The efficiency of a carnot heat engine is 0.6 when the absolute temperatures of the source and sink are `T_(1) and T_(2)` respectively.
The efficiency of another carnot engine is also 0.6 but the source and sink temperatures are different. What are the temperatures of the source and the sink of the other engine ?

Options

A. `T_(1)+5,T_(2)+5`
B. `T_(1)+10,T_(2)-10`
C. `2T_(1),2T_(2)`
D. `2T_(1)andT_(2)/2`

Answer

C

Explanation

For a carnot heat engine, efficiency `eta=1-T_(2)/T_(1)`
For the first carnot engine, `0.6=1-T_(2)/T_(1)`
`thereforeT_(2)/T_(1)=1-0.6=0.4`
For the second carnot engine,
`0.6=1-(T'_(2))/(T'_(1))`
Since the two engines have the same efficiency
`T_(2)/T_(1)=(T'_(2))/(T'_(1))`
This os possible, if `T_(1)andT_(2)` are changed by the same factor.
i.e. `(T'_(2))/(T'_(1))=(2T_(2))/(2T_(1))or(3T_(2))/(3T_(1))or"even"(T_(2)/2)/(T_(1)/2)`
Hence out of the four options, the correct option is `2T_(1) and 2T_(2)`.