An ideal gas heat engine operates in a Carnot cycle between `27^(@)C` and `127^(@)C`. It absorbs `6 kcal` at the higher temperature. The amount of heat (in kcal) converted into work is equal to

Options

A. 4.5
B. 3.5
C. 1.6
D. 1.2

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the temperatures from Celsius to Kelvin
The temperatures given in the problem are:
- Higher temperature \( T_H = 127^\circ C \)
- Lower temperature \( T_L = 27^\circ C \)

To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]

Calculating the temperatures:
\[ T_H = 127 + 273 = 400 \, K \]
\[ T_L = 27 + 273 = 300 \, K \]

### Step 2: Calculate the efficiency of the Carnot engine
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_L}{T_H} \]

Substituting the values we found:
\[ \eta = 1 - \frac{300}{400} \]
\[ \eta = 1 - 0.75 = 0.25 \]

### Step 3: Calculate the work output
The work output \( W \) can be calculated using the efficiency and the heat input \( Q_H \):
\[ W = \eta \times Q_H \]

Given that \( Q_H = 6 \, kcal \):
\[ W = 0.25 \times 6 \, kcal \]
\[ W = 1.5 \, kcal \]

### Step 4: Conclusion
The amount of heat converted into work is \( 1.5 \, kcal \).

### Final Answer
The amount of heat converted into work is **1.5 kcal**.

---