The maximum amount of work done, which can be obtained by supplying 400 cal of heat will be

Options

A. 1 kJ
B. 1.2 kJ
C. `1.76kJ`
D. `1.68kJ`

Answer

To solve the problem of finding the maximum amount of work done by supplying 400 calories of heat, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Efficiency**:
The efficiency (η) of a thermodynamic system is defined as the ratio of the net work done (W) by the system to the heat supplied (Qs) to the system. Mathematically, it is expressed as:
\[ \eta = \frac{W}{Q_s} \]

2. **Maximum Efficiency**:
The maximum efficiency of a heat engine is 1 (or 100%), which means that all the supplied heat can be converted into work. Therefore, at maximum efficiency:
\[ W = Q_s \]

3. **Given Heat Supply**:
We are given that the heat supplied (Qs) is 400 calories. Thus, at maximum efficiency, the work done (W) will also be:
\[ W = 400 \text{ calories} \]

4. **Convert Calories to Joules**:
To convert calories to joules, we use the conversion factor where 1 calorie is approximately equal to 4.18 joules. Therefore, we can convert 400 calories to joules:
\[ W = 400 \text{ calories} \times 4.18 \frac{\text{joules}}{\text{calorie}} = 1672 \text{ joules} \]

5. **Convert Joules to Kilojoules**:
Since 1 kilojoule (kJ) is equal to 1000 joules, we convert joules to kilojoules:
\[ W = \frac{1672 \text{ joules}}{1000} = 1.672 \text{ kJ} \]

6. **Final Answer**:
Rounding to two decimal places, the maximum amount of work done is approximately:
\[ W \approx 1.68 \text{ kJ} \]

### Conclusion:
The maximum amount of work done by supplying 400 calories of heat is **1.68 kJ**.