Question

A Carnot engine absorbed ${227}^{\circ }C$. Calculate works done for the cycle by an engine of its single is maintained at ${127}^{\circ }C$

Answer 1

Work Input (or Heat Absorbs) =$6\times 10⁴Cal.$

$T₁=127°C$

$=(127+273)K.$ [Changing into kelvin]

$=400K.$

$T₂=227°C$

$=(227+273)K$. [Changing into kelvin]

$=500K$

∵ Efficiency $=1-T₁/T₂$

$=1-400/500$

$=(500-400)/500$

$=100/500$

$=1/5$

Also, Efficiency = Work Output/Work Input

∴$ 1/5 = Work Done 6 × 10⁴$

∴ Work Done by the Carnot Engine $=60000/5$

⇒ Work Done = $12000Cal.$

∴ Amount of Heat Converted into Mechanical Energy is$12000Cal.$