Chapter 9, Problem 34E

Interpretation Introduction

Interpretation:

The kilowatt-hours of energy contained in $2.0\times {10}^3\text{ kcal}$ of food consumed per day and the duration for which a bulb of 40W can light up using that energy is to be calculated.

Concept introduction:

Unit conversion is defined as the process in which multiple steps are used to convert the unit of measurement for the same given quantity. It is determined by multiplication with a conversion factor.

The fraction in which numerator and denominator are the same quantities but are determined in different units, is known as a conversion factor.

The power generated by utilizing the chemical or physical resources for providing heat and light or for carrying out various processes is known as energy. The SI unit of energy is joule.

Since, 1 Kcal is equal to 1000 cal, hence, conversion factor is as:

$\frac{1\text{ Kcal}}{1000\text{ cal}}$

Since, $1\text{ gram calorie}=4.18\text{ joule}$, hence conversion factor is as:

$\frac{4.18\text{ J}}{1\text{ cal}}$

Since, $1\text{ Kilowatt hour}=3.6\times {10}^6\text{ joule}$, hence, conversion factor is as:

$\frac{\text{3}\text{.6}\times {\text{10}}^6\text{ J}}{1\text{ kWh}}$

Since, $1\text{ quads}=\text{1}\text{.6}\times {\text{10}}^{18}\text{ joule}$, hence, conversion factor is as:

$\frac{\text{1}\text{.6}\times {\text{10}}^{18}\text{ J}}{1\text{ quad}}$