# CHM110H5 Study Guide - Midterm Guide: Joule

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Department
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Professor Assignment
Part A:
1. For one mole of sucrose, C12H22O11, 5640 kJ of energy is produced.
2. The amount of energy needed to produce 20 kg of C12H22O11 is:
Find the number of moles of C12H22O11:
n=mass
molar mass =20,000 g
342.30 gmol =58.428 mol
58.428 mol x 5640 kJ
mol =3.295 x105kJ
Therefore, the amount of energy needed to produce 20 kg of C12H22O11 is
3.295 x105kJ
. This is equivalent to the amount of energy used to produce
C12H22O11 per hour per hectare:
3.295 x105kJ
h ∙ ha
3. A watt is a unit of power equal to one joule per second. Determining how many
kJ per hour per hectare are equal to 1.0 kw per square meter:
1ha=10,000 m
1kw
mx10,000 m
1ha =10,000 kw
ha
10,000 kw
ha =10,000 kJ
s ∙ha
10,000 kJ
s ∙ha x3600 s
h=3.6 x107kJ
h ∙ha
Therefore,
is the amount of solar energy
4. Compare the amount of energy needed to produce of C12H22O11 (2) and the amount
of solar energy to calculate the efficiency of photosynthesis (3):
energy required
percent efficiency=¿produce sucrose ¿
solar energy
¿
3.295 x105kJ
h ∙ha
3.6 x107kJ
h ∙ ha
¿0.
00915 x 100%
= 0.915%
Therefore, the percent efficiency of photosynthesis is 0.915%
Part B:
A kilowatt-hour is a unit of energy that is equal to 1 kilowatt used for 1 hour. We also
know that, one kilowatt-hour is equal to 3600 kJ.
1. A typical house uses 40kWh, determining the amount of electrical energy needed
in joules by a typical home per day:
40 kWh=1.44 x105kJ
Therefore, the amount of energy needed by a typical home is
1.44 x105kJ
2. As a result, the panels need to convert and provide 144,000 kJ per day. Determine
the original total energy required for the panels to convert:
X kJ of 15% = 144,000 kJ
x=960,000 kJ
Therefore, 960,000 kJ is the total energy needed for conversion.
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## Document Summary

C12h22o11 per hour per hectare: 3. 295 x 105 kj h ha: a watt is a unit of power equal to one joule per second. 10,000 kj s ha x 3600 s h. Therefore, 3. 6 x107 kj h ha is the amount of solar energy: compare the amount of energy needed to produce of c12h22o11 (2) and the amount of solar energy to calculate the efficiency of photosynthesis (3): energy required percent efficiency= produce sucrose solar energy. = 3600kj h m: to calculate the square meters needed, compare the amount of kj per hour (3) and the amount of kj per hour per square meter (4): sqare meters needed= In this study, the minimum amount of surface area for a solar panel is determined to provide electricity for one house. On average household uses about 40 kwh of electricity per day. 1kw/m2 is equal to 1kj/sm2. So in one hour, 1kj/sm2 * 3600s will be equal to.