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Chapter 27: Problem 36

A standard flashlight battery can deliver about \(2.0 \mathrm{~W} \cdot \mathrm{h}\) of energy before it runs down. (a) If a battery costs US\$ \(0.85\), what is the cost of operating a \(100 \mathrm{~W}\) lamp for \(8.0 \mathrm{~h}\) using batteries? (b) What is the cost if energy is provided at the rate of US\$0.06 per kilowatt-hour?

Short Answer

Expert verified
Using batteries costs US$340; using utility power costs US$0.048.

Step by step solution

01

Calculate Total Energy Required

First, determine the total energy required to run a 100 W lamp for 8 hours. The power of the lamp is 100 W and it needs to run for 8 hours. Using the formula: \( \text{Energy (Wh)} = \text{Power (W)} \times \text{Time (h)} \), we calculate the total energy required: \( 100 \text{ W} \times 8 \text{ h} = 800 \text{ Wh} \).
02

Calculate Number of Batteries Needed

Next, find out how many batteries are required to supply 800 Wh of energy. A single battery provides 2 Wh, so we divide the total energy by the energy per battery: \( \frac{800 \text{ Wh}}{2 \text{ Wh per battery}} = 400 \text{ batteries} \).
03

Calculate Cost of Batteries

Determine the cost of using 400 batteries. Each battery costs \(0.85. Multiply the number of batteries by the cost per battery: \( 400 \times 0.85 = 340 \text{ US\\)} \).
04

Calculate Cost for Energy from Utility

Now, compute the cost of 800 Wh of energy if provided by an energy utility at USD 0.06 per kilowatt-hour. First, convert Wh to kWh: \( 800 \text{ Wh} = 0.8 \text{ kWh} \). Then multiply by the cost rate: \( 0.8 \times 0.06 = 0.048 \text{ US\$} \).

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Key Concepts

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

Cost Efficiency
When considering the operational cost of running appliances, it's essential to think about cost efficiency. This means making sure you're getting the most output or utility for the least amount of money spent. In our example, we are examining the cost involved with running a 100 W lamp using batteries versus using electricity from the power grid.
  • If you're using batteries, you would need many of them for a high-power device like a 100 W lamp.
  • You would need 400 batteries, each costing $0.85, leading to a total cost of $340.
This shows that while convenient, batteries may not always be the cost-efficient option for high energy demands. It's often much cheaper to draw power from the electric grid for such needs.
Battery Energy Capacity
Understanding battery energy capacity is crucial when determining how many batteries you need for a task. A battery's energy capacity is the total amount of energy it can store and deliver when needed.
  • In our scenario, each battery can supply 2 Wh (watt-hours) of energy.
  • To run a 100 W lamp for 8 hours, you need a total of 800 Wh of energy.
  • Dividing the total energy required (800 Wh) by the energy capacity of one battery (2 Wh) shows you need 400 batteries.
This illustrates that for high-energy tasks, batteries may not be practical and other energy sources might have better capacity.
Power Calculation
Power calculation is a fundamental concept for determining the energy needs of any device and how long it can run on a given power source. The key formula here is:\[\text{Energy (Wh)} = \text{Power (W)} \times \text{Time (h)}\]In this problem, we calculated that running a 100 W lamp for 8 hours requires:- \(100 \text{ W} \times 8 \text{ h} = 800 \text{ Wh}\)This means you need a source capable of providing 800 Wh for the lamp to operate for the desired time, demonstrating how important accurate power calculations are in planning energy usage.
Utility Rate Comparison
Choosing between energy from batteries and from the utility company requires a good understanding of utility rate comparisons. In our example, the cost efficiency of these differing energy sources was highlighted.
  • Utility companies provide energy at specific rates. In this case, it was $0.06 per kWh.
  • The total energy needed (800 Wh) converts to 0.8 kWh.
  • At $0.06 per kWh, the energy cost from the utility would be $0.048.
This is dramatically cheaper than using 400 batteries at $340. Therefore, comparing utility rates and understanding energy conversions helps make cost-effective energy choices.

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Most popular questions from this chapter

A car battery with a \(12 \mathrm{~V}\) emf and an internal resistance of \(0.030 \Omega\) is being charged with a current of \(40 \mathrm{~A}\). What are (a) the potential difference \(V\) across the terminals, (b) the rate \(P_{r}\) of energy dissipation inside the battery, and (c) the rate \(P_{\mathrm{cmf}}\) of energy conversion to chemical form? When the battery is used to supply 40 A to the starter motor, what are (d) \(V\) and (e) \(P_{r}\) ?

Displays two circuits with a charged capacitor that is to be discharged through a resistor when a switch is closed. In Fig. 27-38a, \(R_{1}=20.0 \Omega\) and \(C_{1}=5.00 \mu \mathrm{F}\). In Fig. \(27-38 b, R_{2}=10.0 \Omega\) and \(C_{2}=8.00 \mu \mathrm{F}\). The ratio of the initial charges on the two capacitors is \(q_{02} / q_{01}=1.75\). At time \(t=0\), both switches are closed. At what time \(t\) do the two capacitors have the same charge?

A voltmeter of resistance \(R_{\mathrm{V}}=300 \Omega\) and an ammeter of resistance \(R_{A}=3.00 \Omega\) are being used to measure a resistance \(R\) in a circuit that also contains a resistance \(R_{0}=100 \Omega\) and an ideal battery with an emf of \(\mathscr{\&}=18.0 \mathrm{~V}\). Resistance \(R\) is given by \(R=V / i\), where \(V\) is the potential across \(R\) and \(i\) is the ammeter read- ing. The voltmeter reading is \(V^{\prime}\), which is \(V\) plus the potential difference across the ammeter. Thus, the ratio of the two meter readings is not \(R\) but only an apparent resistance \(R^{\prime}=V^{\prime} / i\). If \(R=85.0 \Omega\), what are (a) the ammeter reading, (b) the voltmeter reading, and (c) \(R^{\prime} ?\) (d) If \(R_{A}\) is decreased, does the difference between \(R^{\prime}\) and \(R\) increase, decrease, or remain the same?

What multiple of the time constant \(\tau\) gives the time taken by an initially uncharged capacitor in an \(R C\) series circuit to be charged to \(89.0 \%\) of its final charge?

In an \(R C\) series circuit, emf \(\&=12.0 \mathrm{~V}\), resistance \(R=1.40 \mathrm{M} \Omega\), and capacitance \(C=2.70 \mu \mathrm{F}\). (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to \(16.0 \mu \mathrm{C} ?\)
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