Physics, asked by preithibagya2378, 5 months ago

Efficiency of the carnot engine change from 1/6 to 1/3 when source temperature is raised by 100k . The temperature of sink is?

Answers

Answered by Ekaro
6

Given :

Efficiency of the carnot engine changes from 1/6 to 1/3 when source temperature is raised by 100K.

To Find :

Temperature of sink.

Solution :

❒ Efficiency of carnot's engine (an ideal heat engine), is given by

  • η = 1 - T'/T

where,

  • T' denotes temp. of the sink
  • T denotes temp. of the source

Case - I :

➛ 1/6 = 1 - T'/T

➛ T'/T = 1 - 1/6

➛ T'/T = 5/6

T = 6T'/5 .......... (I)

Case - II :

In this case temperature of the sink is raised by 100K.

➛ 1/3 = 1 - T'/(T + 100)

➛ T'/(T + 100) = 2/3

➛ 3T' = 2T + 200

From the equation (I),

➛ 3T' = 12T'/5 + 200

➛ 15T' - 12T' = 1000

➛ T' = 1000/3

T' = 333.33k

Answered by gggggufcycyyccyc
0

Explanation:

Given :

Efficiency of the carnot engine changes from 1/6 to 1/3 when source temperature is raised by 100K.

To Find :

Temperature of sink.

Solution :

❒ Efficiency of carnot's engine (an ideal heat engine), is given by

η = 1 - T'/T

where,

T' denotes temp. of the sink

T denotes temp. of the source

Case - I :

➛ 1/6 = 1 - T'/T

➛ T'/T = 1 - 1/6

➛ T'/T = 5/6

➛ T = 6T'/5 .......... (I)

Case - II :

In this case temperature of the sink is raised by 100K..

➛ 1/3 = 1 - T'/(T + 100)

➛ T'/(T + 100) = 2/3

➛ 3T' = 2T + 200

From the equation (I),

➛ 3T' = 12T'/5 + 200

➛ 15T' - 12T' = 1000

➛ T' = 1000/3

➛ T' = 333.33k

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