A \(1.5-\mathrm{V}\) flashlight battery can maintain a current of $0.30 \mathrm{A}\( for \)4.0 \mathrm{h}$ before it is exhausted. How much chemical energy is converted to electrical energy in this process? (Assume zero internal resistance of the battery.)

Short Answer

Expert verified
Answer: 6480 Joules

Step by step solution

01

List down the given values

We are given the following values: - Voltage, V = 1.5 V - Current, I = 0.30 A - Time, t = 4.0 h We need to find the total amount of chemical energy converted to electrical energy, which can be found using the formula: Energy = Voltage × Current × Time.
02

Convert time to seconds

In order to calculate the energy, we need to convert the time unit from hours to seconds: 1 hour = 60 minutes = 60 × 60 seconds So, 4.0 hours = 4.0 × 60 × 60 seconds = 14400 seconds Now, the time t = 14400 s
03

Calculate the energy

Now, we can use the formula to calculate the total amount of chemical energy converted to electrical energy: Energy = Voltage × Current × Time Energy = V × I × t Energy = (1.5 V) × (0.30 A) × (14400 s) Energy = 6480 Joules So, the total amount of chemical energy converted to electrical energy in this process is 6480 Joules.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

During a "brownout," which occurs when the power companies cannot keep up with high demand, the voltage of the household circuits drops below its normal $120 \mathrm{V} .\( (a) If the voltage drops to \)108 \mathrm{V},$ what would be the power consumed by a "100-W" light-bulb (that is, a light-bulb that consumes \(100.0 \mathrm{W}\) when connected to \(120 \mathrm{V}\) )? Ignore (for now) changes in the resistance of the light-bulb filament. (b) More realistically, the light-bulb filament will not be as hot as usual during the brownout. Does this make the power drop more or less than that you calculated in part (a)? Explain.
(a) What are the ratios of the resistances of (a) silver and (b) aluminum wire to the resistance of copper wire $\left(R_{\mathrm{Ag}} / R_{\mathrm{Cu}} \text { and } R_{\mathrm{A}} / R_{\mathrm{Ca}}\right)$ for wires of the same length and the same diameter? (c) Which material is the best conductor, for wires of equal length and diameter?
In the physics laboratory, Oscar measured the resistance between his hands to be \(2.0 \mathrm{k} \Omega .\) Being curious by nature, he then took hold of two conducting wires that were connected to the terminals of an emf with a terminal voltage of \(100.0 \mathrm{V} .\) (a) What current passes through Oscar? (b) If one of the conducting wires is grounded and the other has an alternate path to ground through a \(15-\Omega\) resistor (so that Oscar and the resistor are in parallel), how much current would pass through Oscar if the maximum current that can be drawn from the emf is \(1.00 \mathrm{A} ?\)
Two copper wires, one double the diameter of the other, have the same current flowing through them. If the thinner wire has a drift speed \(v_{1},\) and the thicker wire has a drift speed \(v_{2},\) how do the drift speeds of the charge carriers compare?
A piece of gold wire of length \(L\) has a resistance \(R_{0}\) Suppose the wire is drawn out so that its length increases by a factor of three. What is the new resistance \(R\) in terms of the original resistance?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free