How does one turn coal into gasoline? Write a single chemical equation that balances for the reaction you use as an example. Gasoline is typically C 10 H 22 . Inputs are only carbon and water. The C and H do not directly match the gasoline molecule, so something
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single chemical equation that balances for the reaction
you use as an example. Gasoline is typically C10H22.
Inputs are only carbon and water. The C and H do not
directly match the gasoline molecule, so something
else is left over on the right side. Take it to be CO2.
X C + Y H2O C10H22 + Z CO2
So there are three unknowns X, Y, and Z.
Start with hydrogen, since H appears only once on
each side. 2Y=22, so Y=11.
Then balance oxygen, once on each side. Y=2Z, or
since Y=11, so Z=5.5.
Now do carbon X=10 + Z, so X=15.5.
Our equation is then 15.5 C + 11 H2O C10H22 + 5.5 CO2
Or 31 C + 22 H2O 2 C10H22 +11 CO2.
language, with the given 50,000 Btu/kg of C10H22.
We will thus need 27 x 1015 Btu / 50,000 Btu/kg
= 0.00054 x 1015 kg=5.4x1011 kg of gasoline.
15.5 x 12 =186 kg of carbon will make 10 x 12 + 22 x 1
= 142 kg of gasoline, from the masses in the equation.
So we have the ratio
186 :142 = XX : 5.4x1011 kg
So XX=7.07 x 1011 kg of carbon, which is 14.1 x 1011 kg of
coal (at the given ratio). At 1000 kg per metric tonne,
we will need to dig 1.41 x 109 metric tonnes,
in addition to the 109 or so tonnes we dig anyway.
We will also need even more coal to run the reaction above.
This would be a very, very big deal! Feel free to edit your
answer to part b) in relation to this number.
Use heat energy to do work.
Water at height Hhigh
Falls to Hlow
CHANGE in potential energy = m g (Hhigh-Hlow)
With g =9.8 m/sec2
Be orthodox in your units. If m is in kg, heights in meters, and g=9.8 m/sec2, then the energy will be in Joules.
And we can use that fall to do useful work.
Heat in = heat out + work done
Qin = Qout + W
Or---QHot = QCold + W
(careful! Heat in calories or Btu, work in Joules or kW-hr)
I burn one ton of coal to generate 2.66 x 107 Btu, and use this energy to do 10 10 Joules of useful work.
A) how much heat was exhausted?
Qout = Qin – W = 2.66 x 107 Btu – 1010 J
= 2.81 x 1010 J – 1 x 1010 J= 1.81 x 1010 J
B) What is the efficiency of my engine?
e=W/Qin = 1010 J / 2.66 x 107 Btu
=1010 J / 2.81 x 1010 J = 0.3559 = 35.59%.
Heat Power in
= Heat power out +(work) Power out.
=work power out/Rate of heat energy in.
I burn one bbl of oil per hour at 25% efficiency. How many kW can I make?
I bbl/hr=6.12x109 J/3600 sec
=1.7 x 106 watts = 1700 kW
Useful power = 0.25 x 1700 kW= 425 kW
At 25%425 kW
or 0.75 x 1700 kWh=1275 kWh
=4351575=4.35 million Btu
into a cooling stream or the air.
with GWe = efficiency x GWt
degrees C. If a power plant at this temperature is next
to a cold lake, just at freezing, to dump the waste heat,
what is the best efficiency possible for this plant to make
3.(3) Here’s another idea, with free !! energy. Cold, deep
ocean water is at 4 deg C, while surface water in the tropics
can be at 30 deg. C. What is the best efficiency you could
obtain with some sort of heat engine operating between
4. (2) Draft a one-sentence (!!!) question (on energy and the environment only!!)you would like to ask of your representative in Washington. Then, add a brief answer that you would hope to hear.