An ideal gas heat engine operates in Carnot cycle between `227^@C` and `127^@C`. It absorbs `6.0 xx 10^4 cal` of heat at high 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

Answer

To solve the problem of finding the amount of heat converted to work in a Carnot cycle heat engine, we can follow these steps:

### Step 1: Identify the temperatures
The temperatures given in the problem are:
- High temperature, \( T_1 = 227^\circ C \)
- Low temperature, \( T_2 = 127^\circ C \)

### Step 2: Convert temperatures to Kelvin
To use the Carnot efficiency formula, we need to convert the temperatures from Celsius to Kelvin using the formula:
\[ T(K) = T(°C) + 273 \]
Thus,
\[ T_1 = 227 + 273 = 500 \, K \]
\[ T_2 = 127 + 273 = 400 \, K \]

### Step 3: Calculate the efficiency of the Carnot engine
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = \frac{T_1 - T_2}{T_1} \]
Substituting the values we found:
\[ \eta = \frac{500 - 400}{500} = \frac{100}{500} = \frac{1}{5} = 0.2 \]

### Step 4: Calculate the work done
The work done \( W \) by the engine can be calculated using the relation:
\[ W = \eta \times Q \]
where \( Q \) is the heat absorbed at high temperature. Given \( Q = 6.0 \times 10^4 \, \text{cal} \):
\[ W = 0.2 \times 6.0 \times 10^4 = 1.2 \times 10^4 \, \text{cal} \]

### Conclusion
The amount of heat converted to work is \( 1.2 \times 10^4 \, \text{cal} \).