An engine absorbs 1.68 kJ from a hot reservoir at 277°C and expels 1.15 kJ to a

Question

An engine absorbs 1.68 kJ from a hot reservoir at 277°C and expels 1.15 kJ to a cold reservoir at 27°C in each cycle. (a) What is the engine's efficiency? % (b) How much work is done by the engine in each cycle? J (c) What is the power output of the engine if each cycle lasts 0.298 s?

An engine absorbs 1.68 kJ from a hot reservoir at 277 C and expels 1.15 kJ to a cold reservoir at 27°C in each cycle (a) What is the engine's efficiency? (b) How much work is done by the engine in each cycle? (c) What is the power output of the engine if each cycle lasts 0.298 s? kW

Explanation / Answer

here,

Q1 = 1680 J

Q2 = 1150 J

T1 = 277 degree C

T2 = 27 degree C

a)

engine's efficiency , n = W/Q1

n = (Q1 - Q2)/Q1

n = (1680 - 1150)/1680

n = 0.32

n = 32 %

the engine's efficiency is 32 %

b)

W = Q1 - Q2

W = 0.53 kJ

work is done by engine in each cycle is 0.53 kJ

c)

t = 0.298 s

P = W/t

P = 0.53/0.298

P = 1.78 W

the power output of the engine is 1.78 W

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