Questions
A heat engine receives 50 kcal of heat from the source per cycle, and operates with an efficiency of 20%. The heat rejected by engine to the sink per cycle is
a
40 kcal
b
25 kcal
c
30 kcal
d
50 kcal
detailed solution
Correct option is A
Given Q1 = 50 kcal and η = 20%Using η = WQ1, we have W = η ×Q1i.e., work obtained per cycle W = 20% × 50 kcal = 10 kcal = 42kJSince, Q1 = Q2+W so, Q2 = Q-Wi.e., heat rejected to the sink per cycle Q2 = 50 kcal -10 kcal = 40 kcalTalk to our academic expert!
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For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum?
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