Chapter 6, Problem 87GP

To determine

a. To Show: 1 kilowatt-hour = $\text{3}.\text{6}\times \text{1}{0}^{\text{6}}\text{J}$ .

b. The energy consumed by a family in a month.

c. To find this energy in joules.

d.The monthly bill in dollars and whether this bill depends on the rate of energy.

Answer to Problem 87GP

Solution:

a. 1 kilowatt-hour = $\text{3}.\text{6}\times \text{1}{0}^{\text{6}}\text{J}$

b. The energy consumed is $418\text{ kW}-\text{h}$ .

c. The energy in joules is $1.5\times {10}^9\text{J}$ .

d. The monthly bill is 50 dollars.

Explanation of Solution

Explanations:

a.

  $\begin{array}{l}1\text{kWh}=1000\text{W}\times 3600\text{s}\\ 1\text{kWh=3600000}\text{W}\cdot \text{s}\\ \text{1}\text{kWh=3}\text{.6}\times {\text{10}}^6\text{J}\end{array}$

b.

The given power is,

  $\begin{array}{l}P=580\text{W}\\ P=0.580\text{kW}\end{array}$

Time,

  $\begin{array}{l}t=1\text{month}\\ t=30\text{days}\\ t=30\times 24\text{h}\\ t=720\text{h}\end{array}$

Energy consumed,

  $\begin{array}{l}E=Pt\\ E=0.580\times 720\text{kW-h}\\ E=418\text{kW-h}\end{array}$

c.

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

So, 418 kW-h is

  $\begin{array}{l}E=418\times 3.6\times {10}^6\text{J}\\ E=1.5\times {10}^9\text{J}\end{array}$

d.

The cost per kW-h is

  $0.12$ dollars

The cost of $418$

kW-h is:

  $\begin{array}{l}=418\times 0.12\text{dollar}\\ \text{=50}\text{dollars}\end{array}$

The monthly bill depends on the unit of energy. The more is the energy consumed, greater is the bill. It does not depend on the rate of electric energy.

Conclusion:

a. To Show: 1 kilowatt-hour = $\text{3}.\text{6}\times \text{1}{0}^{\text{6}}\text{J}$ .

b. The energy consumed by a family in a month.

c. To find this energy in joules.

d.The monthly bill in dollars and whether this bill depends on the rate of energy.