Question

A computer that is 87% efficient consumes 375 kWh of energy. How much useful energy does it provide?

Answer 1

326.25 kWh

Step-by-step explanation:

Efficiency of a machine is defined as the ratio of useful energy to that of the energy consumed by the machine.

Here, efficiency is given as 87% and the energy consumed by the computer is 375 kWh.

Efficiency,
`\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}`

Plug in the values of
`\eta=0.87` and 375 kWh for energy consumed. Solve for useful energy. This gives,

Efficiency,
`\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}\\ 0.87=\frac{\textrm{Useful energy}}{375}\\ \textrm{Useful energy}=0.87* 375=326.25 \textrm{ kWh}`

Therefore, the useful energy provided by the computer is 326.25 kWh.