Phys 239
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Quantitative Physics
Problem Set 2 Solutions
Getting Real with Electric Cars
Electric cars typically travel about 6 km for each kWh of energy held in the car’s battery. Compute how much
PV panel area would be needed to deliver enough daily energy to sustain a typical worker’s commute. How
much would the system cost if solar installations run about $4 per installed peak Watt of power capacity (a
full-sun figure, not average)? If the car is away from home all day, then there must be local battery storage as
well, which can knock total efficiency down by a factor of two (so need more panels). Figure out how much
the system would cost now, including the price of a large enough battery (or battery set) to fully charge the
car for the commute, using online prices for lead-acid batteries (note: a 100 Amp-hour battery at 12 volts
provides 1200 watt-hours of energy, etc.). Oh—and the charging efficiency of the car tends to be 80%.
Let’s imagine that a typical commute is 15 km, for a daily round trip of 30 km. Note that we’re not going
to extremes here on estimating a commute distance. This requires 5 kWh of charge in the battery, which
requires delivery of about 6 kWh from the source, given imperfect charge efficiency.
A photovoltaic panel in full sun converts 15% of the ∼ 1000 W/m2 solar flux into electricity, so that each
square meter of panel contributes 150 W in full sun. But the panel is not always in full sun, 24 hours per
day. How many hours of full-sun equivalent might we get, on average? Well, it’s certainly no more than 12
(night), and we take another factor of two for geometry (projected sun angle) for fixed panels, and a bit more
for weather to arrive at 5 hours per day. This is typical for American locations: 5 full-sun-equivalent hours
per day, or you might say 5 kWh incident on each square meter per day (15% of which the panel converts to
electricity). So in the end, we get 0.75 kWh/day/m2 . Thus we need 8 m2 of PV panel.
We saw that each square meter of panel in full sun will deliver 150 W, so the array here would make 1200 W
in full sun (peak power). At the rate of $4 per peak watt, the system will cost about $5000.
But if the car is away during daylight hours, we must store the energy and re-release at night. By the time
we figure round-trip energy efficiency in batteries and other losses (like having to overbuild the system a
bit to properly overcharge batteries for long-life conditioning), we need a factor of two more panel: 2400 W
peak costing $10,000. For batteries, we probably want at least twice the capacity of the car so that we are
not deep-discharging all the time and cutting down battery life dramatically.
Trojan T-105 batteries are something of a mainstay in renewable systems: 6 V and 200 Ah, so 1.2 kWh of
storage apiece. We aim for a 12 kWh system, which will require ten cells in some configuration (depends on
what voltage your inverter requires). At a cost of $150 each, we add $1500 to the cost, plus a bit more for
charge control, etc.—so maybe $2500.
We therefore add $12,500 to the cost of the vehicle. And likely we find the batteries give out after 2–3 years
(my experience), so that becomes a recurring charge (and hassle).
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Cold Feet
It’s winter and your feet are cold. You can think of several options for heating them up.
a. Fill a sink with hot water to stick your feet in;
b. Put your feet/legs under a blanket with a 1500 W space heater (fire hazard!);
c. Put your feet on a heating pad (50 W), possibly also wrapped;
d. Bundle up (hat, coat, warm socks, down bootie slippers) and let your metabolism do the work.
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As a physicist, you want to compute the amount of energy each method would take. In some cases you need
to estimate the appropriate timescale as well. Compare each to the amount of actual energy it takes to bring
your feet back up to body temperature.
Part a: A sink will probably require about 10 L of water to cover your feet. At a heat capacity of 4200 J/kg/K,
and heating the water from an ambient 5 ◦ C (winter) to 45◦ C, we require 4200 × 10 × 40, or 1.68 MJ. More
straightforwardly, this is 400 kcal (heat capacity of water is 1 kcal/kg/K).
Part b: I would estimate that it will take about 6 minutes for this to do the job, in which case we run 1.5 kW
for one-tenth of an hour, accumulating 0.15 kWh, or 0.54 MJ (130 kcal).
Part c: A 50 W heating pad wrapped under/around your feet with a blanket might take half-an-hour to get
your feet warm. So we have 0.05 kW for 0.5 h, resulting in 0.025 kWh, or 21 kcal (90 kJ).
Part d: For this, I really think I have to evaluate how much actual energy is needed. Let’s say your feet are
each a liter of volume, and have cooled to 20 ◦ C, preferring to be at 37 ◦ C. We’ll assume the heat capacity
is roughly half that of water (mix of tissue/bone/water), so that we need 0.5 kcal/kg/K, resulting in 17 kcal
or 70 kJ. We can assume that a bundled person would accomplish this heat transfer efficiently. If you want
to start with metabolism, and guess that some fraction will go to heating feet (10% of 100 W is 10 W), and
guess it may take two hours, then we get 0.02 kWh, or 72 kJ, and land in the same ballpark. I swear I did
not cook things to line up so well!
For the numbers we chose, the water in the sink is clearly the biggest expenditure, overshooting the actual
need by a factor of 20 or so. Then when you’re done, all that thermal energy just goes down the drain. The
space heater is an order-of-magnitude more than is needed: most of the heat does not couple efficiently into
what’s cold. The heating pad is surprisingly close to the actual need: the heat efficiently transfers to the
place you want it.
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Electromagnetism vs. Gravity
I sometimes explain how weak gravity is by pointing out that a magnet you hold in your hand can defeat the
pull of the entire Earth in picking up a paperclip. Or a balloon you rub on your head will stick to the wall,
also thwarting the entire Earth’s pull. Let’s play around this theme.
a. Compute the ratio of electrostatic to gravitational force between an electron and proton.
b. If we took a section of Earth’s crust, removed an electron from each atom, and placed these electrons on
the Moon, how many tons of rock would have to be ionized in this way in order to double the attraction
between Earth and Moon? What is the physical dimension of this volume of rock, expressed as the side
length of a cube? Why is the answer different from just dividing the 6 × 1024 kg mass of the Earth by
the number from part (a) of this problem? Reflect (to yourself ) on what this says about charge balance
of the planets.
c. Based on the example and demonstration in class of the pull of a neodymium magnet, how much more
powerful would you say (practical) magnetism is compared to gravity (here the length scales differ;
pretend for this exercise that magnetism is 1/r2 )?
Part a: The forces are:
GM m
e2
and
,
4πε0 r2
r2
so that the ratio is
e2
2.6 × 10−38
≈
≈ 2 × 1039 .
−12
4πε0 GM m
12.5 × 9 × 10
× 6.7 × 10−11 × 1.7 × 10−27 × 9 × 10−31
This is rather a large number, worth remembering (just 1040 will do). The ratio comes out differently by
three orders-of-magnitude in either direction for two protons or two electrons, so this one is in the middle of
the range.
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Part b: So we want to find the charge that would balance forces, which we get when Q2 = 4πε0 GM m, where
M is the mass of Earth, and m is the mass of the Moon. We note that whatever charge, Q, is removed from
the Earth is placed on the Moon, so the same charge (magnitude) is on both. This computes to a charge of
6 × 1013 C (rather large), requiring 3.6 × 1032 electron charges to amass1 . But Avogadro’s number is large,
too. If we take the average molecular weight of rock to be 30 g/mol, then every 0.03 kg of rock will donate
6 × 1023 electrons to the cause, so that we need about 18,000 tons of rock. At a density of 2.5 times that of
water (2.5 tons/m3 ), we need 7,000 m3 , or about 20 m on a side. Not so big, after all (easily visualized as a
Urey Hall).
If we had just said that the electric force is 2 × 1039 times as powerful as gravity, and that we just need
to divide the Earth’s mass by this factor (saying that each separated proton–electron counts almost 40orders-of-magnitude as much as gravitational mass), then we would have ended up with a tiny 10−15 kg,
which is substantially different. What went wrong is mainly that the force goes like charge squared, so we
have an N 2 factor thrown in, where N is the (large) number of electron/charges involved—while scaling the
single electron–proton forces is a linear operation in N . Also note the smaller corrections that gravitational
attraction is between protons/neutrons (so 2000 times stronger than the electron–proton attraction) and
also that we only donate one electron per about 30 protons in this game.
Part c: The easiest way to compare magnetism to gravity, perhaps, is to forget about comparing to Earth
and note that the magnet exerted a roughly 50 N force on 5 kg at a distance of ∼ 1 cm. Meanwhile,
the magnet has a volume of about a cubic centimeter and about the density of steel, and therefore has a
mass of about 10 g. How much gravitational force would the magnet exert: Gm1 m2 /r2 computes to about
6.67 × 10−11 × 0.01 × 5/10−4 ≈ 3 × 10−8 N, and 50 N is 0.15 × 1010 times larger than this.
An alternative approach: the magnet mass, at 0.01 kg, is 6 × 1026 times smaller than Earth’s mass. But
the magnetic pull defeats Earth’s pull at a scale of 1 cm, which is 6 × 108 times closer. Pretending that
the magnetic attraction is 1/r2 (otherwise no way to sensibly scale for distance), then the proximity gives
the magnet an advantage to the tune of 36 × 1016 . Applying this corrective factor, I might say the magnet
is 61 × 1010 times stronger. So it’s the same result (not fundamentally different). But it’s a somewhat
hokey/meaningless characterization since the force laws differ in functional form.
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Natural Units
Use dimensional analysis to combine fundamental constants ~, c, G, and kB to create combinations having
units of length, time, mass, energy, and temperature. Present the symbolic combinations, then numerically
evaluate them in terms of m, s, kg, J, K. These are the most fundamentally "natural" scales in physics, even
if inaccessible to modern experiments.
Putting everything into basic units:
• ~ = 1.05 × 10−34 kg · m2 /s
• c = 3.0 × 108 m/s
• G = 6.67 × 10−11 m3 /kg · s2
• kB = 1.38 × 10−23 kg · m2 /s2 · K
Noting that kB is the only one containing Kelvin, we concentrate first on the first three. Let’s first go for
length. We must therefore eliminate seconds and kilograms. Multiplying ~G is an obvious way to cancel
kilograms, leaving m5 /s3 . We must get rid of seconds, so we divide by c3 , leaving m2 . So finally we take the
square root:
r
~G
`=
≈ 1.6 × 10−35 m.
c3
1 Once you get the number of electron charges, you might also approach by calling this 330 kg of electrons (!), and each
proton/neutron is 2000 times more so the average atom has 60,000 times more mass than electrons, bringing us to 18 tons by
another route.
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It is pretty easy to then convert between length and time in the natural units world: c connects those two:
r
~G
≈ 5.4 × 10−44 s.
τ = `/c =
c5
How about mass? If we divide ~ by G, we get kg2 , which we can easily square-root. But we also have s/m
left in the division so we multiply by c:
r
~c
m=
≈ 2.2 × 10−8 kg.
G
This is a laboratory-measurable mass (a cube of water 0.3 mm on a side). Energy? We just use E = mc2 or
~/τ to get:
r
~c5
E=
≈ 1.96 × 109 J.
G
Finally, we get a temperature by taking the natural energy scale and dividing by kB :
s
~c5
32
T =
2 ≈ 1.4 × 10 K.
GkB
That’s hot!
These are known as Planck units: Planck length, Planck time, etc. You might say these describe conditions at
the Big Bang, when physics was (presumably) unified and all these natural constants were equally important.
What emerges is a tiny scale, an incredibly brief time, and a gigantic temperature.
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