Question

A Carnot engine takes $3\times {10}^8$ cal of heat from a reservoir at ${627}^{o}C$, and gives it to a sink at ${27}^{o}C$. The work done by the engine is :

• A. $4.2\times {10}^{6}J$
• B. $8.4\times {10}^{6}J$
• C. $16.8\times {10}^{6}J$
• D. zero

Answer 1

Answer: (B)

$8.4\times {10}^{6}J$

Given: Heat taken by the engine= $3\times {10}^{8}cal$

Initial temperature $T_1=$ ${627}^{0}C$ = 900K

Final temperature $T_2=$ ${27}^{0}C$ = 300K

Solution: As we know, according to Carnot theorem,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}$

$Q_2=\frac{T_2}{T_1}\times Q_1$

$Q_2=\frac{300}{900}\times 3\times {10}^6$

$=1\times {10}^{6}cal$

Work by the engine is given by,

$Q_1-Q_2$

$3\times {10}^6-1\times {10}^6$

$=2\times {10}^6$

Or,

$2\times {10}^6\times 4.2J$

$=8.4\times {10}^6$

Hence, the correct option is (B).