Chapter 9, Problem 35E

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

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:

$\frac{1\text{ kW}}{1000\text{ W}}$

Answer to Problem 35E

Solution:

$\$2.25$

$\$65$

$\$54$

$\$0.75$

Explanation of Solution

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

The electricity cost is $15\text{ cents per kilowatt}-\text{hour}$.

Conversion of watt into kilowatt is as follows:

$\begin{array}{c}\text{Energy}=\left(1\text{ W}\times \frac{1\text{ kW}}{1000\text{ W}}\right)\\ =\left(100\text{ W}\times \frac{1\text{ kW}}{1000\text{ W}}\right)\\ =0.100\text{ kW}\end{array}$

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

$\begin{array}{c}\text{For }1\text{day}=5\text{ hours}\\ \text{For 30 days}=\frac{5\text{ hours}}{1\text{ day}}\times \frac{30\text{ days}}{1\text{ month}}\\ =150\text{ hours per month}\end{array}$

In one day, the consumption is $100\text{ W}$. For $150\text{ hours}$, it is calculated as follows:

$\begin{array}{c}\text{Energy consumption}=\left(\text{0}\text{.100 kW}\times 150\text{ h}\right)\\ \text{=15 kWh}\end{array}$

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

$\begin{array}{c}\text{monthly cost}=15\cancel{\text{kWh}}\times \frac{\$0.15}{\cancel{\text{kWh}}}\\ =\$2.25\end{array}$

Therefore, the monthly cost is $\$2.25$.

b) $\text{A 600 W refrigerator, 24 h/day}$

The electricity cost is $15\text{ cents per kilowatt}-\text{hour}$.

Conversion of watt into kilowatt is as follows:

$\begin{array}{c}\text{Energy}=\left(1\text{ W}\times \frac{1\text{ kW}}{1000\text{ W}}\right)\\ =\left(600\text{ W}\times \frac{1\text{ kW}}{1000\text{ W}}\right)\\ =0.600\text{ kW}\end{array}$

In one month, there are $30\text{ days}$. The per month usage of refrigerator in hours is calculated as follows:

$\begin{array}{c}\text{For }1\text{day}=24\text{ hours}\\ \text{For 30 days}=\frac{\text{24 hours}}{1\text{ day}}\times \frac{30\text{ days}}{1\text{ month}}\\ =720\text{ hours per month}\end{array}$

In one day, the consumption is $600\text{ W}$. For $720\text{ hours}$, it is calculated as follows:

$\begin{array}{c}\text{Energy consumption}=\left(\text{0}\text{.600 kW}\times 720\text{ h}\right)\\ =\text{432 kWh}\end{array}$

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

$\begin{array}{c}\text{monthly cost}=432\cancel{\text{kWh}}\times \frac{\$0.15}{\cancel{\text{kWh}}}\\ =\$65\end{array}$

Therefore, the monthly cost is $\$65$.

c) $\text{A 12000 W electric range, 1 h/day}$

The electricity cost is $15\text{ cents per kilowatt}-\text{hour}$.

Conversion of watt into kilowatt is as follows:

$\begin{array}{c}\text{Energy}=1\text{ W}\times \frac{1\text{ kW}}{1000\text{ W}}\\ =\left(12000\text{ W}\times \frac{1\text{ kW}}{1000\text{ W}}\right)\\ =12\text{ kW}\end{array}$

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

$\begin{array}{c}\text{For }1\text{day}=1\text{ hours}\\ \text{For 30 days}=\frac{\text{1 hours}}{1\text{ day}}\times \frac{30\text{ days}}{1\text{ month}}\\ =30\text{ hours per month}\end{array}$

In one day, the consumption is $\text{12000 W}$. For $30\text{ hours}$, it is calculated as follows:

$\begin{array}{c}\text{Energy consumption}=\left(12\text{ kW}\times 30\text{ h}\right)\\ =\text{360 kWh}\end{array}$

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

$\begin{array}{c}\text{monthly cost}=360\cancel{\text{kWh}}\times \frac{\$0.15}{\cancel{\text{kWh}}}\\ =\$54\end{array}$

Therefore, the monthly cost is $\$54$.

d) $\text{A 100 W toaster, 10 min/day}$

The electricity cost is $15\text{ cents per kilowatt}-\text{hour}$.

Conversion of watt into kilowatt is as follows:

$\begin{array}{c}\text{Energy}=1\text{ W}\times \frac{1\text{ kW}}{1000\text{ W}}\\ =\left(1000\text{ W}\times \frac{1\text{ kW}}{1000\text{ W}}\right)\\ =1\text{ kW}\end{array}$

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

$\begin{array}{c}\text{For }1\text{day}=10\text{ min}\\ \text{For 30 days}=\frac{10\text{ minutes}}{60\text{ minutes}}\times \frac{\text{1 hours}}{1\text{ day}}\times \frac{30\text{ days}}{1\text{ month}}\\ =5\text{ hours per month}\end{array}$

In one day, the consumption is $\text{100 W}$. For $\text{5 hours}$, it is calculated as follows:

$\begin{array}{c}\text{Energy consumption}=\left(1\text{ kW}\times 5\text{ h}\right)\\ =\text{5 kWh}\end{array}$

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

$\begin{array}{c}\text{monthly cost}=5\cancel{\text{kWh}}\times \frac{\$0.15}{\cancel{\text{kWh}}}\\ =\$75\end{array}$

Therefore, the monthly cost is $\$75$.