A Carnot engine takes 9000 cal of heat from a reservoir at `627^@`C and gives it to a sink at `27^@C`. The work done by the angine is

Options

A. 6 kcal
B. 3 kcal
C. 4 kcal
D. 8 kcal

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the temperatures from Celsius to Kelvin
The temperatures of the hot and cold reservoirs need to be converted to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]

- For the hot reservoir:
\[ T_1 = 627 + 273.15 = 900.15 \, K \approx 900 \, K \]

- For the cold reservoir:
\[ T_2 = 27 + 273.15 = 300.15 \, K \approx 300 \, K \]

### Step 2: Apply the Carnot efficiency formula
The efficiency (\( \eta \)) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]

Substituting the values:
\[ \eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 3: Calculate the work done by the engine
The work done (\( W \)) by the engine can be calculated using the formula:
\[ W = \eta \cdot Q_1 \]

Where \( Q_1 \) is the heat absorbed from the hot reservoir. Given \( Q_1 = 9000 \, \text{cal} \):
\[ W = \frac{2}{3} \cdot 9000 \]

Calculating this gives:
\[ W = 6000 \, \text{cal} \]

### Step 4: Convert the work done into kilocalories
Since \( 1 \, \text{kcal} = 1000 \, \text{cal} \):
\[ W = \frac{6000}{1000} = 6 \, \text{kcal} \]

### Final Answer
The work done by the engine is \( 6 \, \text{kcal} \).

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