Question

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?

Answer 1

`8.6kJ`

Explanation:

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

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

Energy supply, `Q=8.6kJ`

Answer 2

`8.6kJ`

Explanation:

Mechanical Work done by the pump`=8.3kJ`
Heat generated in the pump, due to internal resistance both electrical and friction`=300J`
Also given that change in energy of pump `DeltaE=0`

`:.` Total electrical energy was required`="Mechnical work done"+"Heat generated and lost"+Delta E`

Inserting giving values we get

Total electrical energy was required`=8.3+0.3+0=8.6kJ`