An ideal gas heat engine operates in a Carnot's cycle between `227^(@)C and 127^(@)C`. It absorbs `6 xx 10^(4) J` at high temperature. The amount of heat converted into work is

Options

A. `4.8 xx 10^(4)J`
B. `3.5 xx 10^(4)J`
C. `1.6 xx 10^(4)J`
D. `1.2 xx 10^(4)J`

Answer

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

### Step 1: Identify the temperatures
The temperatures given in the problem are:
- High temperature (T1) = 227°C
- Low temperature (T2) = 127°C

### Step 2: Convert temperatures to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
- For T1:
\[ T1 = 227 + 273.15 = 500.15 \, K \approx 500 \, K \]
- For T2:
\[ T2 = 127 + 273.15 = 400.15 \, K \approx 400 \, K \]

### Step 3: Write the formula for efficiency of a Carnot engine
The efficiency (η) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T2}{T1} \]

### Step 4: Calculate the efficiency
Substituting the values of T1 and T2 into the efficiency formula:
\[ \eta = 1 - \frac{400}{500} \]
\[ \eta = 1 - 0.8 = 0.2 \]

### Step 5: Relate efficiency to work done
The efficiency can also be expressed in terms of work done (W) and heat absorbed (Q):
\[ \eta = \frac{W}{Q} \]
Where Q is the heat absorbed by the engine, which is given as:
\[ Q = 6 \times 10^4 \, J \]

### Step 6: Calculate the work done
Rearranging the efficiency formula to find work done:
\[ W = \eta \times Q \]
Substituting the values:
\[ W = 0.2 \times 6 \times 10^4 \]
\[ W = 1.2 \times 10^4 \, J \]

### Conclusion
The amount of heat converted into work by the Carnot engine is:
\[ W = 1.2 \times 10^4 \, J \]

---