Question

a carnot engine operates between a temperature 500 k and 300k the system absorbs 1500j of heat Fromm a hot reservoir find the efficiency of the engine​

Answer 1

Answer:

Correct option is

A

600

Refer image.

We have to find Q

2

from first law of thermal dynamic for a cyclic process of heat engine

∮Q=∮W

Q

1

−Q

2

=W

⇒Q

2

=Q

1

−W

Now from efficiency of cannot cycle.

η=

Q

1

W

=1−

T

1

T

2

(

T

2

=300k

T

1

=900k

)

=1−

900

300

=

900

600

=

3

2

⇒

Q

1

W

=

3

2

⇒Q

1

=

2

3×W

=

2

3×1200

=1800J/cycle

∴ Heat transferred to low temperature reservoir

⇒Q

2

=Q

1

−W=1800−1200

Q

2

=600J/cycle

solution