A heat engine rejects 600 cal to the sink at `27^(@)C` . Amount of work done by the engine will be
(Temperature of source is `227^(@) C` & J = 4.2 J/cal )

Options

A. 1680 J
B. 840 J
C. 2520 J
D. None of these

Answer

To solve the problem, we need to find the amount of work done by the heat engine that rejects 600 calories of heat to a sink at 27°C, while the source temperature is 227°C. We will use the principles of thermodynamics and the efficiency of the heat engine.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- The temperature of the source (T1) is given as 227°C. To convert this to Kelvin:
\[ T1 = 227 + 273 = 500 \, K \]
- The temperature of the sink (T2) is given as 27°C. To convert this to Kelvin:
\[ T2 = 27 + 273 = 300 \, K \]

2. **Identify Given Values:**
- Heat rejected (Q2) = 600 calories
- We need to find the work done (W) by the engine.

3. **Use the Efficiency Formula:**
- The efficiency (η) of a heat engine can be expressed as:
\[ \eta = 1 - \frac{T2}{T1} \]
- Substituting the values of T1 and T2:
\[ \eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 \]

4. **Relate Efficiency to Work Done:**
- The efficiency can also be defined as the ratio of work done (W) to the heat absorbed (Q1):
\[ \eta = \frac{W}{Q1} \]
- Rearranging gives:
\[ W = \eta \times Q1 \]

5. **Use the First Law of Thermodynamics:**
- According to the first law of thermodynamics, the heat absorbed (Q1) is equal to the heat rejected (Q2) plus the work done (W):
\[ Q1 = Q2 + W \]
- Substituting Q2 = 600 calories:
\[ Q1 = 600 + W \]

6. **Substitute Q1 in the Efficiency Equation:**
- Now substituting Q1 in the work equation:
\[ W = \eta \times (600 + W) \]
- Substituting η = 0.4:
\[ W = 0.4 \times (600 + W) \]
- Expanding this:
\[ W = 240 + 0.4W \]

7. **Solve for W:**
- Rearranging gives:
\[ W - 0.4W = 240 \]
\[ 0.6W = 240 \]
\[ W = \frac{240}{0.6} = 400 \, \text{calories} \]

8. **Convert Work Done to Joules:**
- Since 1 calorie = 4.2 joules, we convert the work done:
\[ W = 400 \times 4.2 = 1680 \, \text{joules} \]

### Final Answer:
The amount of work done by the engine is **1680 joules**.