Question

Heena engine gives out 500 joule of heat energy and useful for determine the energy supplied to it as input of its efficiency is 40%​

Answer 1

Answer:

hello mate..

Explanation:

Given : For Carnot engine,

⇒T

1

=600K T

2

=300K

⇒W=800J

We know,

Efficiency (n)=1−

T

1

T

2

=

Q

W

⇒

Q

W

=1−

600

300

⇒Q=

1−

2

1

W

⇒

2

1

800

⇒Q=1600J

i.e, heat given to system is 1600J.

Hence,

the answer is 1600.

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Answer 2

Answer:

1250 J

Explanation:The heat engine gives out=500 J

                      Efficiency=40%

                      Efficiency=output energy/input energy

40/100=500/energy supplied

energy supplied=500*100/40 = 1250 J