An engine takes heat from a reservoir and converts adds 1/6 part into work.By decreasing temperature of sink by 62°C its efficiency becomes double. The temperatures of source and sink must be
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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
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MARK BRAINLIEST MY ANSWER DEAR FRIEND
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✦Solution:-
▶If T1 is temperature and T2 the temperature of the sink, the efficiency of engine
⇒η = [work done (W) / Heat taken (Q1)]
⇒η = 1 -(T2/T1)
∴ 1 - T2/T1 = 1/6 _[i]
▶When temperature of sink is reduced by 62°C, then temperature of sink
T'2 = T2 - 62
∴ η' = 1 - T'2/T1
▶As according to question efficiency becomes double so,
η' = 2η = 2/6 = 1/3
∴ 1/3 = 1 - (T2-62/T1) _[ii]
▶From Eq [i] T2-T1 = 5/6 _[iii]
▶From Eq [ii] (T2-62)/T1 = 2/3 _[iv]
▶Dividing Eq [iii] by Eq [iv]
•T2/(T2-62) = 5/4
⇒ 4T2 = 5T2-310
⇒ T2 = 310 K
and from Eq [iii], we have
•310 /T1 = 5/6 ⇒T1=372 K
▶Hence, T1= 372K = 372-273 = 99°C
and T2= 310K = 310 - 273 = 37°C
As kinetic energy of a gas depends on its atomicity!
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