Question

Efficiency of a Carnot’s cycle changes from 1/6 to 1/3 when source temperature is raised by 100K.

Calculate the temperature of the sink.​

Answer 1

Answer :

Efficiency of carnot's cycle changes from 1/6 to 1/3 when source temperature is raised by 100K.

We have to find temperature of the sink$.$

★ Efficiency of carnot's engine is given by

• n = 1 - T'/T

T' denotes temperature of sink

T denotes temperature of source

★ First case :

→ 1/6 = 1 - T'/T

→ T'/T = 1 - 1/6

→ T'/T = 5/6

→ T = 6T'/5

★ Second case :

→ 1/3 = 1 - T'/(T + 10)

→ T'/(T + 10) = 1 - 1/3

→ T'/(T + 10) = 2/3

→ T' = 2/3 (T + 10)

→ T' = 2/3 (6T'/5 + 10)

→ T' = 2/3 (6T' + 50)/5

→ T' = (12T' + 100) / 15

→ 15T' - 12T' = 100

→ 3T' = 100

→ T' = 33.33K

Answer 2

The answer is 33.33K

Efficiency of carnot's cycle changes from 1/6 to 1/3 when source temperature is raised by 100K.

Formula for Efficiency of carnot's engine is given by

n = 1 - T'/T

Case 1

→ 1/6 = 1 - T'/T

→ T'/T = 1 - 1/6

→ T'/T = 5/6

→ T = 6T'/5

Case 2

→ 1/3 = 1 - T'/(T + 10)

→ T'/(T + 10) = 1 - 1/3

→ T'/(T + 10) = 2/3

→ T' = 2/3 (T + 10)

→ T' = 2/3 (6T'/5 + 10)

→ T' = 2/3 (6T' + 50)/5

→ T' = (12T' + 100) / 15

→ 15T' - 12T' = 100

→ 3T' = 100

→ T' = 33.33K

Therefore temperature of the sink is 33.33K

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