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PHY 102 Test Compilation PHY 102 10 Objective Question
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Question

A light bulb has an input power consumption of 50 watts. The light bulb was activated for 60 seconds and produced heat of 2400 joules. Find the efficiency of the light bulb.
Options
A 20%
Correct Answer
B 80%
C 12.5%
D 60%
Correct Answer

Option A is the correct answer.

Detailed Explanation

Given:Input power (Pin) = 50 WTime (t) = 60 sHeat produced (Qheat) = 2400 JFirst, calculate the total energy input:Energy input (Ein) = Pin * t = 50 W * 60 s = 3000 JThe useful output of a light bulb is light, while heat is a form of wasted energy. So, the useful output energy (Eout) is the total input energy minus the wasted heat energy:Eout = Ein - Qheat = 3000 J - 2400 J = 600 JNow, calculate the efficiency (η):η = (Eout / Ein) * 100%η = (600 J / 3000 J) * 100% = (1/5) * 100% = 20%

Hint

Efficiency is the ratio of useful output energy to total input energy. For a light bulb, heat is generally considered wasted energy, while light is the useful output.

Question Info

Course: PHY 102
Topic: Test Compilation PHY 102 10
Type: Objective
Difficulty: Not specified
Question 921 of 1819
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