An ideal gas heat engine operates in Carnot cycle between `227^(@)C` and 127^(@)C`. It absorbs `6 xx 10^(4) cals` of heat at higher temperature. Amount of heat converted to work is

Options

A. `2.4 xx 10^(4) cal`
B. `6 xx 10^(4) cal`
C. `1.2 xx 10^(4) cal`
D. `4.8 xx 10^(4) cal`

Explanation

To solve the problem of determining the amount of heat converted to work by an ideal gas heat engine operating in a Carnot cycle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Temperatures**:
- The temperatures given are \( T_1 = 227^\circ C \) (higher temperature) and \( T_2 = 127^\circ C \) (lower temperature).
- Convert these temperatures to Kelvin:
\[ T_1 = 227 + 273 = 500 \, K \]
\[ T_2 = 127 + 273 = 400 \, K \]

2. **Calculate the Efficiency of the Carnot Engine**:
- The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
- Substitute the values:
\[ \eta = 1 - \frac{400}{500} = 1 - 0.8 = 0.2 \]

3. **Determine the Heat Absorbed**:
- The amount of heat absorbed at the higher temperature \( Q_1 \) is given as:
\[ Q_1 = 6 \times 10^4 \, \text{calories} \]

4. **Calculate the Work Done**:
- The work done \( W \) by the engine can be calculated using the efficiency:
\[ W = \eta \times Q_1 \]
- Substitute the values:
\[ W = 0.2 \times 6 \times 10^4 = 1.2 \times 10^4 \, \text{calories} \]

### Final Answer:
The amount of heat converted to work is \( 1.2 \times 10^4 \, \text{calories} \).

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