Q124 / 180 · COMEDK 2022 · Physics
If a Carnot heat engine has a sink at a temperature of $27^\circ\text{C}$ and takes $100\text{ kcal}$ of heat from a source at a temperature of $627^\circ\text{C}$, the work done (in J) is:
- Correct answerA$0.28 \times 10^6$
- B$2.8 \times 10^6$
- C$28 \times 10^6$
- D$0.028 \times 10^6$
Solution
$T_2 = 27 + 273 = 300\,\text{K}$, $T_1 = 627 + 273 = 900\,\text{K}$
$\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{900} = \dfrac{2}{3}$
$Q_1 = 100\,\text{kcal} = 10^5 \times 4.2\,\text{J} = 4.2 \times 10^5\,\text{J}$
$W = \eta \times Q_1 = \dfrac{2}{3} \times 4.2 \times 10^5 = 2.8 \times 10^5 = 0.28 \times 10^6\,\text{J}$
Correct option: (1)
$\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{900} = \dfrac{2}{3}$
$Q_1 = 100\,\text{kcal} = 10^5 \times 4.2\,\text{J} = 4.2 \times 10^5\,\text{J}$
$W = \eta \times Q_1 = \dfrac{2}{3} \times 4.2 \times 10^5 = 2.8 \times 10^5 = 0.28 \times 10^6\,\text{J}$
Correct option: (1)
Hence, the correct answer is option: 'A'.