In a Carnot engine when `T_(2) = 0^(@)C` and `T_(1) = 200^(@)C` its efficiency is `eta_(1)` and when `T_(1) = 0^(@)C` and `T_(2) = -200^(@)C`. Its efficiency is `eta_(2)`, then what is `eta_(1)//eta_(2)`?

Options

A. 0.577
B. 0.733
C. 0.638
D. Cannot be calculated

Answer

To solve the problem, we need to calculate the efficiencies of the Carnot engine in two different scenarios and then find the ratio of these efficiencies.

### Step-by-Step Solution:

1. **Understanding the Carnot Efficiency Formula**:
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \(T_1\) is the temperature of the hot reservoir and \(T_2\) is the temperature of the cold reservoir. Note that temperatures must be in Kelvin.

2. **Calculating η₁**:
For the first case, we have:
- \(T_1 = 200^\circ C = 200 + 273 = 473 \, K\)
- \(T_2 = 0^\circ C = 0 + 273 = 273 \, K\)
Now, substituting these values into the efficiency formula:
\[ \eta_1 = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{473} \]
Simplifying this gives:
\[ \eta_1 = \frac{473 - 273}{473} = \frac{200}{473} \]

3. **Calculating η₂**:
For the second case, we have:
- \(T_1 = 0^\circ C = 0 + 273 = 273 \, K\)
- \(T_2 = -200^\circ C = -200 + 273 = 73 \, K\)
Now, substituting these values into the efficiency formula:
\[ \eta_2 = 1 - \frac{T_2}{T_1} = 1 - \frac{73}{273} \]
Simplifying this gives:
\[ \eta_2 = \frac{273 - 73}{273} = \frac{200}{273} \]

4. **Finding the Ratio η₁ / η₂**:
Now we need to find the ratio of the two efficiencies:
\[ \frac{\eta_1}{\eta_2} = \frac{\frac{200}{473}}{\frac{200}{273}} = \frac{200}{473} \times \frac{273}{200} \]
The \(200\) in the numerator and denominator cancels out:
\[ \frac{\eta_1}{\eta_2} = \frac{273}{473} \]

5. **Calculating the Final Value**:
Now, we can compute the numerical value of the ratio:
\[ \frac{273}{473} \approx 0.577 \]

### Final Answer:
Thus, the ratio of the efficiencies \( \frac{\eta_1}{\eta_2} \) is approximately **0.577**.