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

• The thermal efficiency is 0.4113.

Step-by-step explanation:

We know that the thermal efficiency is the ratio of work done by the engine over the heat taken

`\eta = (W_(made))/(Q_(in))`

Now, how much work the engine do in a cycle?

We know that the work done in a cycle must be equal to the heat taken minus the heat rejected

`W{made} = Q_(in) - Q_(rejected)`

So, the thermal efficiency will be:

`\eta = (Q_(in) - Q_(rejected))/(Q_(in))`

`\eta = (Q_(in))/(Q_(in)) - (Q_(rejected))/(Q_(in))`

`\eta = 1 - (Q_(rejected))/(Q_(in))`

Putting the values of the problem

`\eta = 1 - (259.8 kJ )/(441.3kJ)`

`\eta = 0.4113`