Question

A vacuum cleaner has an energy input of 6.5 kWh and an energy output of 5.8 kWh. The efficiency of the vacuum cleaner, to the nearest percentage, is

Answer 1

89%

Step-by-step explanation:

Given:

Input energy to a vacuum cleaner
`(E_(in))` = 6.5 kWh

Output energy to a vacuum cleaner
`(E_(out))` = 5.8 kWh

The efficiency of a machine is a measure of useful work provided by the machine.

When a machine is operated, some of its input energy is lost in friction and other wear and tear. So, the total input energy is not converted to output energy. In order to measure this loss, we make use of the term "efficiency" which is defined as the ratio of output energy to input energy expressed as a percentage.

Thus, efficiency of a vacuum cleaner is given as:

`\eta=(Out put\ energy)/(In put\ energy)* 100\\\\\eta=(E_(out))/(E_(in))* 100`

Plug in the given values and solve for
`\eta`. This gives,

`\eta=(5.8\ kWh)/(6.5\ kWh)* 100\\\\\eta=0.8923* 100\\\\\eta=89.23\approx 89\%`

So, 89% of the input energy is converted to useful work by the vacuum cleaner.

Therefore, the efficiency of the vacuum cleaner is 89%.