Q.

A heat engine absorbs 2000 cal of heat and rejects 6300 J to sink. What is its efficiency?

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a

60%

b

25%

c

40%

d

50%

answer is B.

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Detailed Solution

efficiency of heat engine is η=Work doneheat given(Q1)=WQ1heat given is Q1 =2000calorie heat rejected is Q2= 6300 joule=63004.2calorie;(conversion 1calorie =4.2joule); work done=Q1−Q2; substitute in efficiency equation, η=Q1−Q2Q1⇒η=2000−63004.22000 ⇒η=2000−15002000⇒η=5002000=14⇒percentage efficiency =η=14×100=25%.
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