Question

the efficiency of carnot engine is 1/6 if on reducing the temperature the temperature of sink by 65k the efficnecy becomes 1/3 find the intial and final temperature btween wich the cycle is workng

Answer 1

The efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .
$\eta=1-\frac{T_2}{T_1}$

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

Answer 2

Formula :

We know that :

Efficiency = 1 - T2 / T1

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Given information :

η = 1/6 initially

when T2 is reduced by 65°C then η becomes 1/3

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Solution

η = 1/6

==> 1 - T2 / T1 = 1/6 [ using the Formula ]........................(1)

When η = 1/3 :

η = 1 - ( T2 - 65 ) / T1

==> 1/3 = 1 - (T2 - 65)/T1.........................(2)

==> T2 - T1 = -65 [ because T2 is reduced by 65°C ]

==> T2 = T1 - 65...........................(3)

Put this in (2) :

==> 1/3 = 1 - ( T1 - 65 - 65 ) / T1

==> 1/3 = 1 - (T1 - 130 ) /T1

==> (T1 - 130) / T1 = 1 - 1/3

==> ( T1 -130 ) / T1 = 2/3

==> 3 ( T1 - 130 ) = 2 T1

==> 3 T1 - 390 = 2 T1

==> T1 = 390

The temperature initially was 390°C

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