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 to work is

• A. $4.8\times {10}^{4}cal$
• B. $6\times {10}^{4}cal$
• C. $2.4\times {10}^{4}cal$
• D. $1.2\times {10}^{4}cal$

Answer 1

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

We know that efficiency of Carnot engine = $1-\frac{T_L}{T_H}$
Also, Efficiency of Heat engine = $\frac{\text{Work output}}{\text{Heat input}}$
$\therefore$$1-\frac{T_L}{T_H}=\frac{W}{Q_s}$
$\Rightarrow$$W=Q_s(1-\frac{T_L}{T_H})$
=$6\times {10}^4(1-\frac{127+273}{227+273})$
= $1.2\times {10}^{4}cal$