Question

An ideal gas heat engine operates in a Carnot cycle between ${227}^{o}C$ and ${127}^{o}C$. It absorbs 6 kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal to :

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

Answer 1

Answer: (B)

$1.2$

The efficiency of heat engine is
$\eta =\frac{W}{Q_1}=1-\frac{T_2}{T_1}$
Here, $Q_1$ = heat absorbed from the source of heat = $6kcal$ $T_1$ = temperature of source = $227+273=500K$ and $T_2$ = temperature of sink = $127+273=400K$
Hence $\frac{W}{6}=1-\frac{400}{500}or\frac{W}{6}=\frac{1}{5}$
or $W=1.2kcal$