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Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692

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Section
BuyFindarrow_forward

Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 10, Problem 33E
Textbook Problem
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Use the energy conversion tables in Chapter 9 to determine how many quads of solar energy fall on Earth each year, assuming that average solar power on Earth is 1017 W. How does this number compare with world energy consumption— approximately 558 quad?

Interpretation Introduction

Interpretation:

The comparison of the average solar power on Earth with the world energy consumption is to be determined.

Concept Introduction:

The relation between quads and joules units is given as follows:

1 quad = 1.0× 1018joules, conversion factor is as:

1.0× 1018joules1 quad

The relation between kilowatt-hour and joules units is given as follows:

1 kilowatt-hour =3.× 106joules, conversion factor is as:

3.× 106joules1 kilowatt-hour

Power is given by the expression as follows:

power=energytime

Explanation of Solution

Given information:

The average solar power on Earth is 1017W.

The world energy consumption is 558 quad.

The conversion of one year into hours is done as follows:

1 year=(1 year×365 days1 year×24 hours1 day)=8760 hours

The conversion of energy into kWh is done as follows:

1017 W=energy8760 hourenergy=8.76×1020 Wh=8.76×1017 kWh

Therefore, the solar energy that falls on the Earth in a year is 8.76×1017 kWh.

Now, 1 kilowatt-hour = 3.6 × 106 joules

The conversion of energy into joules is done as follows:

8.76 × 1017 kWh=(8.76 × 1017 kWh×3.6 × 106 J1 kWh)=3

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Chemistry In Focus
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