Question

A carnot engine absorbs 12,000 J from the hot reservoir and discards 6,000 J into the cold reservoir at 300 K. what is the temperature of the hot reservoir and the entropy change of the carnot cycle?

Answer 1

Answer:

Given:

Heat absorbed, Q1 =12,000J

Heat rejected, Q2=6,000J

Temperature of reservoir, T1

=127°C=127+273=300K

To find:

Efficiency of the engine, η=?

Temperature of sink, T

2

=?

Amount of useful work done per cycle, W=?

We know,

Efficiency of Carnot engine can be found be using the following formula,

η=

Q

1

Q

1

−Q

2

=

1000

1000−600

=0.4 or 40%

Hence, the efficiency of the engine is 40%

We know, Efficinecy of carnot engine is also equal to

η=

T

1

T

1

−T

2

∴0.4=

400

400−T

2

⟹400−T

2

=160

⟹T

2

=400−160=240K or 240−273=−33°C

Hence, the temperature of the sink is -33°C.

We know,

W=Q

1

−Q

2

=1000−600=400J

Hence the amount of useful work done during each cycle is 400J.