(bottom of
hill 2) = 2 x 10m/s^{2}x (23.0m - 5.0m);

v(bottom of hill 2) = 19.0 m/s

(f) KE(top of hill 3) + PE(top
of hill 3) = PE(top of hill 1); so (1/2)v^{2}(top of hill 3) =10m/s^{2}(23.0m-12.0m)

and v(top of hill 3) = 10.6m/s.

(g) KE(end) = PE (beginning);
(1/2)mv^{2}(end) = mgx23.0m; v^{2}(end) = 2x10m/s^{2}x23.0m;
v(end)= 21.4m/s.

#40. (a)According to #39c, 1 glazed doughnut = 250 kilocalories x(4186kcal/J)=
1.0465x10^{6}J; work required

to lift mass m through height h = weight xh = mgh = 70.0kg x 10m/s^{2 }
x3.00m = 2.1x10^{3}J. Thus,

# of doughnuts = 2.1x10^{3}J / 1.0465x10^{6}J = (2.1/1.0465)x10^{-3
}=
2.007x10^{-3} = 0.002. (You're not
going to lose weight this way!)

Week 3; Collected 9/27

Chapter 4:

#1. FALSE; charge and mass are physically independent entities.

#4. FALSE. Current is not interrupted as it goes through a complete
circuit so no electrons are lost. Energy is transferred from the
electric circuit to the stove: electrical energy (the ordered kinetic energy
of the electrons of the current) is lost while thermal energy ( the random
kinetic energy of the molecules making up the heating coil) is gained.

#11. P = IV; 100W = I x 120V; I = 100W / 120V = 0.833A; (a).

#17. P =IV = 15Ax120V = 1800W = total power allowed in circuit. Power
used = 1200W + 400W + 150W = 1750W. Trying to add 100W bulb brings
you to 1850W which exceeds maximum current which circuit breaker will allow
to pass, i.e., I = 1850W/120V = 15.4A > 15A. Hence circuit breaker stops
the current (e).

#19. W = QV = 2.00Cx200V = 400J. We are going DOWN in potential,
hence we get work OUT (e).

#30. The bar is at a higher potential than the sphere, hence positive
with respect to it. The 2.00kg object is attracted towards the bar,
hence it is negative. The 2.00kg object goes through a potential
difference of 240V - (-500V) = 740V. It gains a K.E. of (1/2)mv^{2}
and loses a P.E. of QDV,
where Q is the unknown charge and DV
is the potential difference. Energy loss must equal energy gain so
(1/2)mv^{2}
= QDV, from which
Q = (1/2)mv^{2 }/ DV

or Q = (2.00kg)(22.5m/s)^{2} / (2x740V)
= 0.68 Coulombs.

#34. (a) P =IV, thus I = p /V = 100W / 120V =
o.83A

(b)
Twice the resistance, hence half the current (Ohm's Law); thus I = 0.415A

(c)
P = I^{2}R; R same for each bulb, I is now halved for each bulb,
hence P is quartered. Each bulb gives off 0ne-quarter of the original
one bulb case. Both bulbs together give off half as much light as
the original one bulb.

Appendix 1:

#11. 1990 population = 3.888x10^{7},
2.605x10^{7} blacks, 7.776x10^{6}whites, 3.888x10^{6}
colored, 1.166x10^{6} Asians, 1.400x10^{7} English speaking
whites.

#13. person v = 10^{2}m / 10s = 10m/s;
cheetah v = 3x10^{2}m / 10s = 30m/s; jet v = 2.7km / 10s
= 2.67x10^{3}m/10s = 2.7x10^{2}m/s; space v = 20km/10s
= 2x10^{4}m/10s = 2x10^{3}m/s; light v = 3x10^{8}m/s;

person/light = 10 /3x10^{8}
= 3.3x10^{-9}; cheetah/light = 3x10/ 3x10^{8}=10^{-7};
jet/light =2.7x10^{2 }/ 3x10^{8 }= 9x10^{-7};

space/light =2x10^{3} / 3x10^{8}=
6.67x10^{-6}.

#14. F =qvB. Devide both sides by qB; result
is v = F / (qB).

#15. mgh_{1} + (1/2) mv_{1}^{2}
= mgh_{2}; thus (1/2) mv_{1}^{2}= mgh_{2}
- mgh_{1 }= mg(h_{2}-h_{1}); thus v_{1}^{2
}=
2g(h_{2}-h_{1});

finally,
v_{1}^{ }= sq.rt .[2g(h_{2}-h_{1})].

Week 4; Collected 10/4

Chapter 4

#28 (a) PE(A) = 1.00Cx300V = 300joules; PE(B)
= 1.00Cx500V = 500J;

(b) PE(A)
= -5.00Cx300V = - 1500J; PE(B) = -5.00Cx500V = - 2500J;

(c)
In going from B to A, the 1.00C charge loses PE, hence gains KE = 200J

(d)
In going from B to A, the -5.00C charge gains PE, hence either loses KE
= 1000J or has

1000J work done upon it.

#35 (a) Since the current is stepped down, the
voltage is stepped up; so we call it a “step up”.

(b) Power
is current x voltage; author doesn’t give voltage; hence can’t answer question.

#36 (a) Power is current x voltage so current
is power/voltage = 1200W/120V = 10A =

10 Coulombs per second; hence, in 30 seconds, 300C have flowed through
the heater.

(b)
Energy is power x time = 1200Wx30s = 1200J/s x 30s = 36000J.

(c)
Ohm’s Law says V = IR, so R = V/I = 120V/10A = 12 ohms.

Chapter 5

#16 Withdrawal of 4 time the original amount
means it has doubled twice in 10 years so the

doubling time is 5 years; from Table5.3, this corresponds to a growth rate
of 15%. (b)

#17 A growth rate of 3% means a doubling
time of 23.4 years (c).

#19 Going from 1000 to 16000 requires 4
doubling times. Each doubling time (or halving time) is

25 years, so the total required time is 100 years which means the year
2095 (d).

#30 From 1970 to 1990 the population increased
from 203.2m to 248.7m or 22.4%. Thus

electrical
use should increase over these years by the same percentage, 22.4% of 1.37TkWh

which is
0.31 TkWh. So 1990 energy should be 1.68 TkWh.

Chapter 6

#9 True. It uses the cheapest fuel and
uses it most efficiently since the plant runs steadily.

#20 There are 24x365 =8760 hours per year. Of
the option s given, (b), coal or oil fired steam is

cheapest.
However, the graph indicates that hydroelectric would be still cheaper.

#21 (a) gas turbine.

#34 (a) long-distance transport cost =
125km x $0.0002/kWh/km = $0.025/kWh.

(b)
local transport cost = 12km x $0.0016/kWh/km = $0.0192/kWh.

(c)
total cost of delivering electrical power to me =$0.04/kWh + $0.025/kWh
+$0.0192/kWh

= $0.0842/kWh which means the company would lose $0.0042/kWh if it sold
at

$0.08kWh.

#38 Gas turbine has become much less efficient
as we went from 1972 to 1991, i.e., its annual

cost
doubled from $25 in 3000 hours in 1972, in 500 hours in 1991. Efficiency
of fossile-

fuel
steam increased during this time; in 1972 it took about 7000 hours to double
its costs

whereas
in 1991 its cost had not yet doubled in 9000 hours. Nuclear steam became
somewhat

less efficient
during this period, taking about 2000 hours to increase its costs by 10%
in 1991

4500 hours
for the same increase in 1972. In 1972, gas was most economical for less
than

1700 hours;
in 1991 it was cheapest for less than1500 hours. In 1972 nuclear
steam was best

for
base load whereas in 1991 fossil fuel was most economical if you exclude
hydroelectric.

Week 5; Collected 10/11

Chapter 7

#7 False. A typical atom is about 10^{-10}
m in size which is 100,000 larger than nuclear size

#8 True. Just look at Table 7.1

#11 (d); the same amount of energy is involve,
only the rate is different.

#12 (c); 50 - 20 + 10 = 40

#17 (e); its energy increases from -3.4eV
to -1.5eV so it had to ABSORB 1.9eV, which is not one of the choices

#30 Power = current x voltage; thus current
= power/voltage = 4.8MW/120V =

4.8x106W/1.2x102V = 4x104A = 40,000A.

Chapter 8

#4 Second Law says that disorder of entire
universe never decreases, says nothing about decrease
of disorder (increase of order) in some small region of the universe. Increrase
of order (e.g., life) in some region is paid for by energy dissipation,
decrease in order elsewhere.

#27 Air obviously has lot of moving molecules,
hence lots of energy. But Second Law states that

you can't take thermal energy from a substance at some tempreature, to
get work, without returning some of that energy at a lower temperature.
You could use non-thermal energy

to drive your car, i.e., you could have
a wind-driven car.

#32 When they are both at the same temperature,
the average kinetic energies of their molecules are the same. The
ratio of their mean-squared molecular speeds is inverse to the ratio

of their molecular masses, i.e., 32/4.

#35 The ratios of the mean-square
speeds are __inverse__ to the mass ratios: M(H2)/ M(N2)/ M(O2)/M(A)
= (seeFig.7.3) 2/28/32/40. Hence.rms speeds are square roots
of **inverse** of previous numbers: rems(H_{2})/rms(A)
= sq.rt.(40/2) = 4.47; rems(H_{2})/rms(O_{2}) = sq.rt.(32/2)
= 4; rems(H_{2})/rms(N_{2}) = sq.rt.(28/2) = 3.74.

#37 (a) By First Law, Work = 2500J - 200J
= 2300J.

(b)
Efficiency = (work out)/(energy in) = 2300J/2500J = 0.92 = 92%.

(c)
From p.136, Carnot Efficiency = E = (T_{hot} -T_{cold})/
T_{hot }; thus ET_{hot} = T_{hot} -T_{cold}
;

so that (E - 1) T_{hot} =
-T_{cold} ; and T_{hot} = T_{cold} /(1 - E) = (273+100)K/(
1 - 0.92) =373/0.08

= 4662.5K = 4389.5^{o}C.
(Remember, all temperatures must first be converted to K.)

#38 (a) E = (T_{hot} -T_{cold})/
T_{hot} = ([800+273] -[20 +273])/[800+273] =780/1073 = 0.727 =
72.7%.

(b) The river water receives 100% - 72.7% = 27.3% of the energy in the
coal = (Table 7.1)

0.273 x 26.0 MJ/kg x 1000kg/tonne = 7,100
MJ/tonne of average coal.

(c) In this case, the river water receives 100% - 30% = 70% of the
energy in the coal =

0.70 x 26.0 MJ/kg x 1000kg/tonne = 18,200
MJ/tonne of average coal. (Big difference!)

#43 (a) Need 5.00x10.0 kilocalories=50.00Calories
= (42.c; glazed doughnut = 250Calories)

50.00/250 = one-fifth of a glazed doughnut.

(b)
From 42.b, need 50.00/330 = 0.15 of a jelly doughnut.

#44 (a) 1000MW = 10^{9}joule/second
= 3.6x10^{12}joule/hour =

3.6x10^{12}joule/hour x (1Calorie/4186joules)
= 8.60x10^{8} Calories/hr =

8.60x10^{8} Calories x (1 glazed
doughnut/250Calories)/hr = 3.44 x 10^{6} glazed doughnuts/hr

(b) You need three times as much thermal energy, on average, for each amount
of electrical energy produced. Hence, you will need 3 x ( 3.44
x 10^{6} glazed doughnuts/hr)

= 1.03x10^{7} glazed doughnuts/hour.

(c) Each jelly doughnut gives 330/250 = 1.32 times as much energy as each
glazed doughnut

Thus you will only need 1.03x10^{7}
/1.32 =7.80x10^{6} jelly doughnuts/hour.

Week 5; Collected 10/18

Chapter 9

#4. True; coal contains many trace heavy
metals, e.g., mercury. They appear in the smoke and ash of a coal-fired
electrical generating plant, and can accumulate in the vicinity of such
a plaant.

#8. True; efficiencies and economies of
scale would lead to corrections of design and manufacturing error and the
lessening of costs.

#17. According to pp.155-6 of text, c,d,
and e are correct answers.

#21. (e)

#27. Since a nuclear plant doesn't work
at as high a temperature as a conventional fossil fuel plant, its Carnot
efficiency is less and hence the amount of waste heat that it puts out,
for a fixed amount of useful electrical energy produced, is greater than
that of the conventional plant. Both type of plants may be cooled by air
or water, both having detrimental effects on the environment, the more
the waste heat, the bigger the detriment.

Chapter 10

#6. False; Wegener was originally viewed
as a "nut".

#7. True; from fig. 10.2, in 1950 we imported
somewhat over half of our aluminum, but only small fractions of our copper,
zinc, and lead reqirements. From fig. 10.1, we only imported40% of
our iron requirements (our major metal requirement) in 1982, presumeably
much less a half-century ago.

#16. (d)

#25. For a 1% growth rate, the doubling
time is 70 years. The "box size" is 70years x 1.1475(kt/yr)
= 8.03x10^{8} tons; dividing this into the total available, 21,800
(Mt), we get 27.1 boxes available. In the first doubling time, we
use one box; in the second doubling time we have used 1 + 2 = 3;
in the third we have used 3 + 2x2 = 7; in the fourth we have used
7 + 4x4 = 23; in the fifth we would have used 23 + 16x16 =5888 which is
far more than the number available. Thus, actually the supply
of phosphorus would only last a little bit more than four doubling
times; at a one percent growth rate, this is a little over 280 years,
considerably less than the linearly calculated 2.1800x10^{10}tons/1.1475x10^{7}tons/yr
=1.9x10^{3} = 1900 years.

Week 5; Collected 10/25

Chapter 11

#1. Assume the most (or best)! From p.192,
total world supply of petroleum could be as much as 25x10^{11}
barrels while p.181 suggests that 1990 world consumption might be 25x10^{9}b.
Dividing one by the other we get ~100 years of petroleum at current (non-exponential!)
rates. Doing the same for coal, p.199 says that annual production is ~4.8x10^{9}
tonnes whereas the total resources are 10x10^{13} tonnes, which
means that, at current rates of use, coal would last for 20,000 years.

#11.(d)

#24. No, Hubbert makes major assumptions about
how people will exploit and use the resources. Hence he is making
sociological, economic, and psychological assumptions as well as physical
ones.

#33. From p.197, each barrel costs 12 (19784$)
x (1 1992$0 / (.37 1974$) = 32 (1992$) >> $18 spot price.

#34. From p.196, each barrel of shale oil ( 30
gallons) requires about 2 tonnes of rock so that 6x10^{9}
barrels requires 12 Gtonnes.

Chapter 12

#1. False; its manufacture and junking produce
pollution, as do its tires and brakes; its fueling and lubrication, and
the various vents in the engine.

#12. According to Table 12.7, the intercity bus
(c) is the most efficient choice, at 622kJ/passenger mile.

#13. (b)

#16. The 28% full car represents 2300(Btu/pass.mi)/.28
= 8214 (Btu/pass.mi); similarly the 20% full mass transit requires 760(Btu/pass.mi)/.20
= 3800(Btu/pass.mi). Thus for the entire city, we get:

0.03 x 3800(Btu/pass.mi) + 0.97 x 8214(Btu/pass.mi)
= 8082(Btu/pass.mi).

#23."Not invented here", difficult and expensive
to manufacture, no competitive need to do so.

#27. Page 214 indicates average car efficiency
of 15mi/g in 1978; there are ~4 L per gallon; p.223 says that CA cars emitted
9.0 gCO/mi thus we get

9.0gCO/mi x15mi/g x 1g/4L = 33.75 gCO/L,
which is 3375 times as much CO per liter as the given 0.1g lead/L.
On p.221 it says that southern CA emitted 100.9 million tonnes of CO per
year; dividing this by 3375, we get 300,000 tonnes of lead emitted per
year in that region.

#31.From p.115 we see that the energy content
of gasoline is 46.9MJ/kg In the back of the text, "Selected Units
and their Meanings", we see that one barrel of oil (bbl) contains 6.12x10^{9}
J of energy. At 4.3x10^{6} J/km driven, one bbl would supply
enough energy for vehicles to drive

6.12x10^{9} J/4.3x10^{6}
J/km =1423km. Thus per day, there would be 7359000bbl x 1423km/bbl = 1.05x1010
km driven or 1.05x10^{10} km /137.3million cars = 76 km per vehicle
per day.

Week 8; Collected 11/1

Chapter 13

#10. True; they have helped spread sulfur
oxides very far from their industrial/power sources.

#30. (a) Mass = density x volume = 0.1785kg/m3
x 2.20m3 = 0.3927kg; weight = mass x g =

0.3927kg x 9.8m/s2 = 3.848 newtons.

(b) Buoyant force = weight of air - weight of helium; weight of air
= 2.20x1.29x9.8 = 27.81newtons; thus buoyant force = 27.81 N - 3.85 N =
23.96 newtons.

#33. All answers based upon Table 13.4

(a) 40,000 micrograms per cubic meter.

(b) 19,300/2.1 = 9190 micrograms per cubic
meter.

(c) 1430/28 = 51 micrograms per cubic meter.

(d) 514/78 = 6.6 micrograms per cubic meter.

(e) 375/107 = 3.5 micrograms per cubic
meter.

#38. (a) From page 115, burning average
coal releases 26MJ/kg; in the back of the book we learn that 1Btu=1055J
and 1kg=2.2lb. Thus we get 1.12 x 10^{3} BTU/lb. of av. coal.
Burning 10^{3} tones of coal per day gives us 2.2 x 10^{9}BTU/day.
On page 206 we see that burning this coal releases 0.03 lb. of particulates
per MBTU.

Thus we have(2.2 x 10^{9} BTU/day)
x (0.03lb. part./ 10^{6} BTU) = 67lbs particulates/day. This
is equivalent to an "exposure" of (67/2000) x 100 micrograms per cubic
meter, which is 3.36 micrograms per cubic meter.

(b) From page 262, top right: #additional
lung cancers=(0.02 - 0.60)x10^{-4} x 3.36 = (0.067
- 2.02)x10^{-4} per day.

(c) 55% x 3.36 micrograms per cubic
meter = 1.85 micrograms per cubic meter of PM_{10}

from page 255, right middle, we see that
the death rate would increase by 16% x(1.85/100) =0. 30% per day
= 108% per year!

Chapter 14

#3. True; ice reflects more radiant energy
than does bare earth.

#17. (d) Two to the fourth power
is sixteen.

#19. (e)

#25. As warm surface water flows away from
the equator (north or south), it cools and sinks. Since it has no
other place to go, as it is pushed by the following waters, it then flows
back towards the equator, where it rises, again due to the push of the
following waters, to be heated again and continue the cycle.

#27. Dissolved in seawaters, becoming parts
of sea vegetation, sea shells, and corals, eventually sinking to
sea bottom, eventually becoming part of rocky bottom. Becoming part
of land vegetation, covered by earth, compressed and eventually becoming
coal or similar materials.

Chapter 15

#1. False; just the opposite. From
page 301, "the CO2 content of the atmosphere has risen only about half
as much as it would have if all CO2 from fossil-fuel production had been
released to the air."

#11. (a); see page 327.

#19. (e)

#21. More carbon in the atmosphere means
atmospheric warming. More coal soot on ground means lower albedo
(lower reflectivity) means rising temperatures. Also more biological and
health effects such as acid rain, radioactivity, lung particulates, etc.

#22. Not a good idea to increase dependence
upon a diminishing resource on the small chance that new and unusual supplies
of it may be found.

#33. (a) According to Table 15.2, there
are about 2.2 gigatons more carbon emitted than absorbed annually; hence
we need enough trees to take up 2.2gt/year. On page 343 we learn
that one hectare of tropical trees takes up 10 tons of carbon per year.
Hence we need an additional

2.2gt/10t = 220 million hectares of tropical
trees per year.

(b) Temperate trees take up 7.5 tons per hectare per year, so we would
need 2.2 x10^{9} / 7.5

= 2.93 x 10^{8} hectares of new
temperate trees per year.

Week 9; Collected 11/8

**NOTE TO STUDENTS: I am writing to you this way because I don’t have
all of your e-mail addresses and I want this note to reach you all.**
**I have been very disappointed in the large amount of absenteeism
in this course and the lack of effective class participation. I presume
that this apparent lack of interest on your part is an indicator of your
disappointment with the class/course. Rather than just having the
class drift to an unsatisfactory conclusion, I hope that , with your help,
it can be changed so as to reach out effectively to more of you.
Please see me individually, in my office or after class (on Wed. or Fri.)
so that I can get some insight into the source of your apparent dissatisfaction
and so that we can generate ideas to improve things for the rest of the
semester. Thanks - A.M.S.**

Chapter 16

#2. False There are unstable isotopes which have more protons than
neutrons. An, of course, the dominent stable isotope of hydrogen
has one proton and no neutrons!

#4. False. The strong force has a very short range compared to electromagnetic
forces. Out side of this short distance, you can ignore the strong
nuclear forces in comparison with the electromagnetic forces.

#22.In order to fuse the light nuclei, you must get them close enough
to each other for the strong attractive nuclear force to take over and
bond then together (fusion). But the light nuclei are all positively
charged and so repel each other by the long-ranged repulsive electric force.
So you must force them together against this repulsion which means you
must have them moving towards each other with very high speeds, i.e., they
must be at a very high temperature - millions of K. How do you keep
them together (high density) while moving at high relative speeds (high
temperature? Any “ordinary” walls would vaporize or slow the nuclei
down.

#25.(a) From page 360, mass of proton + mass of neutron = 1.00727640u
+ 1.008664904u =

2.0159289u. Subtracting the deuteron mass, 2.014u, from this,
we get 0.0019u ~0.002u.

(b) We can find the value of u in energy units by noting that
the neutron mass is given, in the same section, as 939.6 MeV/c2 = 1.008664904u,
so that

u= 939.6 MeV/c^{2} /1.008664904 = 931.5 MeV/c^{2}.
Therefore, the deuteron binding energy is 0.002 u x c2 = 0.002 x 931.5
MeV = 1.86 MeV.

#29. From Fig.16.2, we see that the binding energy of iron-56
is about 8.8 MeV per nucleon. Multiplying this by the number of nucleons
= 56, we get 492.8 MeV.

#32.(a) 235 + 1 - (154+81) = 236 -235 = 1. So there is one nucleon
left over; is it a proton or a neutron? Add the proton numbers, Z, to find
out: 92 + 0 - (61 + 31) = 0; there are no protons left over so it is a
neutron which is produced in this fission.

(b) Following the argument on page 362, the uranium binding energy
is 1794 MeV. Estimating from Fig. 16.2, the binding energies of promethium-154
and gallium-81 are 8.2 and 8.7 MeV per nucleon, so that the final energy
is 154 x 8.2 + 81 x 8.7 = 1967.5 MeV. the energy released is the
difference 1967.5 - 1794 = 173.5 MeV.

Chapter 17

#4. True since the PWR works at a somewhat higher temperature than
does the BWR.

#10. False, hydrogen bombs have, unfortunately, released a great deal
of energy on earth.

#19. a, c,d, and e.

#24. WE came out 2% ahead for civilian energy; in addition, there was
all of the enrichment done for weapons - which was a societal choice.
Hence, we did come out ahead, i.e., it was not a bad bargain.

#26. 5.6 micro-curies = 5.6 x 10^{-6} x 3.7 x 10^{10}
disintegrations per second (back page of text) = 2.1 x 10^{5} disintegrations
per second.

#30.Page 414 indicates that Chernobyl released about 1.85 EBq =1.85
x 10^{18} Bq which is about twice as much as the 925PBq =925 x
10^{15} Bq = 0.925 x 10^{18} Bq released by the nuclear
weapon tests.

Week 10; Collected 11/15

Chapter 18

#1. Incorrect. The lowest probability mentioned
is roughly one in ten-thousand per reactor per year. But there are
many reactors in the world and we have to live many years. For example,
assuming roughly 300 reactors, the probability per year of an accident
in some one of them is

1 - ( 1 - 10-4 )300 = 0.03 ( already up
to 3% !). Over 10 years, the probability of an accident becomes 1 - ( 1
- 0.03)10 = 26%; for 20 years, you get a probability of 46% for an accident,
etc. thus, you wouldn’t want to bet against there being an accident
in one of the reactors during your lifetime.

#31. By definition (page 410) 7.00Gy =7J/kg
absorbed energy, so she will absorb 55kg x 7J/kg =385J.

#33. (a) From text, p.406-10, 3.3 x 103
Bq = 3.3 x 103 decays per second; 1 MeV = 1.6 x10-19 x 106 J.
Thus, the 6.00kg raccoon receives an energy of 3.3 x 103 x6.05 MeV/second
x

1.6 x 10-13 J/MeV = 31.9 x 10-10 J/s. Multiply
this by the number of seconds in a day to get its exposure: 24x60x60 s/day
x 31.9 x 10-10 J/s = 2.8 x10-4 J. Dividing this by its mass, we get
its exposure: 4.2milligray.

(b) Multiplying exposure by quality
factor, we get a dose of 20 x 4.2 x10-3 = 0.084Sv=8.4rem.

Chapter 19

#3. Its true that climate changes could
result - there is no free lunch - but we will run out of fossil fuels long
before the sun runs out of energy and the use of fossil fuels has already
produced climate changes and will likely continue to do so.

#9. True since the earth’s surface temperature
is considerably less than that of the sun and the mean wavelength of the
electromagnetic waves emitted by any body is roughly proportional to it
absolute temperature.

#16. d

#21. d

#24. Just like the outer wall of a house,
in contact with the cold external air, is colder than the interior of the
house, warmed by the furnace, so the stellar exterior, in contact with
the cold of “outer space” has a lower temperature than the stellar interior,
heated by its “nuclear furnace”. The energy gets from interior to
exterior via conduction (minor), convection, and radiation.

#25. Wavelength x frequency = wave speed
so frequency = 3500km/s / 2.5km = 1400/s =1.4kHz. Period is one over
frequency so period = 1/ 1400/s =6.9 ms.

#29. From p.443, power radiated = area
x S-B constant x fourth power of absolute temperature = (1.4x10-3 m2
)(5.67x10-8 W/m2K4) x (700K)4 = 19.1W.

#31. From page 446, wind power density
= P/A = (1/2) (1.29kg/m2) (6m/s)3 = 139J/s=139W.

Week 11; Collected 11/22

Chapter 20

#1. False -see page 478; just the Gulf
Stream alone could produce many times the US energy requirements.

#20. d; see page 477.

#21. Certainly! Mexico is very unhappy
with the US damming up the Colorado River and removing most of its water
for irrigation before it reaches Mexico. Similarly, Turkey is planning
big dams on the Tigris and Euphrates Rivers which will cause severe hardships
to Iraq and Iran.

#25. Sea waves are very unpredictable and
undependable. Small waves can give very little energy; very big waves
can carry a great deal of energy, so much that they may wreak the devices
intending to harness them. It becomes very expensive to construct
devices sturdy enough to withstand the large destructive waves yet still
efficiently tap the small ones.

#28. From page 477, stored energy = Mgh
= (density x area x height) gh = 1030kg/m^{3} x [15.5km^{2 }x
(10^{3} m/km)^{2}] x4.50 m x 9.8 m/s^{2} x 4.50
m = 3.17 x 1012 Joules.

Chapter 21

#4. Correct; they are much more likely
to use cheap human labor to grow and process the corn or sugar cane rather
than the fossil fuels used in the US. It doesn't make much sense
to use agricultural fuels to replace fossil fuels if you need fossil fuels
in order to obtain the agricultural fuels.

#25. 7.1Gbbl of oil required annually.
From page 498, one acre of these trees could supply 25 bbl of oil per year
so you would need 7.1 x10^{9} / 25 = 2.8 x 108 acres. Currently,
the US uses about 1.6 x10^{8} acres for its entire production
of cereals for animals and humans. So we might have oil for transportation,
etc., but we'd be very, very hungry!

Chapter 22

#3. False - there are lands not suitable
for growing crops for human food (p.511) but which can be used for animal
grazing, e.g., the only human food we could get from such lands is meat.

#10. True. Large scale agriculture,
which is most appropriate for large scale applications of fertilizer (p.515),
is most likely to bring about massive erosion. Small scale farming
has been able to farm the same land for thousands of years without significant
erosion (e.g., Europe).

#20. d; see p.516

#21. (a). Farm employment would decrease
significantly.

(b) Oil imports would sky rocket as would
deficit in the balance of payments.

(c) More demand on petroleum, more erosion,
enhanced shift from small to large farmers, more rural unemployment.

(d) They'd pay for food now; the starving
man doesn't worry about nest year!

#27. As long as US petroleum stocks last or,
if we are economically and militarily strong enough to guarantee petroleum
imports from the rest of the world, as long as world stocks hold out. Turning
to Fig 11.10, maximum production would be about the year 2040 with a total
stock of about 170Gbll. Assuming steady use at the rate given in
#21.25, 7.1Gbbl/year in the US (ignoring growth and previous use), this
means we could continue after that for only 170/7.1 = 24 years - not very
long!!!

Week 13; Collected 12/6

Chapter 23

#2. False. There is air pollution from
the gases contained in the hot subteraneum fluids and steam; it is very
difficult to dispose of the fluids, after the energy has been extracted
from them, without polluting water supplies and the land, due to thelarge
amounts of salts which they contain.

#12. Our population of 250 million people would
require an average power of 250 x 10^{9} Watts. Acoording
to p.535, this would require 250 x 10^{9} Watts/ 240MW per
nuclear weapon = 1042 weapons per day. Thus 20,000 weapons would
last about 20 days:b.

#15. b

#23.Could us periodic pumping of low-pressure
gas(hence cheap pipes) rather than high voltage transmission cables, which
are very expensive, to minimize resisitive losses;also lose about 2/3
energy in converting thermal energy into electrical energy . Qe concerned
aqbout: Have to count energy costs of producing the hydrogen; have to lay
down entire new, hydrogen-leak-free transmission system; have to either
put fuel-cell into each electrical user's establishment to convert hydrogen
energy back into usual electrical form, or produce entirely new set of
devices to get the hydrogen to do the jobs which we are accustomed to getting
from electricity.

#26. (a) According to Table 23.2, it would cost
11.5 x 0.01 = 0.115 cents per kWhr to pump the methane to the power plant
and (17.2/100) x 0.85 = 0.146 cents to deliver the same amount of energy
in electrical form from the power plant to your home. Hence, the gas delivery
is cheaper.

(b) the total cost is 0.261 cents
per kWhr.

(c) This is 3.27% of the total cost.

Chapter 24

1. False (p.555).

#11. The ovens would be used for 700hrs/yr
x 10 yrs = 7000 hours. The difference bertween the power requirements
of the two ovens is 1000 W - 450 W = 550 W. Thus the efficient oven
saves 7000hrs x 550 W = 3850 kWhrs x 6 cents/kWhr = $231 (b) in energy
costs which could be used to purchase the better oven (capital costs).

#23.. It saves energy at least two ways: 1) the
water doesn't cool down on the long way through cold pipes from the
basement hot-water tank to the point of use; the water doesn't lose
energy via conduction through the walls of the hot-water tank while
it sits, hot in the tank, waiting to be called upon.

#25. Each of you has his/her own surprise here.
I'm eager to read them

#29. Assume an energy cost of $0.08 per kWhr.
The total cost of the GE bulb is 26 W x 10 kilohours x $0.08 per
kWhr plus $15 = $35.80. The total cost of the ordinary light bulb
is 90 W x 0.750 kilohours x $0.08 per kWhr + $1 = $6.4 for 750 hours;
but you will need 10,000/750 = 13.3 bulbs, which costs $6.4 x 13.3
= $85.33, to get the same total hours of light! If you use the long
life incandescent bulbs, you will need 10,000/1500 = 6.67 bulbs which will
cost you 6.67 x ( 90 W x 1.500 kilohours x $0.08 per kWhr + $1.49)
= $81.97 for the same number of light hours.

#31. The savings in operating costs is 19yrs
x (979kWhrs/yr - 515kWhrs/yr) x $0.08/kWhr =$705.28 which is much
more than the $838 - $635 = $203 difference in capital cost.

Week 14; Collected 12/13

Chapter 25

#4. True; recycling, reuse, and reproduction jobs as well as
collection, separation, and transport

#16. e; all of them

#17. b.

#23. Recycling requires separation: plastics from glass, etc;
different plastics from each other. All of this easier done in the
small quantities where they are generated (low entropy) than in the big
collection plants (high entropy). It required energy to do the necessary
reduction in entropy characteristic of recycling.

#25. If we could send all bottles back to the same cleaning plant,
where they could then be refilled with almost anything, less energy would
be required than if wqe had to separate the bottles into differnt types,
each type destined for a different cleaning and refilling plant.
Separation implies decrease in entropy and decrease in entropy requires
expenditure of energy.

#26. Paper: (65.3-27.2)/((1/2) (65.3 + 27.20)) = 27.2/46.25 =
58.8% in 28 years 2% growth per year.

Ferrous metals: (10.5 - 9.0)/(10.5 + 9.0)/2/28 = 0.14% per year.

Total: (163.3 - 80.0)/ (163.3 + 80.0)/2/28 = 0.6% per year on average.

#28. From page 610, it costs 0.55/0.42 = 1.3 times as much
energy per use for aluminum as for steel beverage containers, which makes
the stell that much cheaper. But, the aluminum cans are lighter,
hence easier to carry, and are better heat conductors, making them easier
and quicker to chill.

#29. They made $29.40 on the aluminum [200lbs x (29c - 14.3c)], $4.22
0n the HDPE, and $2.25 on the PET; they lost $57.60 on the steel,
$94.60 on the clear glass, $115.50 on the green glass, $122.40 on the amber
glass. The total is a loss of $354.23.

Chapter 26

#4. Climate change due to both natural as well as human
causes. Humans not responsible for first, but certainly the latter
has had major impacts: over grazing, “over farming”, “over deforestation”,
“over mining”, “over industrialization”, “over transportation”, “under
cleaning up” after themselves.

#8. Ignorance? Laziness? Cupidity? Immorality? I’m looking
forward to seeing your opinions on this.

#16. As well as coming across our borders from other countries, toxic
chemicals and metals are also buried in our land, in the mud under our
waters, and dissolved in our waters, from which they gradually are
returned to the human environment.

**END OF SEMESTER**
**JOYFUL WINTER SOLSTICE**
**HAPPY NEW YEAR**
**PLEASANT NEW CENTURY**
**HAVE A FRUITFUL AND ENJOYABLE NEW MILLENIUM**