Carnot engine takes one thousand kilo calories of heat from a reseervoir at `827^@C` and exhausts it to a sink at `27^@C`. How, much work does it perform? What is the efficiency of the engine?

Explanation

To solve the problem regarding the Carnot engine, we will follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
The temperatures given in Celsius need to be converted to Kelvin for use in calculations.

- \( T_1 = 827^\circ C = 827 + 273 = 1100 \, K \)
- \( T_2 = 27^\circ C = 27 + 273 = 300 \, K \)

### Step 2: Calculate the heat rejected (Q2)
Using the Carnot engine relation, we can find the heat rejected to the sink (Q2) using the formula:

\[ \frac{Q_1}{Q_2} = \frac{T_1}{T_2} \]

Rearranging gives us:

\[ Q_2 = Q_1 \cdot \frac{T_2}{T_1} \]

Substituting the values:

- \( Q_1 = 1000 \, \text{kcal} = 10^6 \, \text{cal} \)
- \( T_1 = 1100 \, K \)
- \( T_2 = 300 \, K \)

Calculating \( Q_2 \):

\[ Q_2 = 10^6 \cdot \frac{300}{1100} = 10^6 \cdot \frac{3}{11} \approx 272727.27 \, \text{cal} \approx 2.72 \times 10^5 \, \text{cal} \]

### Step 3: Calculate the work done (W)
The work done by the engine can be calculated using the formula:

\[ W = Q_1 - Q_2 \]

Substituting the values:

\[ W = 10^6 - 2.72 \times 10^5 = 727272.73 \, \text{cal} \approx 7.27 \times 10^5 \, \text{cal} \]

### Step 4: Calculate the efficiency (η)
The efficiency of the Carnot engine can be calculated using the formula:

\[ \eta = 1 - \frac{T_2}{T_1} \]

Substituting the values:

\[ \eta = 1 - \frac{300}{1100} = 1 - 0.272727 \approx 0.727273 \]

To express this as a percentage:

\[ \eta \approx 72.73\% \]

### Final Answers
- Work performed by the engine: \( W \approx 7.27 \times 10^5 \, \text{cal} \)
- Efficiency of the engine: \( \eta \approx 72.73\% \)

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