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Grade 12th passDiscuss with colleagues and IITians

a carnot engine absorbs 200J of heat from a reservoir at the temperature of normal boiling point of water and rejects heat to a reservoir at the temperature of the triple point off water. Find the heat rejected, the work done by the engine and the thermal efficiency.

MANOJ CHAUHAN

8 Years agoGrade 12th pass

1 Answer

Arun

6 Years ago

Q1 = 1000 J taking from source

Q2 = 600 J to sink

W = useful work done per cycle = Q1 - Q2 = 400 J

η = efficiency of the Carnot engine = W / Q1 =(Q1 - Q2)/ Q1 *100 = 40%

T1 = 127° C = 400°K

Q1 / Q2 = T1/T2 => 1000/600 = 400/T2

T2 = 240°K = - 33 °C

Alternately, η = 0.4 = (T1 - T2)/T1

So T1 - T2 = 400*0.4 = 160 K

T2 = 240 K

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