Physics, asked by Teriyaki, 22 hours ago

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​

Answers

Answered by ExoticQuail
6

HEY FRIEND HERE YOUR ANSWER

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

I HOPE IT'S HELP YOU

MARK BRAINLIEST MY ANSWER DEAR FRIEND

THANKS..

Answered by Csilla
20

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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