Question

An engine that has an efficiency of 25% takes in 200 [J] of heat during each cycle. Calculate the amount of work this engine performs.

Answer 1

`W=50J`

Step-by-step explanation:

The work done
`W` and the energy taken
`Q_(in)` by a heat engine are related to the efficiency
`\eta` by the expression

`\eta=(W)/(Q_(in))`

The efficiency is
`\eta`=25%, the numerical form of this percentage is 0.25 and the energy taken is
`Q_(in)`=200J. Replacing in the formula:

`0.25=(W)/(200)\\W=200*0.25\\W=50J`