Energy and Society
Week 3 Section Solution
3. ENERGY BASICS AND UNIT ANALYSIS (ONLY FOR MONDAY SECTIONS)
We will not cover in class certain fundamental concepts, including significant digits and scientific
notation. If you have any doubts or hesitations with these concepts, please see Toolkit 1 or come to
office hours soon.
Energy versus Power
− Work refers to an activity involving a force and movement in the direction of the force. A
force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules
of work.
− Energy is quite tricky to define, but commonly defined as the capacity for doing work. You
must have energy to accomplish work - it is like the “currency” for performing work.
− Power is the rate of doing work or the rate of using energy, which are numerically the same.
The “flow” of energy. If you do 100 joules of work in one second (using 100 joules of
energy), the power is 100 watts.
A key relationship to learn: Energy = Power x time, or Power = Energy/time.
Practice: How much energy is produced by a wind turbine running at 7 kW for 20 minutes?
7 kW * (20min) * (1hr/60min) = (7 * 1/3) kWh = 2.33 kWh = 2 kWh
How many significant digits? 1 Why? (7 kW has one s.f.)
More examples of significant digits.
0.007  1 s.f.
0.0070  2 s.f.
70.0  3 s.f.
70  1 or 2 s.f., but assume 1. Usually scientific notation is used to be precise
7.0x101 or 70.  2 s.f.
4. ENERGY CONVERSION AND EFFICIENCY
Practice Problem 1: Suppose a family spends $100/month on their electricity bill. How much coal
(in kilograms) went into producing their yearly electricity consumption? Assume the price of
electricity is 10 cents per kWh and the power plant has a conversion efficiency of 30%. Coal has an
energy density of 29.3x106 J/kg. Before solving the question, guess the order of magnitude of the answer. This will
help hone your intuition.
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 $100   1kWhelectricity

  
$0.1
 month  
kWhelectricity
  12months 
  
  12,000
yr
yr

 
  1kWhcoal   3,600,000 J coal 
J

  1.44  1011 coal
  



kWhcoal
yr
  0.3kWhelectricity  



J  
1kg coal
kg
kg
  4,915 coal  5 x103 coal
1.44  1011 coal   
6
yr   29.3  10 J coal 
yr
yr

kWhelectricity

12,000
yr

Practice Problem 2: Could the chemical energy content in one teaspoon of sugar be sufficient to
heat a cup of water to prepare tea or coffee? Assume that there are 4 grams of sugar in a teaspoon
and that 1/20 of a gram of sugar contains about 1,000 J of chemical energy. One cup ~ 250 ml and
one teaspoon ~5ml. It takes 4.2 J (or one calorie) to raise the temperature of 1 gram of water by 1
°C.
Before solving the question, take a guess at the answer. Roughly speaking, does this amount of sugar have abundant
energy for heating a cup of water, about the right about of energy, or nowhere near enough?
One approach is to use the Cp of water = 4.18 J/g•ºC and remind them of Q=Cpm∆T
 1,000 J 
Energy in a teaspoon of sugar = 4 g  
  80,000 J
 0.05 g 
Energy needed to raise one cup by 1 °C = 250 ml x 1 g/ml x 4.2 J/g•ºC = 1050 J/ºC
Temperature change in cup = 80,000 J x 1 ºC/1050 J = 76 ºC = 80 ºC
How hot could we make the water by using the sugar for fusion? Use the famous mass-energy
equivalence formula (E=mc2), and assume it takes 2 J to raise the temperature of 1 gram of water by
1 °C1. What kind of units do you need to make this equation work properly if energy is expressed in terms of joules?
A joule, defined with SI units, is a kg-m2/s2. If c, the speed of light, is 2.99x108 m/s, then our
mass value should be expressed in kg. 4 g = 0.004 kg
E = 0.004 kg x (2.99x108 m/s)2 = 3.576x1014 J
From above:
Energy needed to raise one cup by 1 °C = 250 ml x 1 g/ml x 4.2 J/g•ºC = 1050 J/ºC
Temperature change in cup = 3.576x1014 J x 1 ºC/1050 J = 300,000,000,000 ºC (nb: Blow
on it first or you might burn your mouth.)
The specific heat of water changes depending on its temperature, and in its gaseous form, the value is lower than that
of liquid water. With the galactic temperatures we’re contemplating here, there isn’t really a single sensible number to
plug in. Given the extreme scenario we’re considering here, there are lots of factors that we’re not considering with this
simple calculation.
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5. COMBUSTION CHEMISTRY
What is a mole? 1 mole of a substance always contains 6.02×1023 or Avogadro’s number of
representative particles.
Mass ( grams )
Moles 
MolecularWeight ( grams / mol )
1 mole of an ideal gas occupies 22.4L or 22.4×10-3m3 at STP. Alternatively, there are 44.6 mol of
ideal gas in 1 m3.
Simple combustion (in a pure oxygen environment)
CH4 + 2O2  CO2 + 2H2O
Real combustion:
CH4 + α(O2 + 3.78N2)  CO2 + (n/2) H2O + 3.78αN2
Why is air represented as O2 + 3.78N2?
Although air has other components in addition to oxygen and nitrogen, approximate volumetric
shares of oxygen and nitrogen can be considered to be O 2 = 21% and N2 = 79%.
0.21/0.78 ≈ 3.78
Volumetric ratio is ratio of moles, not mass or molecular weight.
Example: The most common method of producing hydrogen today is by steam reforming of
methane. What are the value of x, y and z in the reaction below?
CH4 + x H2O  y CO + z H2
CH4 + H2O  CO + 3 H2
Practice Problem 1: What mass of carbon dioxide would be produced if 100g of butane (C4H10) are
completely oxidized to carbon dioxide and water?
CnHm + (n + m/4) (O2 + 3.78 N2)  nCO2 + m/2 H2O + 3.78 (n + m/4) N2
C4H10 + (n + m/4) (O2 + 3.78 N2)  4CO2 + 5H2O + 3.78 (n + m/4) N2
C4H10 + 6.5 (O2 + 3.78 N2)  4CO2 + 5 H2O + 3.78 *6.5 N2
C4H10 + 6.5 (O2 + 3.78 N2)  4CO2 + 5 H2O + 24.57 N2
58𝑔 𝐶4 𝐻10
100𝑔 𝐶4 𝐻10
=
4 ∗ 44𝑔 𝐶𝑂2
𝑋𝑔 𝐶𝑂2
𝑋𝑔 =
4∗44𝑔∗100𝑔
58𝑔
= 303𝑔 = 3 𝑋 102 𝑔 𝑜𝑓 𝐶𝑂2 (only 1 significant figure)
Note 1: Weight of CO2 is about three times that of butane
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Note 2: Although 100g can have 1,2, or 3 significant figures, use the lowest number of
significant figures for problem sets.
Practice Problem 2: Worldwide combustion of methane CH4 (natural gas) provides about 1.3×1017
kJ of energy per year (BP estimates for 2013). If methane has an energy content of 39×103kJ/m3 (at
STP), what mass of CO2 is emitted into the atmosphere each year? Also, express that emission rate as
metric tons of carbon (not CO2).
CH4 + 2 O2  CO2 + 2H2O
𝑘𝐽
1𝑚3 𝐶𝐻4
44.6𝑚𝑜𝑙𝐶𝐻4 1𝑚𝑜𝑙𝐶𝑂2 44𝑔𝐶𝑂2
1𝑡𝐶𝑂2
×
×
×
×
×
𝑦𝑟 39 × 103 𝑘𝐽
1𝑚3 𝐶𝐻4
1𝑚𝑜𝑙𝐶𝐻4 1𝑚𝑜𝑙𝐶𝑂2 1 × 106 𝑔𝐶𝑂2
𝑡𝐶𝑂
2
= 6.5 × 109
𝑦𝑟
𝑡𝐶𝑂
12𝑡𝐶
𝑡𝐶
6.5 × 109 2 ×
= 1.8 × 109
1.3 × 1017
𝑦𝑟
44𝑡𝐶𝑂2
𝑦𝑟
Note: 1 mole of ideal gas has a volume of 22.4 L at STP i.e. 44.6 moles/ m3
Practice Problem 3: Diesel (CH1.8) engines are used to generate electricity on a stand by basis (i.e.
when there is a power cut) or in remote areas where there is no access to electricity.
Trivia: Number of people in the world without access to electricity is approximately 1.2 billion.
That’s 20% of the world’s population. Most don’t have access to diesel engines, and use kerosene for
lighting. In addition, millions have unreliable access to electricity, especially in the developing
world, and diesel engines are common as a backup source of electricity. How polluting AND
expensive!!!
Balance the combustion equation for diesel. How much carbon is emitted for every kg of diesel
burned? What is the mole fraction of CO2 in the exhaust?
CH1.8 + 1.45 O2  CO2 + 0.9H2O
1000. 𝑔 𝐶𝐻1.8 ×
12𝑔𝐶
= 870𝑔𝐶
13.8𝐶𝐻1.8
Note: Period after 1000 signifies 3 significant figures. Hence, 870 has 3 significant figures
1𝑚𝑜𝑙𝐶𝑂2
= 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 0.53
1.9𝑚𝑜𝑙(𝑒𝑥ℎ𝑎𝑢𝑠𝑡 𝑔𝑎𝑠)
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Practice Problem 4:2 Claim: a regular-sized vehicle, driven an average amount, will emit its own
weight (about 2 tonnes) in CO2 each year. Perform a back-of-the-envelope calculation to verify (or
refute!) this claim. Assume the vehicle is fueled by octane (C 8H18) and use basic combustion
chemistry to inform your calculations.
Assume:
10,000 mi/yr driven
40 mpg fuel economy (a new car, as opposed to an average-aged car)
C8H18 + O2 -> CO2 + H2O is the general combustion reaction
C8H18 + 12.5O2 -> 8CO2 + 9H2O is the balanced combustion equation
104 mi/yr x 1 gal gas/40 mi = 250 gal gas/yr
250 gal gas/yr x 3.78 L/gal x 0.75 kg gas/L gas x 1 mol C8H18/0.114 kg C8H18 x 8 mol CO2/1 mol
C8H18 x 44 g CO2/1 mol CO2 x 1 tonne/106 g = 2 t
This question is slightly more advanced that material covered on the current problem set. It’s here as a challenge
problem, so something to come back to in a week or two.
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