Question

A perfect Carnot engine utilizes an ideal gas. The source temperature is 500 K and sink temperature is 375 K. If the engine takes 600 kcal per cycle from the source, then calculate a. The efficiency of the engine b. Work done per cycle c. Heat rejected to the sink per cycle

Answer 1

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

Answer 2

Answer:

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

Explanation: