Karisma loves to cook. She wants to buy a new electric beater and is considering two models:Model A converts every 1000 J of electrical energy into 600 J of kinetic energy, 300 J of heat, and 100 J of sound.Model B converts every 1500 J of electrical energy into 800 J of kinetic energy, 400 J of heat, and 300 J of sound.Calculate the energy efficiency of the two models and determine which one is more energy efficient.
Question
Karisma loves to cook. She wants to buy a new electric beater and is considering two models:Model A converts every 1000 J of electrical energy into 600 J of kinetic energy, 300 J of heat, and 100 J of sound.Model B converts every 1500 J of electrical energy into 800 J of kinetic energy, 400 J of heat, and 300 J of sound.Calculate the energy efficiency of the two models and determine which one is more energy efficient.
Solution
The energy efficiency of a device can be calculated by dividing the useful output energy (or power) by the total input energy (or power). The result is usually expressed as a percentage.
For Model A: The useful output energy is the kinetic energy, which is 600 J. The total input energy is 1000 J. So, the energy efficiency is (600 J / 1000 J) * 100% = 60%.
For Model B: The useful output energy is the kinetic energy, which is 800 J. The total input energy is 1500 J. So, the energy efficiency is (800 J / 1500 J) * 100% = 53.33%.
Therefore, Model A is more energy efficient than Model B.
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