# An electrically powered pump is used to pull water out of a well. In the course of an afternoon, the pump does 8.3 kJ of work lifting water out of a well?

## An electrically powered pump is used to pull water out of a well. In the course of an afternoon, the pump does 8.3 kJ of work lifting water out of a well. It also heats up, and releases 300 J of heat to the environment before returning to its original temperature. If there was no net change in the energy of the pump, how much electrical energy was required?

May 9, 2016

$8.6 k J$

#### Explanation:

In thermodynamics, you have the famous equation;
$Q = U + W$
Which is; $\text{Energy supply"="Internal energy"+"Work done by the system}$

The system heats up hence $U = 300 J$
Work is done by the system hence $W = 8.3 k J$

Energy supply, $Q = 8.6 k J$

May 9, 2016

$8.6 k J$

#### Explanation:

Mechanical Work done by the pump$= 8.3 k J$
Heat generated in the pump, due to internal resistance both electrical and friction$= 300 J$
Also given that change in energy of pump $\Delta E = 0$

$\therefore$ Total electrical energy was required$= \text{Mechnical work done"+"Heat generated and lost} + \Delta E$

Inserting giving values we get

Total electrical energy was required$= 8.3 + 0.3 + 0 = 8.6 k J$