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
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
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