The efficiency of Carnot's heat engine is 0.5 when the temperature of the source is `T_(1)` and that of sink is `T_(2)`. The efficiency of another Carnot's heat engine is also 0.5 Then temperature of source and sink of the second engine are respectively

Options

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

Answer

A

Explanation

Efficiency of Carbot.s heat engine,
`eta=1-(T_(2))/(T_(1))`
For first Carnot.s heat engine, `0.5=1-(T_(2))/(T_(1))`
For second heat engine, `0.5=1-(T_(2)^(.))/(T_(1)^(.))`
Efficiency remains the same when both `T_(1)` and `T_(2)` are increased by same factor
Therefore, the temperature of source and sink of second engine are
`T_(1)^(.)=2T_(1^(.))T_(2)^(.)=2T_(2)`