# A 1.5 kilowatt electric motor on a water well pumps water from 10 meters below the surface. The...

## Question:

A 1.5 kilowatt electric motor on a water well pumps water from 10 meters below the surface. The density of the water is 1 kilogram per liter. How many liters of water does the motor pump in one hour?

## Conservation Of Energy

Energy can only be converted from one form of energy to another form of energy for an isolated system but we can neither create energy nor we can destroy energy. For example, during free fall gravitational potential energy gets converted into kinetic energy of the falling body.

Data Given

• Power of the motor {eq}P = 1.5 \ kW = 1500 \ W {/eq}
• Height up to which water has to be raised {eq}h = 10 \ m {/eq}
• Time {eq}t = 1 \ h = 3600 \ s {/eq}

Hare the electrical energy gets converted into gravitational potential energy, so the electrical energy consumed by the electric motor as electric power is given by

{eq}\begin{align} \color{red}{P = \frac{E}{t}} \end{align} {/eq}

{eq}\begin{align} E = P.t \end{align} {/eq}

{eq}\begin{align} E = 1500 \ W \times 3600 \ s = 5.40 \times 10^6 \ J \end{align} {/eq}

Now this electrical energy will change the gravitational potential energy of the water

{eq}\begin{align} \color{red}{E = PE = mgh} \end{align} {/eq}

{eq}\begin{align} m = \frac{E}{gh} \end{align} {/eq}

{eq}\begin{align} m = \frac{5.40 \times 10^6 \ J}{9.8 \ m/s^2 \times 10 \ m} = 55102 \ kg \end{align} {/eq}

As we know that {eq}\rho_w = 1 \ kg/L {/eq} so the volume of the water pumped

{eq}\begin{align} \color{blue}{\boxed{V =55102 \ L}} \end{align} {/eq} 