Question

An ideal gas heat engine operates in carnot cycle between ${227}^{o}C$ and ${127}^{o}C$. It absorbs $6\times {10}^{4}cal$ of heat at higher temperature. Amount of heat converted into work, is

• A. $1.2\times {10}^{4}cal$
• B. $2.4\times {10}^{4}cal$
• C. $6.0\times {10}^{4}cal$
• D. $4.8\times {10}^{4}cal$

Answer 1

The correct option is A $1.2\times {10}^{4}cal$

For a Carnot engine, efficiency $e=1-\frac{T_2}{T_1}=1-\frac{127+273}{227+273}=1-\frac{400}{500}=\frac{1}{5}$
Now, $e=\frac{Workoutput}{Heatinput}=\frac{W}{6\times {10}^4}$
$W=e\times 6\times {10}^4=\frac{1}{5}\times 6\times {10}^4=1.2\times {10}^{4}cal$