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:
$\eta= \frac{W}{Q_{in}} = \frac{0.49 \cdot 10^5 J}{1.98 \cdot 10^5 J}=0.247 = 24.7\%$