Art, asked by malik2313, 1 year ago

the efficiency of carnot cycle is 1 by 6 if on reducing the temperature of the sink by 65 degree Celsius the efficiency become 1 by 3 find the initial and final temperatures between which the cycle is working​

Answers

Answered by microsky219
3

Answer:

The efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .

condition 1 :- when η ( efficiency of engine ) = 1/6

then, 1/6 = 1 - T₂/T₁ ---------(1)

condition 2 : when temperature of sink is reduced by 65K then, η(efficiency of engine ) = 1/3

e.g, 1/3 = 1 - (T₂ - 65)/T₁ -----------(2)

solve equations (1) and (2)

2(1 - T₂/T₁) = 1 - (T₂ - 65)/T₁

2 - 1 = 2T₂/T₁ - (T₂ - 65)/T₁ = (T₂ + 65)/T₁

T₁ = T₂ + 65

Put it in equation (1)

1/6 = 1 - T₂/(T₂ + 65)

1/6 = (65)/(T₂ + 65)

T₂ + 65 = 390 ⇒ T₂ = 325K or 52°C

now T₁ = 390K or 117°C

Explanation:

make it the brinliest answer

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