Question

In a Carnot engine, when the temperatures are \( T_2 = 0^\circ C \) and \( T_1 = 200^\circ C \), its efficiency is \( \eta_1 \), and when the temperatures are \( T_1 = 0^\circ C \) and \( T_2 = -200^\circ C \), its efficiency is \( \eta_2 \). Then the value of \( \frac{\eta_1}{\eta_2} \) is:

• A. \( 0.58 \)
• B. \( 0.73 \)
• C. \( 0.64 \)
• D. \( 0.42 \)

Hint:

The efficiency of a Carnot engine depends on the absolute temperatures \( T_H \) and \( T_C \). When comparing efficiencies, use \( \frac{\eta_1}{\eta_2} = \frac{(T_H - T_C)_1}{(T_H - T_C)_2} \).

Answer 1

The Correct Option is A

Step 1: Carnot Efficiency Formula
The efficiency of a Carnot engine is given by: \[ \eta = 1 - \frac{T_C}{T_H} \] where: - \( T_C \) is the temperature of the cold reservoir,
- \( T_H \) is the temperature of the hot reservoir.
Step 2: Calculating \( \eta_1 \)
For the first case: \[ T_H = 200^\circ C = 473 K, \quad T_C = 0^\circ C = 273 K. \] \[ \eta_1 = 1 - \frac{273}{473}. \] \[ \eta_1 = 1 - 0.577. \] \[ \eta_1 = 0.423. \] Step 3: Calculating \( \eta_2 \)
For the second case: \[ T_H = 0^\circ C = 273 K, \quad T_C = -200^\circ C = 73 K. \] \[ \eta_2 = 1 - \frac{73}{273}. \] \[ \eta_2 = 1 - 0.267. \] \[ \eta_2 = 0.733. \] Step 4: Finding \( \frac{\eta_1}{\eta_2} \)
\[ \frac{\eta_1}{\eta_2} = \frac{0.423}{0.733}. \] \[ = 0.577 \approx 0.58. \] Step 5: Conclusion
Thus, the required value is: \[ \frac{\eta_1}{\eta_2} = 0.58. \]