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CHM110H5 Study Guide - Quiz Guide: Kilowatt Hour, Joule


Department
Chemistry
Course Code
CHM110H5
Professor
Thottackad Radhakrishnan
Study Guide
Quiz

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NAME: YVONNE EIFEDIYI
STUDENT NO:1001045102
LAB SECTION A 101
DATE : NOVEMBER 22ND 2013
ASSIGNMENT 5
A. Solar energy = 1 kW
m2
1 mol of C12H22O11 = 5640 kJ
20 kg C12H22O11 = 58.42 mol C12H22O11
58.42 mol C12H22O11 = x kJ
__1 mol___ = _5640 kJ_
58.42 mol x kJ
x = 329496.99 kJ of sunlight used to produce 20 kg C12H22O11.
20 kg/h/ha = 329496.99 kJ/(60 x 60)s/10000 m2
= 0.00915 kJ/s/m2
= 0.00915 kW/m2 (since 1 kJ/s/m2 = 1 kW/m2)
Amount of energy needed to produce C 12H22O11 x 100%
Solar energy
= 0.00915 kW/m 2
x 100% = 0.915% of sunlight used to produce the sucrose.
1.0 kW/m2
B. 15% efficient in converting sunlight to electricity
40 kWh/day of that electricity
_1 kWh_ = 3600 kJ
40 kWh x kJ
x = 144000 kJ of electricity energy needed per day
144000 kJ = 0.15 x x (where x represents the solar energy)
x = 960000 kJ of solar energy
960000 kJ = 120000 kJ/h
8 h
120000 kJ/h is needed for conversion.
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