A Carnot engine operating between temperatures T1 and T2 has efficiency 0.2. When T2 is reduced by 50 K, its efficiency increases to 0.4. Then, `T_(1) and T_(2)` are respectively

Options

A. 200 K, 150K
B. 250 K, 200 K
C. 300 K, 250 K
D. 300 K, 200 K

Answer

B

Explanation

By Carnot.s ideal engine
`eta =1 - (T_(2))/( T _(1))`
where `eta _(i) =0.2, eta _(2) = 0.4`
For the first condition
`eta (1) = 1 - ( T _(2))/( T _(1))`
`0.2 =1 - (T_(2))/(T_(1) )`
or `0.2 =(T_(1) - T _(2))/( T _(1)) " "...(i)`
For the second condition
`eta _(2) =1 - (T_(2) - 50)/(T_(1))`
`0.4 = (T _(1) - (T _(2) - 50))/( T _(1))`
`0.4 = (T_(1) - T _(2) + 50)/( T _(1))`
`= 0.4 = (T_(1) - T _(2))/(T_(1)) + (50)/( T _(1))`
`0.4 = 0.2 + (50)/(T _(1))` [From Eq. (i)]
`0.4 - 0.2 = (50)/( T _(1))`
`T _(1) = (50)/(0.2)`
`T _(1) = 250 K`
Putting the value of `T _(1)` in Eq. (i) we get
`0.2 = (T _(1) - T _(2))/( T _(1))`
`T _(2) = T _(1) =0.2 T _(1)`
`T_(2) = 250 - 0.2 xx 250`
`T _(2) = 250 - 50`
`T _(2) = 200 K`
So, `T _(1) = 250 K, T _(2) = 200 K`