# An average person consumes about 2 .0 × 10 3 kcal of food energy per day. How many kilowatt-hours of energy are consumed? How long could you light a 40 W light bulb with that energy?

### Chemistry In Focus

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

Chapter
Section

### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 9, Problem 34E
Textbook Problem
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## An average person consumes about 2 .0 × 10 3 kcal of food energy per day. How many kilowatt-hours of energy are consumed? How long could you light a 40 W light bulb with that energy?

Interpretation Introduction

Interpretation:

The kilowatt-hours of energy contained in 2.0×103 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:

1 Kcal1000 cal

Since, 1 gram calorie=4.18 joule, hence conversion factor is as:

4.18 J1 cal

Since, 1 Kilowatt hour=3.6×106 joule, hence, conversion factor is as:

3.6×106 J1 kWh

Since, 1 quads=1.6×1018 joule, hence, conversion factor is as:

### Explanation of Solution

The calories in the food are given. With the help of the conversion factor, the conversion of kcal into kWh is done.

The kilowatt-hour of energy is calculated as follows:

The conversion of kcal into kWh is as follows:

1 gram calorie=4.18 joule

2.0×103 kcal=(2.0×103 kcal)×(1000 cal1 kcal)×(4.18 J1cal)×(1 kWh3.6×106 J)=2

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