Asked by Tulika
A reversible engine converts 1/6 of the heat input into work. When the temperature of the sink is reduced by 62C, the efficiency of the engine is doubled.What are the temperarure of the source and sink?
Answers
Answered by
Elena
Efficiency η = W/Q = 1/6,
If T1 is the temperature of the source,
and T2 is the temperature of the sink,
η = (T1-T2)/T1 = 1- T2/T1
η = 1/6 = 1- T2/T1
2 η = 2/6 = (T1-(T2- 62))/T1 =1- (T2-62)/T1,
2/6 -1/6 = 1-(T2-62)/T1 – 1 - T2/T1.
1/6 = 62/T1
T1 = 372 K.
T2/T1 =1 – 1/6 = 5/6.
T2 = (5/6) •T1 = 310 K
If T1 is the temperature of the source,
and T2 is the temperature of the sink,
η = (T1-T2)/T1 = 1- T2/T1
η = 1/6 = 1- T2/T1
2 η = 2/6 = (T1-(T2- 62))/T1 =1- (T2-62)/T1,
2/6 -1/6 = 1-(T2-62)/T1 – 1 - T2/T1.
1/6 = 62/T1
T1 = 372 K.
T2/T1 =1 – 1/6 = 5/6.
T2 = (5/6) •T1 = 310 K
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