# Assume that electricity costs 15 cents per kilowatt- hour. Calculate the monthly cost of operating each of the following: a 100 W light bulb, 5 h/day a 600 W refrigerator, 24 h/day a 12,000 W electric range, 1 h/day a 1000 W toaster, 10 min/day

### 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 35E
Textbook Problem
37 views

## Assume that electricity costs 15 cents per kilowatt- hour. Calculate the monthly cost of operating each of the following:a 100 W light bulb, 5 h/daya 600 W refrigerator, 24 h/daya 12,000 W electric range, 1 h/daya 1000 W toaster, 10 min/day

Interpretation Introduction

Interpretation:

The monthly cost of operating each of the given cases is to be determined.

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.

Since, 1 kW is equal to 1000 W, hence, conversion factor is as follows:

1 kW1000 W

### Explanation of Solution

a) A 100 W light bulb, 5 h/day

The electricity cost is 15 cents per kilowatt-hour.

Conversion of watt into kilowatt is as follows:

Energy=(1 W×1 kW1000 W)=(100 W×1 kW1000 W)=0.100 kW

In one month, there are 30 days. The per month usage of light bulb in hours is calculated as follows:

For 1day=5 hoursFor 30 days=5 hours1 day×30 days1 month=150 hours per month

In one day, the consumption is 100 W. For 150 hours, it is calculated as follows:

Energy consumption=(0.100 kW×150 h)=15 kWh

The cost of electricity for one kilowatt-hour is 15 cents. It is divided by 100 to convert in dollars. The number of a kilowatt-hour is multiplied with cost per kilowatt-hour as follows:

monthly cost=15 kWh×$0.15kWh=$2.25

Therefore, the monthly cost is $2.25. b) A 600 W refrigerator, 24 h/day The electricity cost is 15 cents per kilowatt-hour. Conversion of watt into kilowatt is as follows: Energy=(1 W×1 kW1000 W)=(600 W×1 kW1000 W)=0.600 kW In one month, there are 30 days. The per month usage of refrigerator in hours is calculated as follows: For 1day=24 hoursFor 30 days=24 hours1 day×30 days1 month=720 hours per month In one day, the consumption is 600 W. For 720 hours, it is calculated as follows: Energy consumption=(0.600 kW×720 h)=432 kWh The cost of electricity for one kilowatt-hour is 15 cents. It is divided by 100 to convert in dollars. The number of a kilowatt-hour is multiplied with cost per kilowatt-hour as follows: monthly cost=432 kWh×$0.15kWh=$65 Therefore, the monthly cost is$65.

c) A 12000 W electric range, 1 h/day

The electricity cost is 15 cents per kilowatt-hour

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