An enegine is used to lift a 2700 kg truck to a height of 3.0 m at constant speed . In the lifting process , the engine received `3.3 xx 10^5 ` J of heat from the fuel burned in its interior . What is the efficiency of the engine ?

Options

A. `0.19`
B. `0.29`
C. `0.24`
D. `0.34`

Answer

To find the efficiency of the engine used to lift a truck, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Given Data:**
- Mass of the truck (m) = 2700 kg
- Height (h) = 3.0 m
- Heat received from the fuel (Q) = \(3.3 \times 10^5\) J
- Acceleration due to gravity (g) = 9.8 m/s² (standard value)

2. **Calculate the Work Done (W):**
The work done against gravity to lift the truck can be calculated using the formula:
\[ W = mgh \]
Substituting the values:
\[ W = 2700 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3.0 \, \text{m} \]
\[ W = 2700 \times 9.8 \times 3.0 \]
\[ W = 79,380 \, \text{J} \]

3. **Calculate the Efficiency (η):**
The efficiency of the engine is defined as the ratio of useful work output to the heat input:
\[ \eta = \frac{W}{Q} \]
Substituting the values we have:
\[ \eta = \frac{79,380 \, \text{J}}{3.3 \times 10^5 \, \text{J}} \]
\[ \eta = \frac{79,380}{330,000} \]
\[ \eta \approx 0.240 \]

4. **Convert Efficiency to Percentage:**
To express efficiency as a percentage:
\[ \eta \times 100 = 0.240 \times 100 = 24.0\% \]

### Final Answer:
The efficiency of the engine is approximately **24.0%**.