1.

Efficiency of a Carnot's engine is 100% when

Answer»

sing is at`0^(@)C`
sink is at 0K
source is at `273^(@)C`
sourceis at`100^(@)C`

Solution :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`


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