Carnot engine takes one thousand kilo calories of heat from a reservoir 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, we will follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
The temperatures need to be converted to Kelvin because the Carnot efficiency formula requires absolute temperatures.

- **T1 (hot reservoir)**:
\[ T1 = 827^\circ C + 273 = 1100 \, K \]

- **T2 (cold sink)**:
\[ T2 = 27^\circ C + 273 = 300 \, K \]

### Step 2: Calculate the heat rejected (Q2)
Using the relation for a Carnot engine, we can find the heat rejected (Q2) using the formula:
\[ \frac{Q2}{Q1} = \frac{T2}{T1} \]
Given \( Q1 = 1000 \, \text{kcal} = 10^6 \, \text{cal} \), we can rearrange the formula to find Q2:
\[ Q2 = \frac{T2}{T1} \times Q1 \]
Substituting the values:
\[ Q2 = \frac{300}{1100} \times 10^6 \, \text{cal} \]
Calculating Q2:
\[ Q2 = 272727.27 \, \text{cal} \approx 2.72 \times 10^5 \, \text{cal} \]

### Step 3: Calculate the work done (W)
The work done by the engine can be calculated using the formula:
\[ W = Q1 - Q2 \]
Substituting the values:
\[ W = 10^6 \, \text{cal} - 2.72 \times 10^5 \, \text{cal} \]
Calculating W:
\[ W = 727272.73 \, \text{cal} \approx 7.28 \times 10^5 \, \text{cal} \]

### Step 4: Calculate the efficiency of the Carnot engine
The efficiency (η) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T2}{T1} \]
Substituting the values:
\[ \eta = 1 - \frac{300}{1100} \]
Calculating η:
\[ \eta = 1 - 0.2727 \approx 0.7273 \]
To express efficiency as a percentage:
\[ \eta \times 100 \approx 72.73\% \]

### Final Results
- Work performed by the engine: \( W \approx 7.28 \times 10^5 \, \text{cal} \)
- Efficiency of the engine: \( \eta \approx 72.73\% \)