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. The amount of heat converted into work is equal to:

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

Answer 1

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

Efficiency of a Carnot cycle $=1-\frac{T_L}{T_H}$

$=1-\frac{127+273}{227+273}$ $=0.2=\frac{W}{Q_{in}}$

$⟹W=0.2Q_{in}=0.2\times 6\times {10}^{4}cal$ $=1.2\times {10}^{4}cal$