A Carnot engine operating between temperature `T_1 and T_2` has efficiency 1/6. When `T_2` is lowered by 62K its efficiency increase to 1/3. Then `T_1 and T_2` are, respectively:

Options

A. `372 K and 330 K`
B. `330 K and 268 K`
C. `310 K and 248 K`
D. `372 K and 310 K`

Answer

D

Explanation

`eta= 1-(T_(2))/(T_(1))`
case (i):
`1/6= 1-(T_(2))/(T_(1)) or (T_(2))/(T_(1))= 1- 1/6= 5/6` ….(i)
Case(ii):
`1/3= 1-(T_(2)-62)/(T_(1)) or (T_(2)-62)/(T_(1))= 1-1/3= 2/3` ….(ii)
By solving (i) and (ii), we get
`T_(1)= `372K` and `T_(2)= 310K`