1. The engine's efficiency is calculated by using the formula
η=Q2Q2−Q1=1−Q2Q1,where Q1=881J and Q2=2364J are the amounts of heat, respectively, expelled and absorbed by the system in one cycle. We obtain η=1−881/2364≈0.627.
Answer: 0.627
2. The work done in each cycle is W=Q2−Q1=1483J.
Answer: 1483 J.
3. The mechanical power output is W/t, where t=0.32s is the duration of one cycle. We have W/t=1483J/0.32s≈4634W=4.634kW.
Answer: 4.634 kW.
Comments
this is all correct.
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