In a Carnot's engine, the temperature of the heat source and heat sink 500 K and 375 K respectively. If the engine consumes 6x 10^6 J per cycle, find (6) the efficiency of engine (i) work done per cycle by the engine, and (ii) heat rejected by the engine in each cycle of operation.
Do this ques fastly plz :)
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mark brainliest plzzzz.....
Given :
T1=500K
T2=375K
Q1=heat absorbed=600kca
l) Efficiency of engine:
Formula :
Efficiency of engine=n=1-(T2/T1)
n=T1-T2/T1
=500-375/500
=125/500
=0.25
n=0.25 x 100=25%
ii)Let us assume that W be the work done
n=W/Q1
W=nxQ1=0.25x600
=150kcal=150 x 10³ x 4.2 joules
=6.3 x 10⁵ Joules
iii) Let Q2 be the heat rejected to sink
W=Q1-Q2
Q2=Q1-W
=600-150
=450Kcal
Ishan90:
Is 6 × 10^6 = 600 kcal ??
Reply fast plz
no
it means 10x10x10x10x10x10x6
mark brainliest plzzzz.....
No I mean 6 × 10^6 joules ( in ques ) , but u have written it as 600 kcal
Just describe me how u have done this , then I will mark u as brainliest :)
I have already seen this answer before it , m confused only in it ( 6* 10^6)
just convert Joule value to kca
I have converted already but solution is so different than 600 kcal , it seems 1434
Answered by
2
Given :
T1=500K
T2=375K
Q1=heat absorbed=600kca
l) Efficiency of engine:
Formula :
Efficiency of engine=n=1-(T2/T1)
n=T1-T2/T1
=500-375/500
=125/500
=0.25
n=0.25 x 100=25%
ii)Let us assume that W be the work done
n=W/Q1
W=nxQ1=0.25x600
=150kcal=150 x 10³ x 4.2 joules
=6.3 x 10⁵ Joules
iii) Let Q2 be the heat rejected to sink
W=Q1-Q2
Q2=Q1-W
=600-150=450Kcal
T1=500K
T2=375K
Q1=heat absorbed=600kca
l) Efficiency of engine:
Formula :
Efficiency of engine=n=1-(T2/T1)
n=T1-T2/T1
=500-375/500
=125/500
=0.25
n=0.25 x 100=25%
ii)Let us assume that W be the work done
n=W/Q1
W=nxQ1=0.25x600
=150kcal=150 x 10³ x 4.2 joules
=6.3 x 10⁵ Joules
iii) Let Q2 be the heat rejected to sink
W=Q1-Q2
Q2=Q1-W
=600-150=450Kcal
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