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

1 Answer

4 votes

Answer:

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.

User Jacob Saylor
by
8.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.