Question

A perfect Carnot engine utilises an ideal gas and works between the temperature ${227}^{o}C$and ${127}^{o}C$. If the work output of the engine is ${10}^4$ Joule, then the amount of heat received from the source will be

• A. $1\times {10}^{5}Joule$
• B. $2\times {10}^{5}Joule$
• C. $3\times {10}^{5}Joule$
• D. $5\times {10}^{5}Joule$

Answer 1

Answer: (D)

$5\times {10}^{5}Joule$

Efficiency of the Carnot cycle is given by:
$\eta =1-\frac{T_{sink}}{T_{source}}$
$\eta =\frac{W}{Q}$
$T_{source}={227}^{\circ }C=500K$
$T_{sink}={127}^{\circ }C=400K$
$\eta =1-\frac{T_{sink}}{T_{source}}=1-\frac{400}{500}=0.2$
$\eta =\frac{W}{Q}=\frac{{10}^4}{Q}$
i.e.
$Q=\frac{{10}^4}{0.2}=5\times {10}^{5}J$