Question

An engine working on Carnot cycle rejects 40% of absorbed heat from the source, while the sink temperature is maintained at ${27}^{0}C$, then what is the source temperature (in ${}^{0}C$)?

• A. 477
• B. 346
• C. 564
• D. none of the above

Answer 1

The correct option is A 477

$\eta =\frac{60}{100}$

$T_2={27}^{o}C$

$T_2={300}^{o}K$

$T_1=?$

$\eta =\frac{T_1-T_2}{T_1}$

Here, $T_1=Sourcetemperature$

$T_2=sinktemperature$

$\eta =Efficiency$

$\eta =100-rejecion$

$\therefore \frac{60}{100}=\frac{T_1-T_2}{T_1}$

$\therefore 0.6=\frac{T_1-300}{T_1}$

$\therefore T_1(1-0.6)=300$

$T_1=\frac{300}{0.4}=\frac{3000}{4}={750}^{o}K$

$\therefore T_1={(750-273)}^{o}C$

$\therefore T_1={477}^{o}C$