Chapter 10, Problem 33E

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:

$\text{1 quad }=\text{ 1}.0\text{6 }\times \text{ 1}{0}^{\text{18}}\text{joules}$, conversion factor is as:

$\frac{\text{1}.0\text{6 }\times \text{ 1}{0}^{\text{18}}\text{joules}}{\text{1 quad}}$

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

$\text{1 kilowatt-hour }=\text{3}.\text{6 }\times \text{ 1}{0}^{\text{6}}\text{joules}$, conversion factor is as:

$\frac{\text{3}.\text{6 }\times \text{ 1}{0}^{\text{6}}\text{joules}}{\text{1 kilowatt-hour}}$

Power is given by the expression as follows:

$\text{power}=\frac{\text{energy}}{\text{time}}$