Acarnot heat engine receives heat from a heat source at 1200°c and has a thermal efficiency of 40 percent. the heat engine does maximum work equal to 490 kj. determine the heat supplied to the heat engine by the heat source, the heat rejected to the heat sink, and the temperature of the heat sink. (round the final answers to the nearest whole number.)

T₂=610.8⁰C

Qa=1225 KJ

Qr=735 KJ

Explanation:

Given that

T₁ = 1200⁰C

T₁ =1473 K

η= 40% = 0.4

We know that efficiency of Carnot cycle given as

T₁ = temperature of source

T₂= temperature of sink

T₂=883.8 K

T₂=610.8⁰C

Lets take heat supplied is Qa

We know that

W= Net work output

Qa=1225 KJ

From first law of thermodynamics

Qa= W+ Qr

Qr=Heat rejected to sink

Qa= W+ Qr

1225 = 490 + Qr

Qr=735 KJ

graphing motion: mastery test

209be

what is the instantaneous acceleration of

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0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

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