Question

An ideal gas heat engine operates in Carnot cycle between 227${}^{\circ }$C and 127${}^{\circ }$C. It absorbs $6\times {10}^4$ cals of heat at the higher temperature. Amount of heat converted to work is:

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

Answer 1

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

$T_1={227}^{o}C=(227+273)K=500K$

$T_2={127}^{o}C=(127+273)K=400K$

Efficiency of Carnot engine= $1-\frac{T_2}{T_1}=1-\frac{400}{500}=\frac{1}{5}$

Also, efficiency of the heat engine= $\frac{Outputwork}{Inputheat}$

$\Rightarrow \frac{1}{5}=\frac{Outputwork}{6\times {10}^4}$

Therefore, Output work=$1.2\times {10}^4$ cals.