Question

It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you

use an electric water heater that is 60% efficient, how many kilojoules of electrical energy will the heater
actually use by the time the water reaches its final temperature?

Answer 1

we know that

Power efficiency is defined as the ratio of the output power divided by the input power:

so

`efficiency =( Power\ out)/(Power\ in) *100`

in this problem

`Power\ out=56.5\ kJ\\ Power\ in=x\ kJ`

`efficiency=60`%

Substitute in the formula above

`60 =( 56.5)/(x) *100\\ \\ 60*x=56.5*100\\ \\ x=94.17\ kJ`

therefore

the answer is

`94.17\ kJ`