1.

The efficiency of an ideal heat engine working between the freezing point and boiling point of water is

Answer»

`26.8%`
`20%`
`6.25%`
`12.5%`

Solution :Efficiency of an ideal heat engine, `eta = (1- (T_(2))/(T_(1)))`
FREEZINGPOINT of water `= 0^(@)C = 273 K`
Boiling POINT of water `= 100^(C) = (100 + 273)K = 373 K`
`T_(2)` : Sink temperature = 273 K
`T_(1)` : Source temperature = 373 K
`% eta = (1-(T_(2))/(T_(1))) xx 100 = (1-(273)/(373)) xx 100 = 26.8%`


Discussion

No Comment Found