An ideal gas heat engine operates in carnot’s cycle between 227oc and 127oc. If absorbs 6×104 cals of heat at higher temperature amount of heat converted to work is
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a
4.8×104 cals
b
1.2×104 cals
c
2.4×104 cals
d
6×104 cals
answer is B.
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Detailed Solution
Temperature T1=227+273=500 K (converting degree celsius into kelvin) :T2=127+273=400 K effieciency of heat engine is η=1-T2T1---(1) substitute given temperatures in equation (1) ⇒η=1-400500 ⇒ η=15---(2) also efficiency is η=Work done (w)heat given to engine(Q)---(3)substitute given Q=6×104calorie, and equation (2) in (3) ⇒15=W6×104 W=6×1045 W=1.2×104 calamount of heat converted to work =1.2×104 cal