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A \(5.7\) A current is set up in a circuit for \(15.0 \mathrm{~min}\) by a rechargeable battery with a \(6.0 \mathrm{~V}\) emf. By how much is the chemical energy of the battery reduced?

Short Answer

Expert verified
30.78 kJ

Step by step solution

01

Convert Time into Seconds

The time given is 15.0 minutes, but to use it in calculations, we need to convert it to seconds. Since there are 60 seconds in a minute, multiply 15 by 60: \[ 15.0 \text{ minutes} \times 60 = 900 \text{ seconds} \]
02

Calculate Total Charge

The electric current (\(I\) is 5.7 A, and time (\(t\)), is 900 seconds. Use the formula \(Q = I \times t\) to calculate the total charge that flows.\[ Q = 5.7 \times 900 = 5130 \text{ C} \]
03

Calculate Energy Reduced

The energy reduced, \(E\), in the battery can be calculated using the formula \(E = Q \times V\), where \(Q\) is the charge and \(V\) is the voltage (6.0 V):\[ E = 5130 \times 6.0 = 30780 \text{ J} \].
04

Express Final Energy in kJ

Convert the energy from joules to kilojoules by dividing by 1000:\[ 30780 \text{ J} \div 1000 = 30.78 \text{ kJ} \].

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Key Concepts

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

Electric Current
Electric current is the flow of electric charge through a conductor. In most cases, conductors are materials like wires, where electrons move freely to carry the current. Electric current is measured in amperes (A), which quantifies how much charge flows through the conductor per unit time. Think of it as a river of electrons that is driven by a difference in voltage, much like water flows from a higher elevation to a lower one.

In our exercise, the electric current is constant at 5.7 A. This tells us that there are 5.7 coulombs of electric charge flowing through the circuit every second. Imagine turning on a faucet to let water flow at a steady rate—this steady flow is similar to how current moves through a circuit to power a device or recharge a battery.
  • Electric currents are foundational to operating any electronic device.
  • They can carry energy to where it is needed, such as charging batteries or powering appliances.
Charge Calculations
Charge calculations are essential to understanding how much electricity is involved in a particular process. Charges are measured in coulombs (C), and we can calculate the total charge that passes through a circuit by using the formula: \[ Q = I \times t \]where:
  • \( Q \) is the charge in coulombs
  • \( I \) is the current in amperes (A)
  • \( t \) is the time in seconds
In our example, we calculated that the total charge passing through the circuit was 5130 C. This was found by multiplying 5.7 A by 900 seconds, which is the total number of seconds that passed while the current was flowing.

Knowing the amount of charge is vital, as it allows us to determine how much energy is being used or stored, and helps in designing circuits that meet specific energy requirements.
  • Charge is a fundamental property that relates directly to energy usage.
  • Understanding charge helps predict how long a battery will last or how much energy a device needs.
Energy Conversion
Energy conversion in batteries is the process where chemical energy is converted into electrical energy. In this context, it's essential to understand that batteries store chemical energy, which is then converted into an electric form to power devices or circuits when a current is flowing.

This conversion is quantified using the formula:\[ E = Q \times V \]where:
  • \( E \) is the energy in joules (J)
  • \( Q \) is the charge in coulombs (C)
  • \( V \) is the voltage or emf of the battery in volts (V)
In the given exercise, the chemical energy reduced in the battery is calculated to be 30.78 kJ. This calculation comes from multiplying the total charge (5130 C) by the battery's voltage (6.0 V), which yields energy in joules. Dividing by 1000 converts it into kilojoules for easier understanding in bigger scales.

Understanding energy conversion allows us to grasp how efficiently a battery or any energy-storage device operates. It also helps us appreciate the limitations and capabilities of various power sources.
  • Energy conversion efficiency impacts how much stored energy can actually be used.
  • It's a key factor in developing sustainable energy solutions.

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Most popular questions from this chapter

Two identical batteries of emf \(8=10.0 \mathrm{~V} \quad\) Problems and internal resistance \(r=0.200 \Omega\) are to be 5 and 6 . connected to an external resistance \(R\), either in parallel (Fig. 27-20) or in series (Fig. 27-21). If \(R=2.00 \mathrm{r}\), what is the current \(i\) in the external resistance in the (a) parallel and (b) series arrangements? (c) For which arrangement is \(i\) greater? If \(R=r / 2.00\), what is \(i\) in the external resistance in the (d) parallel arrangement and (e) series arrangement? (f) For which arrangement is \(i\) greater now?

A total resistance of \(5.00 \Omega\) is to be produced by connecting an unknown resistance to a \(15.0 \Omega\) resistance. (a) What must be the value of the unknown resistance, and (b) should it be connected in series or in parallel? (c) What is the total resistance if the unknown resistance is connected the other way?

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 capacitor with an initial potential difference of \(80.0 \mathrm{~V}\) is discharged through a resistor when a switch between them is closed at \(t=0 .\) At \(t=10.0 \mathrm{~s}\), the potential difference across the capacitor is \(1.00 \mathrm{~V}\). (a) What is the time constant of the circuit? (b) What is the potential difference across the capacitor at \(t=17.0 \mathrm{~s} ?\)

Six \(18.0 \Omega\) resistors are connected in parallel across a \(12.0 \mathrm{~V}\) ideal battery. What is the current through the battery?

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