Question

Give two expressions to calculate the efficiency of a heat engine.

Answer 1

`\eta=(W)/(Q_(c))\\\\ \eta=(Q_(c)-Q_(f))/(Q_(c))`

Step-by-step explanation:

We have the expression for a heat engine as:

`Q_(c)=W+Q_(f)`

Where Qc is the hot source, or the energy that the engine receives, W is the work produced by the engine and Qf is the heat that is delivered to a cold spot.

If we have a 100% efficiency, all the energy that the engine receives turns into output work, so Qf is zero, therefore, from the original expression we can define a formula for efficiency:

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

Now, by doing some algebra, we can have the following expression:

`\eta=(Q_(c)-Q_(f))/(Q_(c))`

That also can be expressed as:

`\eta=1-(Q_(f))/(Q_(c))`