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,

3.6 x107kJ

h∙ ha

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.

## 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.