Efficiency of Carnot engine is `100%` if

Options

A. sing is at`0^(@)C`
B. sink is at 0K
C. source is at `273^(@)C`
D. source is at `100^(@)C`

Answer

b

Explanation

Efficiency Carnot's cycle is `eta=(T_(2) - T_(1))/(T_(2))`
`T_(1) =` Temperature of sink,
`T_(2) =` Temperature of source
For`100%` efficiency , `eta= 1`
Hence, `1= 1- (T_(1))/(T_(2))` or `(T_(1))/(T_(2))=0` or `T_(1) = 0 K`