Q104 / 125 · BITSAT 2013 · Physics
A Carnot's heat engine works between the temperatures $427^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$. What amount of heat should it consume per second to deliver mechanical work at the rate of $1.0 \mathrm{~kW} ?$
- Correct answerA$0.417 \mathrm{kcal} / \mathrm{s}$
- B$4.17 \mathrm{kcal} / \mathrm{s}$
- C$41.7 \mathrm{kcal} / \mathrm{s}$
- D$0.212 \mathrm{kcal} / \mathrm{s}$
Solution
The efficiency of the heat engine is
$$
\eta=1-\frac{T_{2}}{T_{1}}=1-\left(\frac{273+27 K}{273+427 K}\right)=\frac{4}{7}
$$
But $\quad \eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}$
$$
\therefore \mathrm{Q}_{1}=\frac{\mathrm{W}}{\eta}=\frac{1.0 \mathrm{~kW}}{4 / 7}=1.75 \mathrm{~kW}=0.417 \mathrm{kcal} / \mathrm{s}
$$
Thus, the engine would require 417 cal of heat per second, to deliver the requisite amount of work.
Hence, the correct answer is option: 'A'.