Department

ChemistryCourse Code

CHM110H5Professor

Thottackad RadhakrishnanStudy Guide

QuizThis

**preview**shows half of the first page. to view the full**2 pages of the document.** 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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