Question

A steam engine absorbs 1.98 × 105 j and expels 1.49 × 105 j in each cycle. assume that all of the remaining energy is used to do work.

a. what is the engine's efficiency?
b. how much work is done in each cycle?

Answer 1

b) The heat absorbed by the engine is
`Q_(in) = 1.98 \cdot 10^5 J` while the heat expelled is
`Q_(out) = 1.49 \cdot 10^5 J`, therefore the work done by the engine is the difference between the heat absorbed and the heat expelled:

`W=Q_(in) - Q_(out) = 1.98 \cdot 10^5 J - 1.49 \cdot 10^5 J = 0.49 \cdot 10^5 J`

a) The efficiency of the engine is the ratio between the work done by the engine and the heat absorbed, therefore:

`\eta= (W)/(Q_(in)) = (0.49 \cdot 10^5 J)/(1.98 \cdot 10^5 J)=0.247 = 24.7\%`