Download Quiz 1 from 2012

Massachusetts Institute of Technology
Physics Department
Physics 8.21
Fall 2012
Physics of Energy
October 2, 2012
Quiz 1
Problem
1
2
3
4
5
6
Total
Points
10 points
15 points
25 points
25 points
20 points
(+ 25 points )
100 + 25 points
Instructions
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You must do problems 1-5. You may do problem 6 for extra credit
Do all problems in the white exam booklets
Remember to write your name clearly on every exam booklet you use
Open book, open notes, open 8.21 Energy Information Card
Do not use red ink
No cell phones or computers
Calculators are allowed
The problems have different point values as marked
You may need to make assumptions to solve some problems. Explain your assumptions
Problem summary
• Problem 1 [10 pts] Physical properties
• Problem 2 [20 pts] Energy and power quantities
• Problem 3 [25 pts] Thomas Edison’s Pearl Street Station
• Problem 4 [25 pts] Thermal energy storage
• Problem 5 [25 pts] Entropy and energy of a three-dimensional oscillator
• Problem 6 [EXTRA CREDIT 25 points] A car that runs on water!?
1
MIT
8.21 Physics of Energy
Fall 2012
2
Problems
Problem 1.
Physical properties [10 points]
Here are three quantities in SI units along with a list of physical properties. Identify the property
described by each quantity:
1.
2.
3.
4.
Quantity
joule/meter
coulomb2 /farad
pascal meter/second
volt ampere
A.
B.
C.
D.
E.
Physical property
Energy
Energy flux
Power
Pressure
Force
Answer: 1 ↔ E
2↔A
3↔B
4↔C
Problem 2.
Energy and power quantities [15 points (5 points each)]
How much energy (in joules) or power (in watts) is either consumed or produced by the following
processes...
(a) Burning 1 kg of coal.
Answer:
One ton (metric tonne) of coal is equivalent to 29.3 GJ, so burning 1 kg produces 29.3 MJ of
energy.
(b) Overcoming the air resistance encountered by an automobile at 60 mph if the auto’s cross
sectional area is 2 m2 , its drag coefficient is cd = 1/4, and its engine delivers power to the wheels
at 25% efficiency.
Answer:
1
3
The total rate of energy loss of the car due to air resistance is dE
dt = 2 cd Aρv . The density of
3
air, ρ is 1.275 kg/m . The velocity in SI units is 26.8 m/s. To overcome this, the engine must
1
provide power equal to P = 2ε
cd Aρv 3 . where ε is the engine’s efficiency. So, the engine must
provide P = 0.5/0.25(0.25)(2 m2 )(1.275 kg/m3 )(26.8 m/s)3 = 24.5 kW of power.
(c) Absorbing 1018 photons emitted from hydrogen atoms when they make transitions from their
first excited state to their ground state.
Answer:
MIT
8.21 Physics of Energy
Fall 2012
3
Figure 1: A circuit diagram for Edison’s early power network with many lights connected to a DC generator
The energy states of the hydrogen atom have energies given by En = −13.6 eV/n2 where n is
a positive integer, so the energy of a single photon emitted from the first excited state (n = 2)
to the ground state (n = 1) is E = 13.6(1 − 1/(22 )) = 10.2 eV. There are 1018 of these, so the
total energy is 1.02 × 1019 eV = 1.63 J.
Problem 3.
Thomas Edison’s Pearl Street Station [25 points]
Early in the development of electric power, Thomas Edison built a direct current (DC) power station
at Pearl Street in New York City. The station generated 175 horsepower of electric power at 110 V
(DC). It powered about 10 000 electric lights. For simplicity we can assume that all the lights were
connected on a single circuit at a distance of 1 km from the generating station. The voltage across
the lights was recorded as 100 V. A circuit diagram is shown in Figure 1.
(a) (3 points) How much power in watts did the station generate?
Answer:
P = 175hp × 746W/hp = 131kW
(b) (3 points) What is the ratio of the resistance in the transmission wires (Rw ) to the effective
resistance of the Reff lights?
Answer:
Since these resistances are in series, they share the same current and their resistances are proportional to their voltage drops. The voltage drop across the wires must be 110 − 100 = 10 V,
so Rw /Reff = 10/100 = 0.1.
(c) (8 points) What is the current in the circuit and the resistance of the wires?
Answer:
For DC circuits P = V I, so I = 130 kW/110 V = 1 180 A. The voltage drop across the wires is
Vw = IRw , so Rw = 10/1 180 = 8.5 mΩ.
(d) (8 points) If the wires are made of copper and have a circular cross section, what must their
radius have been? (Remember that the circuit requires two 1 km wires to serve the load.)
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8.21 Physics of Energy
Fall 2012
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Answer:
The resistivity of copper is 1.7 × 10−8 Ωm, and Rw = %Lw /Aw , so Aw = %Lw /Rw = 4 × 10−3 m2 .
Setting A = πr2 , we find r = 3.5 cm.
(e) (3 points) Can you explain why alternating current (AC) won out over DC in the early days of
electric power?
Answer:
These wires are already very massive and the transmission distance is only 1 km. DC transmission over long distances at the delivered voltage seems impractical. AC allows the use of
transformers to step up the voltage to a high value for transmission, reducing losses dramatically, and allowing the transmission wires to be much less massive.
Problem 4.
Thermal energy storage [25 points]
The Bonneville Power Administration in the Pacific Northwest region of the U. S. faces an unusual
challenge: how to deal with violent storms which cause simultaneous surges in wind and hydroelectric
power generation. These surges generated 2 GW of excess power during an incident in 2010, for
example, overwhelming grid capacity.
One solution being explored is to store the energy by heating the water in the hot water tanks of
homes of volunteers living in the area. How well does this work?
(a) (5 points) Let’s begin by computing the energy stored in a hot water tank. Suppose the tank
holds 400 liters of water. It is normally at 50◦ C, but when a surge hits, the temperature is raised
to 80◦ C. How much energy is required to accomplish this, expressed in MJ?
Answer:
Mass of water: 400 l = 400 kg
E = cV m∆T = 4.18 ∗ 400 ∗ 30kJ = 50 MJ
(b) (5 points) Suppose a 1 GW power surge hits, for one hour. How many homes need to participate,
to be able to absorb all the energy in the surge (assuming all the tanks are the same size, in the
same initial 50◦ C state)?
Answer:
E = 50 MJ ×
N=
1 MWh
= 13.9 kWh
3600 MJ
1 GWh
106 kWh
=
= 71, 942 homes
13.9 kWh/home
13.9 kWh/home
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8.21 Physics of Energy
Fall 2012
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(c) (8 points) At what rate is energy lost by conduction to heating the room where the tank is
located? Suppose the tank has surface area A = 0.6 m2 , the room is at 20◦ C, and the water is
at 80◦ C. Around the tank is wrapped insulation with an R-value of 4 (SI). Observations indicate
that under these conditions, the outer surface of the wrapped tank reaches a temperature of
30◦ C.
Answer:
P =
∆T
(80 − 30)
A=
0.6 = 7.2 W
R
4
(d) (7 points) Suppose that the wrapped tank, with outer surface temperature 30◦ C, has emissivity
= 0.1. At what rate does it lose energy through heat transfer by radiation? Is this more or
less than that lost by conduction through the insulation?
Answer:
Correct:
P = σA∆(T 4 ) = (0.1)(5.68 × 10−8 )(0.6) (30 + 273)4 − (20 + 273)4 = 3.6 W
Wrong:
P = σAT 4 = (0.1)(5.68 × 10−8 )(0.6)(30 + 273)4 = 9.8 W
MIT
8.21 Physics of Energy
Problem 5.
Fall 2012
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Entropies and energy of a three dimensional oscillator [25 points]
The quantum energy levels of a particle in a three dimensional harmonic oscillator are El,m,n =
(l + m + n + 3/2)~ω, where l, m, n = 0, 1, 2, 3, .... (You may find the solution to Problem 2 on Pset 3
useful in this problem.)
(a) (15 points) If a single oscillator is known to have energy E = 3~ω/2, what is its entropy? Suppose
the energy is E 0 = 7~ω/2, what then is its entropy?
Answer:
S(E) = kB ln N (E), where N is the number of microstates with energy E. There is only
one micro state with E = 3~ω/2, so S(3~ω/2) = 0, but here are 6 micro states with energy
E 0 = 7~ω/2, so S(7~ω/2) = kB ln 6.
(b) (10 points) Suppose ~ω = 0.1 eV. Consider a collection of N of these oscillators in thermal
equilibrium at room temperature. The oscillators do not interact with one another. What is the
ratio of the number of oscillators with energy E 0 to the number with energy E?
Answer:
The probability to find a particle in a given micro state is proportional to e−E/kB T (the normalizing factor of Z is irrelevant since we’re looking for a ratio). So the ratio is
R=
Problem 6.
6e−7~ω/2kB T
= 6e−2~ω/kB T = 6e−2(.1)/.025) ≈ 0.002.
e−3~ω/2kB T
EXTRA CREDIT: A car that runs on water!? [25 points]
A “car that runs on water” that got a lot of attention in the press recently turned out to be a hoax.
The interior of the cylinders of the car were coated with a thick layer of calcium carbide (CaC2 ), a
solid which reacts with water to form acetylene (C2 H2 ), a gas, and calcium hydroxide (Ca(OH)2 ). The
acetylene was subsequently combusted, and the heat generated was used to run the engine. Remember
that the heat comes both from the reaction of calcium carbide with water and the combustion of the
acetylene. The standard enthalpies and free energies of formation of some relevant materials are listed
in the table.
(a) (5 points) Write balanced chemical reactions for both reactions: (I) calcium carbide plus liquid
water making acetylene and calcium hydroxide, and (II) combustion of acetylene.
Answer:
(I) This reaction is
CaC2 + 2H2 O −→ C2 H2 + Ca(OH)2
(1)
(II) This reaction combines acetylene with oxygen to yield water and carbon dioxide.
2C2 H2 + 5O2 −→ 2H2 O(liquid) + 4CO2 (vapor)
(2)
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8.21 Physics of Energy
Fall 2012
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(b) (5 points) Verify that reaction (I) can proceed spontaneously at 25◦ C and 1 atm.
Answer:
The Gibbs free energy of reaction is
∆G = ∆G(C2 H2 ) + ∆G(Ca(OH)2 ) − ∆G(CaC2 ) − 2∆G(H2 O) = −150 kJ/mol
(3)
for liquid water and -166 kJ/mol for water vapor. Since ∆G is negative, the reaction can occur
spontaneously.
(c) (10 points) Compute the heat released in kJ/kg(CaC2 ) by these two reactions. You can assume
that the water formed when acetylene is burned is produced as a vapor.
Answer:
The heat released in reaction (I) (assuming that liquid water is used since otherwise energy
would be needed to create steam) is
∆H = −987 + 227 − (−63) − 2(−286) = −125 kJ/mol
(4)
Calcium carbide has a molar mass of 0.064 kg/mol, so this is equivalent to -1950 kJ/kg Ca2 .
The heat released in reaction (II) is (assuming that water vapor is released and using the fact
that diatomic gases have ∆H f = 0),
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∆H = 2(−394) + (−242) − (227) − (0) = −1260 kJ/mol C2 H2 = −1260 kJ/mol CaC2
2
(5)
or 19 600 kJ/kg CaC2 .
(d) (5 points) How does this compare with the energy released (kJ/kg) in complete combustion of
gasoline (see the energy card)?
Answer:
A car running on liquid water would then release 21.6 MJ/kg CaC2 . Gasoline provides 120 MJ
per gallon. A gallon of gasoline has a mass of around 6 pounds, so this is 20 MJ/lb = 44 MJ/kg.
So, gasoline provides about twice as much energy per unit mass.
Compound
Water vapor
Water liquid
Carbon dioxide (g)
Calcium hydroxide
Acetylene
Calcium carbide
Chemical Formula
(Molar mass)
H2 O(g) (18)
H2 O(l) (18)
CO2 (g) (44)
Ca(OH)2 (74)
C2 H2 (26)
CaC2 (64)
∆H f (kJ/mol)
∆Gf (kJ/mol)
− 242
− 286
− 394
− 987
+ 227
− 63
−229
− 237
− 394
− 898
+ 209
-65
Table 1: Standard enthalpy, free energy of formation and molar mass for some useful chemical compounds, all
at 25◦ C and 1 atm.