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Chapter 25: Problem 72

A typical cost for electric power is \(\$ 0.120\) per kilowatt- hour. (a) Some people leave their porch light on all the time. What is the yearly cost to keep a \(75-\) W bulb buming day and night? (b) Suppose your refrigerator uses 400 \(\mathrm{W}\) of power when it's running, and it runs 8 hours a day. What is the yearly cost of operating your refrigerator?

Short Answer

Expert verified
(a) $78.84; (b) $140.16

Step by step solution

01

Find Daily Energy Consumption of the Porch Light

The power of the bulb is 75 W which is equivalent to 0.075 kW. Since it runs for 24 hours a day, the energy consumption per day is \(0.075 \times 24 = 1.8\) kWh.
02

Calculate Yearly Energy Consumption of the Porch Light

To find the yearly consumption, multiply the daily consumption by the number of days in a year: \(1.8 \times 365 = 657\) kWh.
03

Compute Yearly Cost of Running Porch Light

Multiply the yearly energy consumption with the cost per kWh: \(657 \times 0.120 = 78.84\). The yearly cost is \$78.84.
04

Determine Daily Energy Consumption of the Refrigerator

The refrigerator uses 400 W, equivalent to 0.4 kW. It runs for 8 hours a day, so the daily consumption is \(0.4 \times 8 = 3.2\) kWh.
05

Calculate Yearly Energy Consumption for Refrigerator

To calculate the yearly energy consumption, multiply the daily consumption by 365 days: \(3.2 \times 365 = 1168\) kWh.
06

Compute Yearly Cost of Operating Refrigerator

Multiply the yearly energy consumption with the cost per kWh: \(1168 \times 0.120 = 140.16\). The yearly cost is \$140.16.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Consumption
Energy consumption refers to the amount of energy that an appliance or device uses over a period of time. Understanding how energy consumption works is crucial in determining the cost of running household appliances. To calculate energy consumption, you multiply the power rating of the device (in kilowatts, kW) by the amount of time it runs (in hours). Once you have this, you get the energy used, which is measured in kilowatt-hours (kWh). For example, a 75-watt bulb running all day will use energy calculated as follows:
  • Convert watts to kilowatts: 75 W = 0.075 kW.
  • Multiply by hours used per day: 0.075 kW x 24 hours = 1.8 kWh/day.
Understanding this concept helps you manage electricity bills by making you aware of how much energy your appliances use and highlighting the impact of running devices continuously.
Kilowatt-Hour
A kilowatt-hour (kWh) is a unit of energy that represents the consumption of 1 kilowatt (kW) of power over the duration of one hour. It is a standard unit used by utility companies to measure energy consumption for billing purposes. For instance, if an appliance rated at 400 watts (0.4 kW) is used for 8 hours, the total energy consumption is calculated as:
  • Power in kilowatts: 400 watts = 0.4 kW.
  • Energy used: 0.4 kW x 8 hours = 3.2 kWh.
By calculating in kilowatt-hours, you can determine the energy cost of running various appliances. Knowing how to do this helps you estimate and control household electricity expenses more effectively.
Power Usage
Power usage refers to the rate at which energy is used by appliances and devices. It is typically measured in watts (W) or kilowatts (kW) and helps in understanding which devices consume the most energy. Power usage can vary widely between devices. For example, a small bulb may use only 75 watts, while larger appliances like refrigerators use around 400 watts. To convert watts to kilowatts, divide by 1000:
  • 75 watts = 0.075 kW.
  • 400 watts = 0.4 kW.
Recognizing the differences in power usage allows you to identify which devices are more energy-intensive and can help you make informed decisions about reducing electricity consumption and costs.

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Most popular questions from this chapter

A resistor with resistance \(R\) is connected to a battery that has emf 12.0 \(\mathrm{V}\) and internal resistance \(r=0.40 \Omega .\) For what two values of \(R\) will the power dissipated in the resistor be 80.0 \(\mathrm{W} ?\)

A \(12.0-\mathrm{V}\) battery has an internal resistance of 0.24 \(\mathrm{s}\) and a capacity of 50.0 \(\mathrm{A} \cdot \mathrm{h}\) (see Exercise 25.47\() .\) The battery is charged by passing a 10 -A current through it for 5.0 \(\mathrm{h}\) . (a) What is the terminal voltage during charging? (b) What total electricalenergy is supplied to the battery during charging? (c) What electrical energy is dissipated in the internal resistance during charging? (d) The battery is now completely discharged through a resistor, again with a constant current of 10 \(\mathrm{A}\) . What is the external circuit resistance? (e) What total electrical energy is supplied to the external resistor? (f) What total electrical energy is dissipated in the internal resistance? (g) Why are the answers to parts (b) and (e) not the same?

A tightly coiled spring having 75 coils, each 3.50 \(\mathrm{cm}\) in diameter, is made of insulated metal wire 3.25 \(\mathrm{mm}\) in diameter. An ohmmeter connected across its opposite ends reads 1.74\(\Omega\) ? What is the resistivity of the metal?

A Nonideal Ammeter. Unlike the idealized ammeter described in Section \(25.4,\) any real ammeter has a nonzero resistance. (a) An ammeter with resistance \(R_{\mathrm{A}}\) is connected in series with a resistor \(R\) and a battery of emf \(\mathcal{E}\) and internal resistance \(r .\) The current measured by the ammeter is \(I_{\mathrm{A}}\) . Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of \(I_{A}, r, R_{\mathrm{A}},\) and \(R .\) The more "ideal" the ammeter, the smaller the difference between this current and the current \(I_{\mathrm{A}}\) . (b) If \(R=3.80 \Omega, \mathcal{E}=7.50 \mathrm{V},\) and \(r=0.45 \Omega,\) find the maximum value of the ammeter resistance \(R_{\mathrm{A}}\) so that \(l_{\mathrm{A}}\) is within 1.0\(\%\) of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a maximum value.

The following measurements were made on a Thyrite resistor: $$\begin{array}{llll}{I(\mathbf{A})} & {0.50} & {1.00} & {2.00} & {4.00} \\\ {V_{a b}(\mathbf{V})} & {2.55} & {3.11} & {3.77} & {4.58}\end{array}$$ (a) Graph \(V_{a b}\) as a function of \(I .(\mathrm{b})\) Does Thyrite obey Ohm's law? How can you tell? (c) Graph the resistance \(R=V_{a b} / I\) as a function of \(I\) .
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