An ideal gas heat engine operates in a carnot cycle between `227^(@)C` and `127^(@)C` .It absorbs 6 kcal at the higher temperature. The amount of heat (in kcal converted into work is equal to

Options

A. 1.6
B. 1.2
C. 4.8
D. 3.5

Answer

B

Explanation

The efficiency of heat engine is `eta=1-(T_(2))/(T_(1))`
`"or" " " (W)/(Q_(1))=1-(T_(2))/(T_(1))`
Here , `Q_(1)` Heat absorbed from the source of heat
= 6 kcal
`T_(1)` = Temperature pf source =227+273=500 K
and `T_(2)` =Temperature of sink =127+273=400 K
`"Hence" " " (W)/(6)=1-(400)/(500)`
`"or" " " (W)/(6)=(100)/(500)"or "W=1.2"kcal"`