Chapter 6, Problem 11STP

To determine

To calculate: the electrical energy saved if the bulb is on only for $4\text{ h}$ instead of $24\text{ h}$ .

Answer to Problem 11STP

The electrical energy saved if the bulb is on only for $4\text{ h}$ is $\underset{\_}{\text{2}\text{.0 kWh}}$ .

Explanation of Solution

Given data:

Power of the lightbulb, $P=100\text{ W}$

Time, $t_1=24\text{ h, }{t}_2=4\text{ h}$

Formula used:

• Electrical Energy Equation:

  $E=Pt$

Where $E$ is the electrical energy provided by the power company in kilowatt-hours, $P$ is the power of an appliance in kilowatts, and $t$ is time in hours for which the energy is provided.

Calculation:

As per the problem,

  $P=100\text{ W=0}\text{.1 kWh}$

When the bulb is left on for $t_1=24\text{ h}$ ,

The electrical energy used in $t_1=24\text{ h}$ can be calculated by substituting the values of power and time in the electrical energy equation:

  $\begin{array}{l}{E}_1=P{t}_1\\ E_1=(0.1\text{ kWh)(24 h)}\\ E_1=2.4\text{kWh}\end{array}$

If the bulb was on only for $t_2=4\text{ h}$ ,

Then electrical energy used in $t_2=4\text{ h}$ can be calculated as:

  $\begin{array}{l}{E}_2=P{t}_2\\ E_2=(0.1\text{ kWh)(4 h)}\\ E_2=0.4\text{kWh}\end{array}$

The electrical energy saved if the bulb is on for $4\text{ h}$ rather than $24\text{ h}$ is calculated as :

  $E=E_1-E_2=2.4\text{ kWh}-0.4\text{ kWh}=2.0\text{ kWh}$

Conclusion:

The electrical energy saved if the bulb is on only for $4\text{ h}$ is $\underset{\_}{\text{2}\text{.0 kWh}}$ .