Question

A certain heat engine does 30.2 kJ of work and dissipates 9.14 kJ of waste heat in a cyclical process.

A) What was the heat input to this engine?
B) What was its efficiency?

Answer 1

a)
`H_(in)=39.34 kJ`

b) Efficiency=76.77%

Step-by-step explanation:

a)

In order to solve this problem, we can use the following formula:

`H_(in)=H_(out)+W`

the problem provides us with all the necessary information so we can directly use the formula:

`H_(in)=9.14kJ+30.2kJ`

`H_(in)=39.34 kJ`

b) In order to find the efficiency, we can use the following formula:

`Efficiency=(W)/(H_(in))*100\%`

so we get:

`Efficiency=(30.2kJ)/(39.34kJ)*100\%`

Efficiency=76.77%