Question

An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) during each cycle. What is the thermal efficiency of the engine?

Answer 1

`\eta = 0.411`

Step-by-step explanation:

As we know that efficiency is defined as the ratio of output useful work and the input energy to the engine

So here we know that the

input energy = 441.3 kJ

energy rejected = 259.8 kJ

so we have

`Q_1 - Q_2 = W`

`W = 441.3 kJ - 259.8 kJ = 181.5 kJ`

now efficiency is defined as

`\eta = (W)/(Q_1)`

`\eta = (181.5)/(441.3)`

`\eta = 0.411`