Lesson 10 The Efficiency Formula
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| Course: | Science 10 [5 cr] - AB Ed copy 1 |
| Book: | Lesson 10 The Efficiency Formula |
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| Date: | Sunday, 7 September 2025, 6:46 PM |
Introduction
Can you determine how effectively a machine converts energy input into useful energy output?
C10.1 oscillating fan
Efficiency is a measure of how much work or energy is converted to useful energy in an energy transformation process. In many processes, work or energy is lost as waste heat.
The efficiency is the energy output, divided by the energy input and is expressed as a percentage.
The efficiency is the energy output, divided by the energy input and is expressed as a percentage.
Targets
By the end of this lesson, you will be able to- explain, using calculations, efficiency as a measure of the “useful” work compared to the total energy put into an energy conversion process or device
- apply concepts related to efficiency of thermal energy conversion to analyze the design of a thermal device
Watch This
Energy Transfer and Efficiency © YouTube DoodleScience
This video provides you with a quick overview of energy transfer and efficiency. It goes through the concept of waste energy and introduces you to the efficiency calculation formula. This will help get you in the right mindset for this lesson.
The Efficiency Formula
To calculate the percent efficiency of any system or device, what do you need to know?
C10.2 moving metronome
To calculate the efficiency of a device, you must identify the useful output energy and the total input energy. You can then analyze the system or machine.
Calculating Efficiency
Efficiency is a ratio, so it has no units. The output and input energies must be in the same units so that they will cancel out.
Calculating Efficiency
Efficiency is a ratio, so it has no units. The output and input energies must be in the same units so that they will cancel out.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#FFFFFF¨»percent«/mi»«mo mathcolor=¨#FFFFFF¨»§#160;«/mo»«mi mathcolor=¨#FFFFFF¨»efficiency«/mi»«mo mathcolor=¨#FFFFFF¨»=«/mo»«mfrac mathcolor=¨#FFFFFF¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#FFFFFF¨»§#215;«/mo»«mn mathcolor=¨#FFFFFF¨»100«/mn»«mo mathcolor=¨#FFFFFF¨»%«/mo»«/math»
Examples
- When a 100.0 W light bulb is on for 2.4 h, it uses 864 kJ of electrical energy. During that time, the light bulb emits 45.6 kJ of light. What is the efficiency of the light bulb in transforming electrical energy into light energy?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»864«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»45«/mn»«mo».«/mo»«mn»6«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«/mtable»«/math»
Note: The information of 100.0 W and 2.4 h is not used in the calculation.
Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»45«/mn»«mo».«/mo»«mn»6«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»864«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/math»
Step 4: Calculate the answer.
percent efficiency = 5.277...% = 5.28% (to three significant digits)
- A bobcat uses 346 kJ of chemical potential energy stored in the fuel to lift 2 725 kg of dirt 3.2 m straight up, to dump it into a truck. What was the efficiency of the bobcat in converting chemical potential energy into gravitational potential
energy of the dirt?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»346«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#160;«/mo»«mn»725«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Useful output energy must first be calculated. Enough information has been provided to determine gravitational potential energy.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#160;«/mo»«mn»725«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mo»=«/mo» «mi»m«/mi» «mi»g«/mi» «mi»h«/mi» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mo»=«/mo» «mfenced» «mrow» «mn»2«/mn» «mo»§#160;«/mo» «mn»725«/mn» «mo»§#160;«/mo» «mi»kg«/mi» «/mrow» «/mfenced» «mfenced» «mrow» «mn»9«/mn» «mo».«/mo» «mn»81«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «mo»/«/mo» «msup» «mi mathvariant=¨normal¨»s«/mi» «mn»2«/mn» «/msup» «/mrow» «/mfenced» «mfenced» «mrow» «mn»3«/mn» «mo».«/mo» «mn»2«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «/mrow» «/mfenced» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mo»=«/mo» «mn»85«/mn» «mo»§#160;«/mo» «mn»543«/mn» «mo».«/mo» «mn»2«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/math»
Energy needs to be in the same units, so we convert Ep to kJ.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfenced»«mrow»«mn»85«/mn»«mo»§#160;«/mo»«mn»543«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«/mfenced»«mfenced»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mtext»kJ«/mtext»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«/mfrac»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»85«/mn»«mo».«/mo»«mn»5432«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»85«/mn»«mo».«/mo»«mn»5432«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»346«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
percent efficiency = 24.723...% = 25% (to two significant digits)
- A pitching machine uses 185 J of electrical energy to pitch a 320 g baseball. If the speed of the ball is 24 m/s, what is the efficiency of the pitching machine in transforming electrical energy into kinetic energy of the baseball?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»185«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»320«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»24«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Useful output energy must first be calculated. Enough information has been provided to determine kinetic energy. Remember, mass needs to be in kg when using the kinetic energy formula, so it needs to be converted.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»320«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«/mtable»«/math»
since mass needs to be in kg, the mass needs to be converted
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfenced»«mrow»«mn»320«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mfenced»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi»g«/mi»«/mrow»«/mfrac»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»320«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»24«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi mathvariant=¨normal¨»v«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«/mtable»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»320«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfenced»«msup»«mfenced»«mrow»«mn»23«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»84«/mn»«mo».«/mo»«mn»64«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»84«/mn»«mo».«/mo»«mn»64«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»185«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
percent efficiency = 45.751...% = 46% (to two significant digits)
- A 3.7 kg steel ball is dropped on a spring, compressing it. As a result, the compressed spring stores 72 J of elastic potential energy. If the gravitational potential energy of the steel ball was converted into elastic potential energy in
the spring with an efficiency of 80.4%, from what height was the steel ball dropped?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»80«/mn»«mo».«/mo»«mn»4«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»72«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo».«/mo»«mn»7«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Total input energy must first be calculated.
To isolate total input energy, you must multiple each side by total input energy. To move total input energy to the other side, you must use the opposite operation. Multiplication is opposite to division.
Then you would have the equation
(total input energy)(percent efficiency) = useful output energy × 100%
To isolate total input energy, you must divide each side by percent efficiency. To move percent efficiency to the other side, you must use the opposite operation. Division is opposite to multiplication.
Then you would have the equation
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»=«/mo»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»72«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»80«/mn»«mo».«/mo»«mn»4«/mn»«mo»%«/mo»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»89«/mn»«mo».«/mo»«mn»552«/mn»«mo»§#160;«/mo»«mn»2«/mn»«mo»§#8230;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«/mtable»«/math»
Then, enough information has been provided to determine height.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»=«/mo»«mn»89«/mn»«mo».«/mo»«mn»552«/mn»«mo»§#160;«/mo»«mn»2«/mn»«mo»§#8230;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo».«/mo»«mn»7«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«/mtd»«/mtr»«/mtable»«/math»
To isolate h, you must divide each side by mg. To move mg to the other side, you must use the opposite operation. Division is opposite to multiplication.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mtable columnspacing=¨0px¨ columnalign=¨right center left¨» «mtr» «mtd» «mfrac» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mrow» «mi»m«/mi» «mi»g«/mi» «/mrow» «/mfrac» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mi»h«/mi» «/mtd» «/mtr» «mtr» «mtd» «mi»h«/mi» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mfrac» «mrow» «mn»89«/mn» «mo».«/mo» «mn»552«/mn» «mo»§#160;«/mo» «mn»2«/mn» «mo»§#8230;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/mrow» «mrow» «mfenced» «mrow» «mn»3«/mn» «mo».«/mo» «mn»7«/mn» «mo»§#160;«/mo» «mi»kg«/mi» «/mrow» «/mfenced» «mstyle displaystyle=¨true¨» «mfenced» «mrow» «mn»9«/mn» «mo».«/mo» «mn»81«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «mo»/«/mo» «msup» «mi mathvariant=¨normal¨»s«/mi» «mn»2«/mn» «/msup» «/mrow» «/mfenced» «/mstyle» «/mrow» «/mfrac» «/mtd» «/mtr» «/mtable» «/math»
Step 4: Calculate the answer.
h = 2.467...m = 2.5 m (to two significant digits)
Do you want a bit more detailed explanation of how to use the efficiency formula in a calculation? Watch this video for more information on this calculation.
https://adlc.wistia.com/medias/qjvymvdj79
Digging Deeper
C10.3 bicycle
The bicycle is considered to be the most efficient machine that humans have ever made. In terms of the amount of energy that a person must use to travel a certain distance, cycling is considered to the most efficient self-powered way to travel.
Even more efficient than walking!
To learn more about the efficiency of a bicycle, click on the link. https://en.wikipedia.org/wiki/Bicycle_performance#Energy_efficiency
Learn More
To learn more about the efficiency of a bicycle, click on the link. https://en.wikipedia.org/wiki/Bicycle_performance#Energy_efficiency
Learn More
Read This
Please read pages 216 to 220 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on how efficiency is calculated, and communicated. Remember, if you have any questions, or do not understand
something, ask your teacher!
Practice Questions
Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses
(where necessary) to study from.
- If a light bulb is 6.34% efficient and it emits a total of 5.27 × 103 J of light energy, how much electrical energy does it use?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»6«/mn»«mo».«/mo»«mn»34«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»5«/mn»«mo».«/mo»«mn»27«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»3«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Total input energy must first be calculated.
To isolate total input energy, you must multiple each side by total input energy. To move total input energy to the other side, you must use the opposite operation. Multiplication is opposite to division.
Then you would have the equation
(total input energy)(percent efficiency) = useful output energy × 100%
To isolate total input energy, you must divide each side by percent efficiency. To move percent efficiency to the other side, you must use the opposite operation. Division is opposite to multiplication.
Then you would have the equation
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»=«/mo»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»27«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»3«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»6«/mn»«mo».«/mo»«mn»34«/mn»«mo»%«/mo»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
total input energy = 83 123.028...J = 8.31 × 104 (to three significant digits)
- An archer does work on a bow by exerting an average force of 173 N over a distance of 0.59 m to stretch the string and bend the bow. In the process, elastic potential energy is stored in the bow. When the archer releases the 0.062 kg arrow, its
initial speed is 51 m/s. What was the efficiency of transforming the archer’s work in the arrow’s kinetic energy? Hint: This question is very similar to the example in the video above.
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»062«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»51«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»F«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»173«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»N«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»d«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Total input energy and useful output energy must first be calculated.
Total input energy is the work done. Remember that work is a measurement of energy.
W = Fd
Useful output energy is the kinetic energy that the arrow has.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «mo»=«/mo» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi»m«/mi» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»062«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfenced»«msup»«mfenced»«mrow»«mn»51«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«mo»=«/mo»«mn»80«/mn»«mo».«/mo»«mn»631«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«mtr»«mtd»«mi»W«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»F«/mi»«mi»d«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»W«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfenced»«mrow»«mn»173«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»N«/mi»«/mrow»«/mfenced»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»=«/mo»«mn»102«/mn»«mo».«/mo»«mn»07«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»80«/mn»«mo».«/mo»«mn»631«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»102«/mn»«mo».«/mo»«mn»07«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
percent efficiency = 78.995 7...% = 79% (to two significant digits)
Efficiency Calculations
C10.4 thermal image of electrical transformers
In this lesson, we have looked at how efficiency is a measure of how much work or energy is conserved in an energy transformation process. The less efficient a process, the more energy that is lost as waste heat.
You learned to calculate efficiency when you know the useful output energy and total input energy of a system or device.
Our last lesson of this unit looks at comparing the benefits and drawbacks of different fuels—efficiency is one of the major considerations in this.
You learned to calculate efficiency when you know the useful output energy and total input energy of a system or device.
Our last lesson of this unit looks at comparing the benefits and drawbacks of different fuels—efficiency is one of the major considerations in this.
Watch This
What Is Efficiency? Efficiency Explained and Calculated © YouTube davenport1947
Watch this video for a detailed explanation of efficiency in a machine. It reviews the concepts of useful output energy, waste energy, and total input energy. A number of examples are also discussed, along with efficiency calculation.
3.6 Assignment
Unit 3 Assignment Lessons 9-11
It is now time to complete the Lesson 10 portion of 3.6 Assignment. This assignment has two parts.
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Online Questions
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