A Carnot engine workds between `200^(@)C` and `0^(@)C` and `.-200^(@)C`. In both caes the working substance absorbes 4 kilocalories of heat from the source. The efficiency of first engine will be

Options

A. `(100)/(173)`
B. `(200)/(473)`
C. `(173)/(273)`
D. `(273)/(373)`

Explanation

To find the efficiency of the first Carnot engine that works between \(200^\circ C\) and \(0^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- The temperature of the hot reservoir \(T_1\) is given as \(200^\circ C\).
\[ T_1 = 200 + 273 = 473 \, K \]
- The temperature of the cold reservoir \(T_2\) is given as \(0^\circ C\).
\[ T_2 = 0 + 273 = 273 \, K \]

2. **Use the Efficiency Formula for a Carnot Engine**:
- The efficiency \(\eta\) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
- Substituting the values of \(T_1\) and \(T_2\):
\[ \eta = 1 - \frac{273}{473} \]

3. **Calculate the Efficiency**:
- First, calculate \(\frac{273}{473}\):
\[ \frac{273}{473} \approx 0.577 \]
- Now, substitute this value back into the efficiency formula:
\[ \eta = 1 - 0.577 \approx 0.423 \]

4. **Convert Efficiency to Percentage**:
- To express the efficiency as a percentage, multiply by 100:
\[ \eta \approx 0.423 \times 100 \approx 42.3\% \]

### Final Answer:
The efficiency of the first Carnot engine is approximately **42.3%**.

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