the efficiency of a carnot cycle is 1/6.If on reducing the temperature of the sink 75 degrees celcius ,the efficiency becomes 1/3,determine he initial and final temperatures between which the cycle is working.

450°C

Explanation: Given that the efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .

η = 1 - T2 / T1

η = 1/6 initially

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

Solution

η = 1/6

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

When η = 1/3 :

η = 1 - ( T2 - 75 ) / T1

1/3 = 1 - (T2 - 75)/T1(2)

T2 - T1 = -75 [ because T2 is reduced by 75°C ]

T2 = T1 - 75(3)

Put this in (2) :

> 1/3 = 1 - ( T1 - 75 - 75 ) / T1

> 1/3 = 1 - (T1 - 150 ) /T1

> (T1 - 150) / T1 = 1 - 1/3

> ( T1 -150 ) / T1 = 2/3

> 3 ( T1 - 150 ) = 2 T1

> 3 T1 - 450 = 2 T1

Collecting the like terms

3 T1- 2 T1 = 450

T1 = 450

The temperature initially was 450°C

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5